MATH 108 - ELEMENTARY AND
INTERMEDIATE ALGEBRA - Intemediate
Value Theorem Question Bank
Question 1
State the Intermediate Value Theorem and provide a step-by-step explana-
tion.
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x) is a continuous function on
a closed interval [a, b] and Nis any number between f(a) and f(b), then there
exists a number cin the interval (a, b) such that f(c) = N.
Explanation: 1. Verify that the function f(x) is continuous on the closed
interval [a, b], meaning that it is continuous at every point within that interval.
2. Determine the values of f(a) and f(b) to establish the range of f(x) on
the interval [a, b]. 3. Identify a number Nsuch that f(a)< N < f(b). 4.
Apply the Intermediate Value Theorem to conclude that there exists a number
cin the open interval (a, b) where f(c) = N. 5. Thus, the Intermediate Value
Theorem guarantees the existence of a solution for f(c) = Nwithin the interval
[a, b].Question 1:
State the Intermediate Value Theorem and provide a step-by-step
explanation.
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on a closed interval [a, b]and Nis any number between f(a)and f(b),
then there exists a number cin the interval (a, b)such that f(c) = N.
Explanation: 1. Verify that the function f(x)is continuous on
the closed interval [a, b], meaning that it is continuous at every point
within that interval. 2. Determine the values of f(a)and f(b)to
establish the range of f(x)on the interval [a, b]. 3. Identify a number N
such that f(a)< N < f(b). 4. Apply the Intermediate Value Theorem
to conclude that there exists a number cin the open interval (a, b)
where f(c) = N. 5. Thus, the Intermediate Value Theorem guarantees
the existence of a solution for f(c) = Nwithin the interval [a, b].
1
Question 2
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c) = 0 by the Intermediate Value Theorem.Question 2: Let f(x) =
x3−2x+ 1 be a continuous function on the interval [0,2]. Show that
there exists a number cin the interval (0,2) such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c)=0by the Intermediate Value Theorem.
Question 3
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].Question 3:
Use the Intermediate Value Theorem to show that there is a root of
the equation x3−2x−5=0in the interval [2,3].
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
2
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].
Question 4
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assis-
tance.Sure, here is question number 4 on the Intermediate Value
Theorem presented in LateX code:
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assistance.
3
Question 5
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c)=0.Question
5: Let f(x) = x3−4x2+ 3x+ 2 be a continuous function on the interval
[0,3]. Show that there exists a value cin the interval (0,3) such that
f(c)=0using the Intermediate Value Theorem.
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c) = 0.
Question 6
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.Question 6: Use the Intermediate Value Theorem to show
that the function f(x) = x3−2x+ 1 has a root in the interval [0,1].
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.
Question 7
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
4
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.Sure! Here’s a question
on the Intermediate Value Theorem along with its solution written
in LateX code:
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.
Question 8
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.Certainly!
Here is a question regarding the Intermediate Value Theorem with a
step-by-step solution in LaTeX code:
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.
Question 9
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
6
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) =
0.Certainly! Here is a question along with step-by-step solutions on
the Intermediate Value Theorem written in LateX code:
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) = 0.
Question 10
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.Question 10:
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
7
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.
Question 11
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
8
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 11:
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 12
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
9
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c)=0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.Question
12:
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c) = 0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.
10
Question 13
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2=0<0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].Question 13: Use the Intermediate Value Theorem to show that
the equation x3−6x2+ 9x−2=0has a solution in the interval [1,2].
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2 = 0 <0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].
Question 14
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c) = 0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].Question 14: Use the Intermediate Value
11
Theorem to prove that the polynomial function f(x) = x3−4x2+ 2x−3
has a root in the interval [−2,0].
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c)=0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].
Question 15
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in the
interval [1, 2].Sure, here is a question and step-by-step solution on
Intermediate Value Theorem for Liberty University in LateX code:
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
12
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in
the interval [1, 2].
