MATH 121 - COLLEGE ALGEBRA -
Rational Functions
Question Bank - Set 4
Liberty University
Question 1
Question
Solve the following rational equation for x:
2
x+ 3 +1
x−1=4
x2+ 2x−3
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation.
2(x−1)
(x+ 3)(x−1) +(x+ 3)
(x+ 3)(x−1) =4
x2+ 2x−3
Step 2: Combine the fractions on the left side of the equation.
2(x−1) + (x+ 3)
(x+ 3)(x−1) =4
x2+ 2x−3
Step 3: Simplify the numerators on the left side of the equation.
2x−2 + x+ 3
(x+ 3)(x−1) =4
x2+ 2x−3
3x+ 1
(x+ 3)(x−1) =4
x2+ 2x−3
Step 4: Find a common denominator for the fractions on both sides of the
equation.
4(x2+ 2x−3) = (3x+ 1)
Step 5: Expand and simplify the terms on both sides of the equation.
4x2+ 8x−12 = 3x+ 1
Step 6: Rearrange the equation to set it equal to zero.
4x2+ 5x−13 = 0
Step 7: Solve the quadratic equation using the quadratic formula.
x=−b±√b2−4ac
2a
x=−5±√52−4(4)(−13)
2(4)
x=−5±√25 + 208
8
x=−5±√233
8
Therefore, the solutions to the equation are x=−5+√233
8and x=−5−√233
8.
Question 2
Question
Find the domain of the following rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: Determine where the denominator is equal to zero to identify any po-
tential restrictions on x. Setting the denominator equal to zero and solving for
x:
x2−4x−5 = 0
(x−5)(x+ 1) = 0
x= 5 or x=−1
Step 2: Identify the values of xthat make the function undefined by exclud-
ing the values found in Step 1 from the domain. The function is undefined when
the denominator is zero, so x= 5 and x=−1.
Step 3: Write the domain of the function. The domain of the function f(x)
is all real numbers except x= 5 and x=−1. Thus, the domain is
(−∞,−1) ∪(−1,5) ∪(5,∞)
2
Question 3
Question
Find the domain of the rational function: f(x) = 1
x2−4x.
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. Therefore, we need to
find the values of xfor which x2−4x= 0.
Step 2: Let’s solve the equation x2−4x= 0 to find the values that make
the denominator zero.
x2−4x= 0
Step 3: Factoring out an xfrom the left side gives:
x(x−4) = 0
Step 4: Setting each factor to zero gives the roots:
x= 0 or x= 4
Step 5: So, the values x= 0 and x= 4 would make the denominator zero.
Therefore, the domain of the function is all real numbers except 0and 4, or in
interval notation: (−∞,0) ∪(0,4) ∪(4,∞).
Question 4
Question
Given the rational function f(x) = 3x2
−6x
x2−4, determine the domain of the func-
tion.
Solution
Step 1: The domain of a rational function is all real numbers except those
that make the denominator equal to zero. This is because division by zero is
undefined. So, to find the domain, we need to find the values of xwhere the
denominator x2−4is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x+ 2 = 0 or x−2 = 0
3
x=−2or x= 2
Step 3: Therefore, the values x=−2,2make the denominator equal to zero,
resulting in the function being undefined at these points. Thus, the domain
of the function is all real numbers except x=−2,2, which can be written in
interval notation as:
(−∞,−2) ∪(−2,2) ∪(2,∞)
Question 5
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: Determine the values of xthat make the denominator of the rational
function equal to zero. These values correspond to the vertical asymptotes of
the function.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
Step 2: Set each factor equal to zero to find the roots:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 3: The vertical asymptotes of the rational function are at x= 5 and
x=−1. Thus, the vertical asymptotes are x= 5 and x=−1.
Question 6
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to determine all
the values of xfor which the function is defined. Since division by zero is not
allowed, we need to find the values that make the denominator zero.
4
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 ⇒x= 5
x+ 1 = 0 ⇒x=−1
Step 5: Therefore, the domain of the function is all real numbers except
x= 5 and x=−1.
Domain: x∈R, x = 5,−1
Question 7
Question
Let f(x) = 2x3+3x2
−2x−3
x2−1and g(x) = x2
−1
x2+2x+1 . Find the domain of f◦g(x).
Solution
Step 1: Find the composition f◦g(x) = f(g(x)) by substituting g(x)into f(x).
f(g(x)) = f(x2−1
x2+ 2x+ 1)
Step 2: Simplify the expression inside fby replacing xwith x2
−1
x2+2x+1 .
f(x2−1
x2+ 2x+ 1)=
2(x2
−1
x2+2x+1 )3+ 3 (x2
−1
x2+2x+1 )2
−2(x2
−1
x2+2x+1 )−3
(x2−1
x2+2x+1 )2
−1
Step 3: Simplify the expression further.
f(x2−1
x2+ 2x+ 1)=
2(x2
−1)3
(x2+2x+1)3+3(x2
−1)2
(x2+2x+1)2−2(x2
−1)
x2+2x+1 −3
(x2−1)2
(x2+2x+1)2−1
Step 4: Determine the domain of f◦g(x)by considering restrictions on the
denominator. Since the denominator cannot be equal to 0, we need to find
where x2+ 2x+ 1 = 0.
x2+ 2x+ 1 = 0
5
(x+ 1)(x+ 1) = 0
(x+ 1)2= 0
x=−1
Therefore, the domain of f◦g(x)is all real numbers except x=−1.
Question 8
Question
Find the domain of the following rational function:
f(x) = 4x2−7x−3
x2−5x+ 6
Solution
To find the domain of a rational function, we need to identify the values of x
for which the function is defined. The function will be undefined at any value
of xthat makes the denominator equal to zero.
Step 1: Find the values of xthat make the denominator zero.
Set the denominator x2−5x+ 6 equal to zero and solve for x:
x2−5x+ 6 = 0
Step 2: Solve for x.
Factor the quadratic equation:
(x−2)(x−3) = 0
Set each factor equal to zero and solve for x:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 3: Determine the domain.
The values of xthat make the denominator zero are x= 2 and x= 3.
Therefore, the domain of the function f(x) = 4x2−7x−3
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Answer: The domain of the function is all real numbers except x= 2 and
x= 3.
Question 9
Question
Let f(x) = x3
−3x2
−4x+12
x2−2x−8. Determine the x-intercepts, y-intercepts, vertical
asymptotes, horizontal asymptotes, and the domain of the function f(x).
6
Solution
Step 1: To find the x-intercepts, set f(x) = 0 and solve for x.
0 = x3−3x2−4x+ 12
x2−2x−8
0 = x3−3x2−4x+ 12
Using synthetic division or polynomial long division, we can find that x= 3 is a
root, which means (x−3) is a factor. Dividing through gives (x−3)(x2+x−4) =
0. Factoring the quadratic factor further, we get (x−3)(x−1)(x+ 4) = 0, so
the x-intercepts are x= 3,x= 1, and x=−4.
Step 2: To find the y-intercept, set x= 0 and evaluate f(0).
f(0) = 03−3(0)2−4(0) + 12
02−2(0) −8=12
−8=−3
2
Thus, the y-intercept is at (0,−3
2).
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x.
x2−2x−8 = 0
(x−4)(x+ 2) = 0
So, the vertical asymptotes are x= 4 and x=−2.
Step 4: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator of the rational function. Since the degree of the numerator
is greater than the degree of the denominator, there is no horizontal asymptote.
Step 5: The domain of the function f(x)is all real numbers except where
the denominator is equal to zero. Thus, the domain is x= 4,−2.
Therefore, the x-intercepts are x= 3,x= 1, and x=−4, the y-intercept is
at (0,−3
2), the vertical asymptotes are at x= 4 and x=−2, the function has
no horizontal asymptote, and the domain of f(x)is x= 4,−2.
Question 10
Question
Let f(x) = 2x3
−x2
−9x
x2−x−6. Determine the domain of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify the values of xthat will
make the denominator of f(x)equal to zero, since division by zero is undefined.
Step 2: Set the denominator x2−x−6equal to zero and solve for x:
x2−x−6 = 0
7
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor equal to zero and solve for x: For x−3 = 0,x= 3
For x+ 2 = 0,x=−2
Step 5: Therefore, the values x= 3 and x=−2make the denominator
of f(x)zero. Thus, the domain of f(x)is all real numbers except x= 3 and
x=−2. In interval notation, this can be written as (−∞,−2)∪(−2,3)∪(3,∞).