Question 16
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘Certainly! Here is
question 16 on the Intermediate Value Theorem in LateX code:
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
13
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘
Question 17
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have any
more questions.Sure, here is a question and its step-by-step solution
related to the Intermediate Value Theorem in LateX code:
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
14
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have
any more questions.
Question 18
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘Sure, here is question number 18 on the Intermediate Value
Theorem in LateX code:
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
15
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘
Question 19
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.Sure, here is a question along with step-by-step solution on
Intermediate Value Theorem in LateX code:
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
16
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.
Question 20
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
2. Since f(1) = −6and f(2) = −8, and f(x)is a continuous func-
tion, by the Intermediate Value Theorem, there exists a root of the
equation x3−4x2+ 3x−6 = 0 in the interval [1,2].
Therefore, there exists a number c∈[1,2] such that f(c)=0, which
means that there is a root of the equation in the interval [1,2]. “‘“‘la-
tex Question 20: Use the Intermediate Value Theorem to show that
there is a root of the equation x3−4x2+ 3x−6=0in the interval [1,2].
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
17
Question 2
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c) = 0 by the Intermediate Value Theorem.Question 2: Let f(x) =
x3−2x+ 1 be a continuous function on the interval [0,2]. Show that
there exists a number cin the interval (0,2) such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c)=0by the Intermediate Value Theorem.
Question 3
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].Question 3:
Use the Intermediate Value Theorem to show that there is a root of
the equation x3−2x−5=0in the interval [2,3].
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
2
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].
Question 4
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assis-
tance.Sure, here is question number 4 on the Intermediate Value
Theorem presented in LateX code:
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assistance.
3
Question 5
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c)=0.Question
5: Let f(x) = x3−4x2+ 3x+ 2 be a continuous function on the interval
[0,3]. Show that there exists a value cin the interval (0,3) such that
f(c)=0using the Intermediate Value Theorem.
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c) = 0.
Question 6
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.Question 6: Use the Intermediate Value Theorem to show
that the function f(x) = x3−2x+ 1 has a root in the interval [0,1].
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.
Question 7
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
4
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.Sure! Here’s a question
on the Intermediate Value Theorem along with its solution written
in LateX code:
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.
Question 8
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.Certainly!
Here is a question regarding the Intermediate Value Theorem with a
step-by-step solution in LaTeX code:
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.
Question 9
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
6
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) =
0.Certainly! Here is a question along with step-by-step solutions on
the Intermediate Value Theorem written in LateX code:
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) = 0.
Question 10
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.Question 10:
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
7
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.
Question 11
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
8
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 11:
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 12
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
9
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c)=0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.Question
12:
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c) = 0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.
10
Question 13
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2=0<0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].Question 13: Use the Intermediate Value Theorem to show that
the equation x3−6x2+ 9x−2=0has a solution in the interval [1,2].
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2 = 0 <0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].
Question 14
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c) = 0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].Question 14: Use the Intermediate Value
11
Theorem to prove that the polynomial function f(x) = x3−4x2+ 2x−3
has a root in the interval [−2,0].
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c)=0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].
Question 15
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in the
interval [1, 2].Sure, here is a question and step-by-step solution on
Intermediate Value Theorem for Liberty University in LateX code:
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
12
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in
the interval [1, 2].
Question 16
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘Certainly! Here is
question 16 on the Intermediate Value Theorem in LateX code:
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
13
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘
Question 17
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have any
more questions.Sure, here is a question and its step-by-step solution
related to the Intermediate Value Theorem in LateX code:
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
14
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have
any more questions.
Question 18
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘Sure, here is question number 18 on the Intermediate Value
Theorem in LateX code:
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
15
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘
Question 19
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.Sure, here is a question along with step-by-step solution on
Intermediate Value Theorem in LateX code:
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
16
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.