Question 11
Question
Find the vertical asymptotes of the rational function given by f(x) = 3x2−x−2
x2−4x−5.
Solution
Step 1: First, factor the numerator and denominator of the rational function.
f(x) = 3x2−x−2
x2−4x−5=(3x+ 2)(x−1)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero. These values will give us the vertical asymptotes of the function. Setting
the denominator equal to zero:
(x−5)(x+ 1) = 0
This gives us two possible vertical asymptotes:
x= 5 and x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−3x−2
x2−4x+ 3 .
8
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero because division by zero
is undefined.
Step 2: Find the values of xthat make the denominator zero by solving the
equation x2−4x+ 3 = 0.
Step 3: Factor the quadratic equation x2−4x+3 = 0 to get (x−3)(x−1) = 0.
Step 4: Set each factor to zero and solve for x.
{x−3 = 0
x−1 = 0
Step 5: Solve the equations to find the values of x.
{x= 3
x= 1
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 3 and x= 1. In interval notation, the domain is (−∞,1) ∪
(1,3) ∪(3,∞).
Question 13
Question
Find the domain of the rational function f(x) = 2x2−x−3
x2−9.
Solution
Step 1: Remember that the domain of a rational function is all real numbers
except for the values of xthat make the denominator zero, since division by zero
is undefined. Hence, we need to find the values of xthat make the denominator
x2−9equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−9 = 0
Step 3: Factor the quadratic expression:
x2−9 = (x+ 3)(x−3) = 0
Step 4: Set each factor equal to zero and solve for x:
{x+ 3 = 0 =⇒x=−3
x−3 = 0 =⇒x= 3
9
Step 5: The values x=−3and x= 3 would make the denominator zero, so
they are not included in the domain of the function f(x).
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers excluding x=−3and x= 3, which can be expressed in interval notation
as (−∞,−3) ∪(−3,3) ∪(3,+∞).
Question 14
Question
Find the domain of the rational function:
f(x) = 3x2−2x−8
x2−5x+ 6
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xthat make the denominator equal to zero, since division by zero is
undefined. So, we need to solve the equation x2−5x+ 6 = 0 to find the values
of xthat would make the denominator zero.
Step 2: Factoring the quadratic equation x2−5x+ 6 = 0, we get (x−2)(x−
3) = 0.
Step 3: Setting each factor to zero, we find the roots: x−2 = 0 ⇒x= 2
and x−3 = 0 ⇒x= 3.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 2 and x= 3. This can be written in interval notation as
(−,2)(2,3)(3,).
Question 15
Question
Solve the following rational equation: 2x+3
x−1=4x−1
x+3 .
Solution
Step 1: Cross multiply to eliminate the fractions.
(x−1)(4x−1) = (x+ 3)(2x+ 3)
4x2−x−4x+ 1 = 2x2+ 3x+ 6x+ 9
4x2−5x+ 1 = 2x2+ 9x+ 9
10
Step 2: Move all terms to one side of the equation.
4x2−5x+ 1 −2x2−9x−9 = 0
2x2−14x−8 = 0
Step 3: Factor the quadratic equation.
2(x2−7x−4) = 0
2(x−8)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x.
x−8 = 0 or x+ 1 = 0
x= 8 or x=−1
Step 5: Check for extraneous solutions.
Since x= 1 would make the denominators zero in the original equation,
x= 1 is an extraneous solution.
Therefore, the solution to the rational equation is x= 8.
Question 16
Question
Simplify the following rational function:
3x2−8x−3
2x2−5x−3
Solution
Step 1: Factor the numerator and denominator:
3x2−8x−3 = (3x+ 1)(x−3)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the rational function with the factored forms:
(3x+ 1)(x−3)
(2x+ 1)(x−3)
Step 3: Cancel out the common factor of (x−3) in the numerator and
denominator: 3x+ 1
2x+ 1
Therefore, the simplified form of the given rational function is 3x+1
2x+1 .
11
Question 17
Question
Solve the rational equation: 3
x−1−1
x+3 =1
2.
Solution
Step 1: To solve the given rational equation, we will first find a common denomi-
nator for the fractions on the left side of the equation. The common denominator
will be (x−1)(x+ 3).
Step 2: Rewrite the equation with the common denominator:
(x+ 3) 3
x−1−(x−1) 1
x+ 3 =1
2(x−1)(x+ 3)
Step 3: Simplify the equation:
3(x+ 3) −(x−1) = 1
2(x2+ 2x−3)
Step 4: Expand and simplify the equation:
3x+ 9 −x+ 1 = 1
2x2+x−3
2
2x+ 10 = 1
2x2+x−3
2
Step 5: Multiply both sides of the equation by 2 to eliminate the fraction:
4x+ 20 = x2+ 2x−3
Step 6: Rearrange the equation into standard form:
0 = x2−2x−23
Step 7: Factor the quadratic equation:
0 = (x−5)(x+ 3)
Step 8: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 3 = 0 =⇒x=−3
Step 9: Check for extraneous solutions by plugging each potential solution
back into the original equation: - For x= 5:3
5−1−1
5+3 =1
2, which is true. -
For x=−3:3
−3−1−1
−3+3 =1
2, which is not valid as it results in division by 0.
Therefore, the only solution to the given rational equation is x= 5.
12
Question 18
Question
Solve for x:
3x
x−4−x
x−2=4
x−4
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation:
3x
x−4−x
x−2=4
x−4
3x(x−2)
(x−4)(x−2) −x(x−4)
(x−2)(x−4) =4
x−4
Step 2: Combine the fractions on the left side of the equation:
3x(x−2) −x(x−4)
(x−4)(x−2) =4
x−4
3x2−6x−x2+ 4x
(x−4)(x−2) =4
x−4
2x2−2x
(x−4)(x−2) =4
x−4
Step 3: Simplify the equation:
2x(x−1)
(x−4)(x−2) =4
x−4
Step 4: Clear the denominators by multiplying both sides by (x−4)(x−2):
2x(x−1) = 4(x−2)
Step 5: Expand and simplify:
2x2−2x= 4x−8
2x2−2x= 4x−8
Step 6: Rearrange the equation in standard form:
2x2−2x−4x+ 8 = 0
2x2−6x+ 8 = 0
Step 7: Factor the quadratic equation:
13
2(x2−3x+ 4) = 0
2(x−2)(x−2) = 0
Step 8: Find the solutions:
x= 2, x = 2
However, the repeating solution, x= 2, is extraneous as it makes the de-
nominator equal to zero. Thus, the solution to the equation is x= 2.
Question 19
Question
Find the domain of the rational function: f(x) = x2−4
x2−9.
Solution
To find the domain of a rational function, we need to identify all real numbers
that make the denominator equal to zero since division by zero is undefined.
Step 1: Identify the denominator In this case, the denominator is x2−9.
Step 2: Find the values that make the denominator zero Set the
denominator equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
The solutions are x= 3 and x=−3.
Step 3: Determine the domain The domain of the function is all real
numbers except for the values that make the denominator zero. Therefore, the
domain of the function is (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 20
Question
Solve the rational inequality: 3x2+7x+2
x2−2x−3>0.
14
Solution
Step 1: Find the critical points by setting the numerator and denominator equal
to zero and solving for x.
3x2+ 7x+ 2 = 0
(x+ 1)(3x+ 2) = 0
x=−1,−2
3
x2−2x−3 = 0
(x−3)(x+ 1) = 0
x= 3,−1
Step 2: Create a sign chart using the critical points −1,−2
3,3.
− −2
3−1 3
3x2+ 7x+ 2 −+ + +
x2−2x−3 + −+ +
Step 3: Analyze the sign chart to determine where the rational function is
positive. Since we are looking for when 3x2+7x+2
x2−2x−3>0, we want to find where this
expression is positive, meaning when the function is above the x-axis. Therefore,
the solution to the inequality is x∈(−∞,−1) ∪(−2
3,3).
Question 21
Question
Simplify the rational expression:
2x2−5x−3
x2+ 2x−3∇ · x2−4
x2+x−6
Solution
To simplify the given rational expression, we will first factor all the quadratic
expressions involved.