Question 20
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
2. Since f(1) = −6and f(2) = −8, and f(x)is a continuous func-
tion, by the Intermediate Value Theorem, there exists a root of the
equation x3−4x2+ 3x−6 = 0 in the interval [1,2].
Therefore, there exists a number c∈[1,2] such that f(c)=0, which
means that there is a root of the equation in the interval [1,2]. “‘“‘la-
tex Question 20: Use the Intermediate Value Theorem to show that
there is a root of the equation x3−4x2+ 3x−6=0in the interval [1,2].
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
17
Question 2
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c) = 0 by the Intermediate Value Theorem.Question 2: Let f(x) =
x3−2x+ 1 be a continuous function on the interval [0,2]. Show that
there exists a number cin the interval (0,2) such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that f(x)is continuous on the closed interval [0,2] and determine
the values of f(0) and f(2):
1. Continuity of f(x): Notice that f(x) = x3−2x+ 1 is a polynomial
function, and polynomial functions are continuous everywhere. Thus,
f(x)is continuous on the interval [0,2].
2. Determine f(0) and f(2): - f(0) = 03−2(0) + 1 = 1 -f(2) =
23−2(2) + 1 = 8 −4 + 1 = 5
Since f(0) = 1 and f(2) = 5 are of opposite signs, by the Inter-
mediate Value Theorem, there exists a number c∈(0,2) such that
f(c)=0.
Therefore, there exists a number cin the interval (0,2) such that
f(c)=0by the Intermediate Value Theorem.
Question 3
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].Question 3:
Use the Intermediate Value Theorem to show that there is a root of
the equation x3−2x−5=0in the interval [2,3].
Step-by-step Solution: 1. Define a function f(x) = x3−2x−5. 2.
Check if f(2) and f(3) have opposite signs: - f(2) = 23−2(2) −5 =
2
8−4−5 = −1-f(3) = 33−2(3) −5 = 27 −6−5 = 16 - Since f(2) is
negative and f(3) is positive, f(x)changes sign on the interval [2,3].
3. Apply the Intermediate Value Theorem, which states that since
f(x)is continuous on [2,3] and changes sign on this interval, there
exists a value cin [2,3] such that f(c) = 0. 4. Therefore, there is a
root of the equation x3−2x−5=0in the interval [2,3].
Question 4
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assis-
tance.Sure, here is question number 4 on the Intermediate Value
Theorem presented in LateX code:
Question 4: Use the Intermediate Value Theorem to show that the
polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the interval [1, 2].
Solution:
To apply the Intermediate Value Theorem, we first need to evalu-
ate f(1) and f(2):
f(1) = 3(1)3−5(1)2+ 2(1) + 7 = 3 −5 + 2 + 7 = 7
f(2) = 3(2)3−5(2)2+ 2(2) + 7 = 24 −20 + 4 + 7 = 15
Since f(1) = 7 and f(2) = 15, and the function f(x)is continuous on
the closed interval [1, 2], the Intermediate Value Theorem guarantees
that there exists a number cin the interval [1, 2] such that f(c)=0,
i.e. there is a root of the polynomial in the interval [1, 2].
Therefore, the polynomial f(x)=3x3−5x2+ 2x+ 7 has a root in the
interval [1, 2].
I hope this helps! Let me know if you need any more assistance.
3
Question 5
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c)=0.Question
5: Let f(x) = x3−4x2+ 3x+ 2 be a continuous function on the interval
[0,3]. Show that there exists a value cin the interval (0,3) such that
f(c)=0using the Intermediate Value Theorem.
Step-by-step Solution: 1. Verify continuity of the function fon
the closed interval [0,3]: Since f(x) = x3−4x2+ 3x+ 2 is a polynomial
function, it is continuous for all real numbers.