Step 1: Factor all the quadratic expressions: For the numerator of the first
fraction, 2x2−5x−3, we have:
2x2−5x−3 = (2x+ 1)(x−3)
For the denominator of the first fraction, x2+ 2x−3, we have:
x2+ 2x−3 = (x+ 3)(x−1)
15
For the numerator of the second fraction, x2−4, we have:
x2−4 = (x+ 2)(x−2)
For the denominator of the second fraction, x2+x−6, we have:
x2+x−6 = (x+ 3)(x−2)
Step 2: Rewrite the expression with factored terms:
(2x+ 1)(x−3)
(x+ 3)(x−1) ∇ · (x+ 2)(x−2)
(x+ 3)(x−2)
Step 3: Simplify by multiplying by the reciprocal of the second fraction:
(2x+ 1)(x−3)
(x+ 3)(x−1) ×(x+ 3)(x−2)
(x+ 2)(x−2)
Step 4: Cancel out the common factors:
(2x+ 1)(x−3)(x+ 3)(x−2)
(x+ 3)(x−1)(x+ 2)
Step 5: Simplify the expression further:
2x+ 1
x+ 2
Therefore, the simplified form of the given rational expression is 2x+1
x+2 .
Question 22
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, we need to
consider the values of xthat would make the denominator equal to zero, since
division by zero is undefined.
Step 2: Set the denominator equal to zero and solve for xto find the values
that will make the function undefined:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
16
Step 4: Set each factor equal to zero:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 5: The values of xthat make the denominator zero are x= 2 and
x= 3. Therefore, the domain of the function excludes these values.
Step 6: So, the domain of the rational function f(x)is all real numbers
except x= 2 and x= 3. In interval notation, the domain can be written as
(−∞,2) ∪(2,3) ∪(3,∞).
Question 23
Question
Solve the inequality 2x−1
x+3 ≥2.
Solution
Step 1: Find the critical points by setting the expression equal to zero:
2x−1
x+ 3 = 2
2x−1 = 2(x+ 3)
2x−1 = 2x+ 6
0 = 7
Since the equation 0 = 7 is false, there are no critical points in this case.
Step 2: Determine the values not in the domain of the inequality: The
expression 2x−1
x+3 is not defined when x=−3. Therefore, x=−3.
Step 3: Choose test points: Select test points in each of the intervals deter-
mined by the critical points and points not in the domain. Let’s choose x=−4
and x= 0.
Step 4: Plug the test points into the inequality: For x=−4:
2(−4) −1
−4+3 = 3 >2
For x= 0:2(0) −1
0+3 =−1
3≥ 2
Step 5: Analyze the results: Since the expression is greater than 2 for x=−4
and not greater than 2 for x= 0, the solution to the inequality is:
x < −3
17
Question 24
Question
Solve the following rational inequality for x:
2x−3
x+ 1 ≥4
x−2.
Solution
To solve the rational inequality, we will first find the critical points where the
inequality may change and then test each interval defined by these critical points
to determine where the inequality holds.
Step 1: Find the critical points The critical points occur where the
numerator or denominator of one fraction is zero. Equating the numerator and
denominator of the left-hand side to zero gives us: 2x−3 = 0 ⇒x=3
2.
Equating the numerator and denominator of the right-hand side to zero gives
us: x−2 = 0 ⇒x= 2. So, we have critical points at x=3
2and x= 2.
Step 2: Test intervals and solution We will test three intervals: (−∞,3
2),
(3
2,2), and (2,∞)by choosing test points within each interval.
1. For x= 0, we have:
2(0) −3
0+1 ≥4
0−2⇒ −3≥ −2,
which is false. So, (−∞,3
2)is not part of the solution.
2. For x=5
2, we have:
2(5
2)−3
5
2+ 1 ≥4
5
2−2⇒4
5≥ −4,
which is true. So, (3
2,2) is part of the solution.
3. For x= 3, we have:
2(3) −3
3+1 ≥4
3−2⇒3≥4,
which is false. So, (2,∞)is not part of the solution.
Therefore, the solution to the inequality is x∈(3
2,2).
Question 25
Question
Solve the following rational equation for x:
2
x+ 3 −3
x−2=1
x
18
Solution
Step 1: Find a common denominator for all fractions on the left side of the
equation. In this case, the common denominator is (x+ 3)(x−2)x.