2. Find the values of f(0) and f(3):f(0) = (0)3−4(0)2+ 3(0) + 2 = 2
f(3) = (3)3−4(3)2+ 3(3) + 2 = 27 −36 + 9 + 2 = 2
3. Apply the Intermediate Value Theorem: Since f(0) = 2 and
f(3) = 2, and fis continuous on [0,3], by the Intermediate Value
Theorem, there exists a value cin the interval (0,3) such that f(c) = 0.
Therefore, there exists a value c∈(0,3) such that f(c) = 0.
Question 6
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.Question 6: Use the Intermediate Value Theorem to show
that the function f(x) = x3−2x+ 1 has a root in the interval [0,1].
Step-by-step Solution: 1. Verify that f(x)is continuous on the
interval [0,1]. 2. Calculate f(0) and f(1). 3. Check if f(0) and f(1)
have opposite signs. 4. Apply the Intermediate Value Theorem to
conclude that there exists a value cin the interval [0,1] such that
f(c)=0.
Question 7
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
4
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.Sure! Here’s a question
on the Intermediate Value Theorem along with its solution written
in LateX code:
Question 7: Let f(x) = x3−3x+ 1. Use the Intermediate Value
Theorem to show that there is a solution to the equation f(x)=0in
the interval [1,2].
Solution: The Intermediate Value Theorem states that if fis a
continuous function on a closed interval [a, b], and Nis any number
between f(a)and f(b), then there exists a number cin (a, b)such that
f(c) = N.
Given f(x) = x3−3x+ 1 is a polynomial function, it is continuous
on all real numbers.
Let N= 0, then we have f(1) = 13−3(1) + 1 = −1and f(2) =
23−3(2) + 1 = 1.
Since f(1) = −1and f(2) = 1 have opposite signs, by the Intermedi-
ate Value Theorem, there exists a number cin the interval (1,2) such
that f(c)=0.
Therefore, there is a solution to the equation f(x)=0in the inter-
val [1,2] by the Intermediate Value Theorem.
Question 8
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.Certainly!
Here is a question regarding the Intermediate Value Theorem with a
step-by-step solution in LaTeX code:
Question 8: Let f(x) = 3x3−2x2+ 4x−5be a continuous function
on the interval [0,2]. Use the Intermediate Value Theorem to prove
that there exists a value cin the interval (0,2) such that f(c)=0.
Solution: Since f(x) = 3x3−2x2+ 4x−5is continuous on [0,2], it
follows that f(x)takes on all values between f(0) and f(2) on the
interval [0,2].
We calculate the values of f(0) and f(2):
f(0) = 3(0)3−2(0)2+ 4(0) −5
= 0 −0+0−5
=−5
f(2) = 3(2)3−2(2)2+ 4(2) −5
= 3(8) −2(4) + 8 −5
= 24 −8+8−5
= 19
Since f(0) = −5and f(2) = 19, by the Intermediate Value Theorem,
there exists a value cin the interval (0,2) such that f(c)=0.
Question 9
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
6
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) =
0.Certainly! Here is a question along with step-by-step solutions on
the Intermediate Value Theorem written in LateX code:
Question 9: Let f(x) = 3x3−5x2+ 2x−7on the interval [0,2]. Use
the Intermediate Value Theorem to determine if there is a value cin
the interval [0,2] such that f(c)=0.
Solution: To apply the Intermediate Value Theorem, we need to
show that there exists a number cin the interval [0,2] such that f(c) =
0.
Step 1: Evaluate f(0) and f(2) f(0) = 3(0)3−5(0)2+ 2(0) −7 = −7
f(2) = 3(2)3−5(2)2+ 2(2) −7 = 17
Step 2: Check the signs of f(0) and f(2) Since f(0) = −7and f(2) =
17, we have f(0) <0and f(2) >0.
Step 3: Conclusion Since the function changes sign from negative
to positive on the interval [0,2], by the Intermediate Value Theorem,
there exists at least one value cin the interval [0,2] such that f(c) = 0.
Therefore, there is a value cin the interval [0,2] such that f(c) = 0.