Step 2: Rewrite the equation with the common denominator:
2x(x−2)
x(x+ 3)(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 3: Combine the fractions on the left side of the equation:
2x(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 4: Simplify the numerator on the left side of the equation:
2x2−4x−3x2−9x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 5: Combine like terms in the numerator on the left side of the equation:
−x2−13x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 6: Cross multiply to eliminate the denominators:
(−x2−13x)(x) = (x+ 3)(x−2)(x)
Step 7: Expand both sides of the equation:
−x3−13x2=x3−2x2+ 3x2−6x
Step 8: Combine like terms and solve for x:
0 = 2x3+ 15x2−6x
Step 9: Factor out an xfrom the equation:
0 = x(2x2+ 15x−6)
Step 10: Solve for xby setting each factor to zero:
x= 0 or 2x2+ 15x−6 = 0
Step 11: Solve the quadratic equation 2x2+ 15x−6 = 0 using the quadratic
formula or factoring. The solutions are x=−3
2and x=2
3.
Thus, the solutions to the original rational equation are x= 0,x=−3
2, and
x=2
3.
19
Question 26
Question
Let f(x) = 5x2
−3x−2
2x2−7x+3 . Find the domain of the function f(x).
Solution
Step 1: In order to find the domain of a rational function, we need to determine
the values of xfor which the denominator is not equal to zero. This is because
division by zero is undefined. Step 2: Set the denominator of the function equal
to zero and solve for x:
2x2−7x+ 3 = 0
Step 3: We can factor the quadratic equation as:
(2x−1)(x−3) = 0
Step 4: Setting each factor to zero gives us the solutions:
2x−1 = 0 =⇒x=1
2
x−3 = 0 =⇒x= 3
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x=1
2and x= 3. In interval notation, the domain is (−∞,1
2)∪(1
2,3) ∪(3,∞).
Question 27
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Determine the values that make the denominator equal to zero, since
division by zero is undefined.
x2−5x+ 6 = 0
Step 2: Factor the quadratic expression.
(x−2)(x−3) = 0
Step 3: Find the roots of the quadratic equation.
x−2 = 0 ⇒x= 2
20
x−3 = 0 ⇒x= 3
Step 4: The domain is all real numbers except for the values that make the
denominator equal to zero. Therefore, the domain of the rational function is all
real numbers except x= 2 and x= 3.
D:x∈R, x = 2,3
Question 28
Question
Find the domain of the function f(x) = x2
−4
x2−9.
Solution
Step 1: The domain of a function is the set of all real numbers for which the
function is defined. In this case, the function f(x)is defined for all xsuch
that the denominator x2−9is not equal to zero, because division by zero is
undefined.
Step 2: We need to find the values of xthat make x2−9 = 0. This can be
factored as (x−3)(x+ 3) = 0.
Step 3: Setting each factor to zero gives us x−3 = 0 and x+ 3 = 0, which
lead to x= 3 and x=−3.
Step 4: Therefore, the domain of the function f(x) = x2
−4
x2−9is all real numbers
except x= 3 and x=−3. This can be written as (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 29
Question
Solve the rational inequality: x−2
x+3 ≤1.
Solution
To solve the rational inequality x−2
x+3 ≤1, we need to find the values of xthat
satisfy the inequality.
Step 1: Rewrite the inequality without the fraction:
x−2
x+ 3 ≤1⇒x−2≤x+ 3
Step 2: Solve the equation x−2 = x+ 3 to find the boundary points:
x−2 = x+ 3
x−x= 3 + 2
0 = 5
21
This equation has no solution, which means the boundary points of the inequal-
ity are undefined.
Step 3: Test a point in each interval to determine the solution set:
• Test x= 0:0−2
0+3 =−2
3≤1(true)
• Test x=−4:−4−2
−4+3 =−6
−1= 6 ≤1(false)
Step 4: Analyze the test points:
•x= 0 satisfies the inequality.
•x=−4does not satisfy the inequality.
Step 5: Express the solution set: The solution to the rational inequality
x−2
x+3 ≤1is x∈(−∞,0].
Therefore, the solution set is x∈(−∞,0] .
Question 30
Question
Simplify the following rational function:
f(x) = 4x3−27
x2−4
Solution
Step 1: Factor the numerator and denominator:
f(x) = 4x3−27
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Step 2: Simplify the expression by canceling out common factors:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
x2−4
Step 3: Notice that x2−4can be factored further as the difference of squares:
f(x) = (2x−3)(2x2+ 6x+ 9)
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Therefore, the simplified form of the rational function f(x)is:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
22
Question 31
Question
Solve the rational inequality: 4x−8
x+ 3 ≥1.