Question 10
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.Question 10:
State and prove the Intermediate Value Theorem for a function
f(x)on an interval [a, b].
7
Step-by-step Solution:
Intermediate Value Theorem (IVT): If f(x)is a continuous function
on the interval [a, b], and dis any number between f(a)and f(b),
then there exists at least one number cin the interval [a, b]such that
f(c) = d.
Proof: Given f(x)is continuous on the interval [a, b]and dis any
number between f(a)and f(b).
Since f(x)is continuous on [a, b], by the Extreme Value Theorem,
f(x)attains its maximum and minimum on this interval.
Let m= min{f(a), f(b)}and M= max{f(a), f(b)}.
Since dis between f(a)and f(b), we have m≤d≤M.
Consider two cases:
Case 1: If d=f(a)or d=f(b), then c=aor c=brespectively, and
therefore the statement holds.
Case 2: If d=f(a)and d=f(b), then m<d<M.
Since f(x)takes on all values between mand Mby continuity, there
exists at least one point cin the interval [a, b]such that f(c) = d.
Thus, the Intermediate Value Theorem is proved.
Question 11
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
8
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 11:
Let f(x) = x3−2x2+3x−4on the interval [1,2]. Use the Intermediate
Value Theorem to show that there exists a value c∈[1,2] such that
f(c)=0.
Solution:
1. First, we need to check if f(x)is continuous on the closed interval
[1,2].
2. f(x)is a polynomial function, and polynomials are continuous
everywhere, including the interval [1,2].
3. Next, we need to find f(1) and f(2) to see if the function changes
sign on the interval.
4. Calculate f(1):
f(1) = 13−2(1)2+ 3(1) −4
= 1 −2+3−4
=−2
5. Calculate f(2):
f(2) = 23−2(2)2+ 3(2) −4
= 8 −8+6−4
= 2
6. Since f(1) = −2and f(2) = 2, and the function is continuous on
[1,2], by the Intermediate Value Theorem, there exists a value
c∈[1,2] such that f(c)=0.
Question 12
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
9
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c)=0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.Question
12:
Let
f(x) = x2−5x+ 6
.
1. Show that there is a root of the function f(x)in the interval
[1,3]. 2. Use the Intermediate Value Theorem to find an approximate
value for this root within an error of 0.01.
Solution:
1. To show that there is a root of f(x)in the interval [1,3], we first
need to verify that f(1) <0and f(3) >0.
f(1) = 12−5(1) + 6 = 1 −5 + 6 = 2 >0
f(3) = 32−5(3) + 6 = 9 −15 + 6 = 0
Since f(1) >0and f(3) <0, we can conclude that there is at least
one root of f(x)in the interval [1,3].
2. To find an approximate value for the root using the Intermediate
Value Theorem, we can choose a value cin the interval [1,3] such that
f(c)=0.
By the Intermediate Value Theorem, since f(1) >0and f(3) <0,
there exists c∈(1,3) such that f(c) = 0.
To find an approximate value for cwithin an error of 0.01, we can
use a numerical method like bisection method or Newton’s method.
Therefore, there is a root of the function f(x) = x2−5x+ 6 in
the interval [1,3], and using the Intermediate Value Theorem, we can
approximate the value of this root within an error of 0.01.
10
Question 13
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2=0<0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].Question 13: Use the Intermediate Value Theorem to show that
the equation x3−6x2+ 9x−2=0has a solution in the interval [1,2].
Step 1: Identify the function f(x)involved in the given equation.
In this case, f(x) = x3−6x2+ 9x−2.
Step 2: Check if the function f(x)is continuous on the closed
interval [1,2]. Since f(x)is a polynomial function, it is continuous
everywhere, including the interval [1,2].
Step 3: Calculate f(1) and f(2) to determine the sign change.
f(1) = 13−6(1)2+ 9(1) −2=1−6+9−2=2>0
f(2) = 23−6(2)2+ 9(2) −2=8−24 + 18 −2 = 0 <0
Step 4: Since f(1) >0and f(2) <0, by the Intermediate Value
Theorem, there exists at least one cin the interval [1,2] such that
f(c) = 0, which implies the equation has a solution in the interval
[1,2].
Question 14
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c) = 0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].Question 14: Use the Intermediate Value
11
Theorem to prove that the polynomial function f(x) = x3−4x2+ 2x−3
has a root in the interval [−2,0].
Step-by-step Solution: 1. Determine the values of f(−2) and f(0):
f(−2) = (−2)3−4(−2)2+ 2(−2) −3 = −8−16 −4−3 = −31
f(0) = 03−4(0)2+ 2(0) −3 = −3
2. Check the sign of f(−2) ·f(0):
f(−2) ·f(0) = (−31) ·(−3) = 93
3. Since f(−2) ·f(0) >0, the function f(x)does not change sign in
the interval [−2,0].
4. By the Intermediate Value Theorem, since f(x)is continuous
and does not change sign in the interval [−2,0], there exists at least
one value cin the interval [−2,0] such that f(c)=0.
Therefore, the polynomial function f(x) = x3−4x2+ 2x−3has a
root in the interval [−2,0].
Question 15
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in the
interval [1, 2].Sure, here is a question and step-by-step solution on
Intermediate Value Theorem for Liberty University in LateX code:
Question 15: Let f(x) = x3+ 2x2−3x−4. Show that f has a real
root in the interval [1, 2].
Step-by-step solution: To show that f(x) = x3+ 2x2−3x−4has
a real root in the interval [1, 2], we can use the Intermediate Value
Theorem.
12
1. Calculate f(1) and f(2).
f(1) = 13+ 2(1)2−3(1) −4
= 1 + 2 −3−4 = −4
f(2) = 23+ 2(2)2−3(2) −4
= 8 + 8 −6−4=6
2. Since f(1) = −4and f(2) = 6, and −4<0<6, by the Intermediate
Value Theorem, there exists a real number cin the interval [1, 2] such
that f(c)=0.
Therefore, the function f(x) = x3+ 2x2−3x−4has a real root in
the interval [1, 2].
Question 16
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘Certainly! Here is
question 16 on the Intermediate Value Theorem in LateX code:
“‘latex Question 16:
Let f(x)=2x3−3x2−4x+ 1 be a polynomial function. Determine if
the Intermediate Value Theorem applies to f(x)on the interval [0,2].
Solution:
To check if the Intermediate Value Theorem applies to f(x)on the
interval [0,2], we need to verify the following conditions:
1. f(x)is continuous on [0,2]. 2. f(0) =f(2).
13
1. Since f(x)is a polynomial function, it is continuous on all real
numbers. Thus, f(x)is continuous on the closed interval [0,2].
2. Calculate f(0) and f(2):
f(0) = 2(0)3−3(0)2−4(0) + 1
= 1
f(2) = 2(2)3−3(2)2−4(2) + 1
= 16 −12 −8+1
=−3
Since f(0) = 1 and f(2) = −3,f(0) =f(2).
Therefore, both conditions are satisfied, and the Intermediate Value
Theorem applies to f(x)on the interval [0,2]. “‘
Question 17
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have any
more questions.Sure, here is a question and its step-by-step solution
related to the Intermediate Value Theorem in LateX code:
Question 17: Let f(x)=2x3−3x2−12x+ 5 be a function defined on
the interval [0,3]. Use the Intermediate Value Theorem to show that
there exists a value c in the interval (0,3) such that f(c) = 0.
Step-by-step Solution: To apply the Intermediate Value Theorem,
we need to show that the function f(x)is continuous on the closed
interval [0,3] and that it takes on both positive and negative values
on that interval.
1. First, find f(0) and f(3):f(0) = 2(0)3−3(0)2−12(0)+5 = 0−0−0+5 =
5
f(3) = 2(3)3−3(3)2−12(3) + 5 = 54 −27 −36 + 5 = −4
14
2. Since f(0) = 5 >0and f(3) = −4<0, by the Intermediate Value
Theorem, there exists a value c in the interval (0,3) such that f(c)=0.
Therefore, the Intermediate Value Theorem guarantees the exis-
tence of a value c in the interval (0,3) such that f(c)=0.
This concludes the solution.
Feel free to reach out if you need any further assistance or have
any more questions.
Question 18
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘Sure, here is question number 18 on the Intermediate Value
Theorem in LateX code:
“‘latex Question 18: Let f(x) = x3−6x2+ 9x−1. Show that there
exist two real numbers cand dsuch that f(c) = f(d).
15
Solution: To apply the Intermediate Value Theorem, we need to
show that the function f(x)is continuous on the interval [a, b]where
a<b.
Here, f(x) = x3−6x2+ 9x−1is a polynomial function, which is
continuous everywhere. Therefore, f(x)is continuous on the interval
(−∞,∞).
Next, we need to find two points cand dsuch that f(c) = f(d).
1. Let f(a) = f(−2) and f(b) = f(3). Calculate f(−2) and f(3).
Substituting x=−2into the function f(x):
f(−2) = (−2)3−6(−2)2+ 9(−2) −1
=−8−24 −18 −1
=−51
Substituting x= 3 into the function f(x):
f(3) = 33−6(3)2+ 9(3) −1
= 27 −54 + 27 −1
=−1
2. Since f(−2) = −51 and f(3) = −1, by the Intermediate Value
Theorem, there exists a number cbetween −2and 3such that
f(c)=0. Similarly, there exists a number dbetween −2and 3
such that f(d)=0.
Therefore, there exist two real numbers cand dsuch that f(c) =
f(d). “‘
Question 19
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.Sure, here is a question along with step-by-step solution on
Intermediate Value Theorem in LateX code:
Question 19: Let f(x) = 2x3−x2+ 3x−1be a function that is
continuous on the interval [0,2]. Show that there exists a value cin
the interval (0,2) such that f(c) = 5.
16
Solution: Since f(x)is continuous on the interval [0,2], we can
apply the Intermediate Value Theorem. Let’s define a= 0 and b= 2.
First, we need to find f(a)and f(b):f(0) = 2(0)3−(0)2+3(0) −1 = −1
f(2) = 2(2)3−(2)2+ 3(2) −1 = 16 −4+6−1 = 17
Since 5lies between f(0) = −1and f(2) = 17, by the Intermediate
Value Theorem, there exists a value cin the interval (0,2) such that
f(c)=5.
Question 20
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
2. Since f(1) = −6and f(2) = −8, and f(x)is a continuous func-
tion, by the Intermediate Value Theorem, there exists a root of the
equation x3−4x2+ 3x−6 = 0 in the interval [1,2].
Therefore, there exists a number c∈[1,2] such that f(c)=0, which
means that there is a root of the equation in the interval [1,2]. “‘“‘la-
tex Question 20: Use the Intermediate Value Theorem to show that
there is a root of the equation x3−4x2+ 3x−6=0in the interval [1,2].
Solution:
Let f(x) = x3−4x2+ 3x−6.
1. Calculate f(1) and f(2):
f(1) = 13−4(1)2+ 3(1) −6
= 1 −4+3−6
=−6
f(2) = 23−4(2)2+ 3(2) −6
= 8 −16 + 6 −6
=−8
17
2. Since f(1) = −6and f(2) = −8, and f(x)is a continuous func-
tion, by the Intermediate Value Theorem, there exists a root of the
equation x3−4x2+ 3x−6 = 0 in the interval [1,2].
Therefore, there exists a number c∈[1,2] such that f(c)=0, which
means that there is a root of the equation in the interval [1,2]. “‘
18