Solution
Step 1: Begin by rewriting the inequality:
4x−8
x+ 3 ≥1 =⇒4x−8≥x+ 3.
Step 2: Simplify the inequality:
4x−8≥x+ 3 =⇒3x≥11.
Step 3: Solve for x:
3x≥11 =⇒x≥11
3.
Therefore, the solution to the inequality is x≥11
3.
Question 32
Question
Simplify the rational function:
4x2−25
x2−9
Solution
Step 1: Factor the numerator and denominator:
4x2−25 = (2x+ 5)(2x−5)
x2−9 = (x+ 3)(x−3)
Step 2: Rewrite the rational function with factored numerator and denomi-
nator: 4x2−25
x2−9=(2x+ 5)(2x−5)
(x+ 3)(x−3)
Step 3: Simplify the rational function by canceling out common factors:
(2x+ 5)(2x−5)
(x+ 3)(x−3) =2x+ 5
x+ 3
Therefore, the simplified form of the rational function is 2x+5
x+3 .
23
Question 33
Question
Find the domain of the rational function:
f(x) = 3x2+ 5x−2
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to exclude any
x-values that make the denominator equal to zero, since division by zero is
undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor or use the quadratic
formula. Here, we will use factoring.
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1make the denominator zero, so they
are not in the domain of the rational function.
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1. In interval notation, the domain can be expressed
as:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 34
Question
Simplify the rational function 3x3
−5x2+2x
2x2−3x−2.
Solution
Step 1: Factor the numerator and the denominator.
Step 2: Simplify by canceling out common factors.
Step 1: First, let’s factor the numerator and the denominator: The numer-
ator 3x3−5x2+2xcan be expressed as x(3x2−5x+2), where 3x2−5x+2 factors
to (3x−2)(x−1). The denominator 2x2−3x−2factors to (2x+ 1)(x−2).
24
Step 2: Now, substitute the factored forms back into the expression and
simplify:
3x3−5x2+ 2x
2x2−3x−2=x(3x2−5x+ 2)
(2x+ 1)(x−2) =x(3x−2)(x−1)
(2x+ 1)(x−2) .
Notice that the terms (3x−2) and (x−2) can be canceled out, leaving us with:
x(x−1)
2x+ 1 .
Therefore, the simplified form of the rational function is x(x−1)
2x+1 .
Question 35
Question
Simplify the rational expression: 3x2+ 5x−2
2x2−3x−2.
Solution
To simplify the rational expression 3x2+ 5x−2
2x2−3x−2, we need to factor both the
numerator and denominator and then cancel out any common factors.
Step 1: Factor the numerator We need to find two numbers that multiply
to 3× −2 = −6and add up to 5. These numbers are 6and −1. So, we can
rewrite the numerator as: (3x+ 6)(x−1).
Step 2: Factor the denominator Similarly, we need to find two numbers
that multiply to 2×−2 = −4and add up to −3. These numbers are −4and 1.
So, we can rewrite the denominator as: (2x+ 1)(x−2).
Therefore, the rational expression becomes (3x+ 6)(x−1)
(2x+ 1)(x−2).
Step 3: Cancel out common factors We can cancel out the common
factor (x−1) in the numerator and denominator.
Therefore, the simplified form of the rational expression is 3x+ 6
2x+ 1 .
This is the final simplified form of the given rational expression.
25
Question 3
Question
Find the domain of the rational function: f(x) = 1
x2−4x.
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. Therefore, we need to
find the values of xfor which x2−4x= 0.
Step 2: Let’s solve the equation x2−4x= 0 to find the values that make
the denominator zero.
x2−4x= 0
Step 3: Factoring out an xfrom the left side gives:
x(x−4) = 0
Step 4: Setting each factor to zero gives the roots:
x= 0 or x= 4
Step 5: So, the values x= 0 and x= 4 would make the denominator zero.
Therefore, the domain of the function is all real numbers except 0and 4, or in
interval notation: (−∞,0) ∪(0,4) ∪(4,∞).
Question 4
Question
Given the rational function f(x) = 3x2
−6x
x2−4, determine the domain of the func-
tion.
Solution
Step 1: The domain of a rational function is all real numbers except those
that make the denominator equal to zero. This is because division by zero is
undefined. So, to find the domain, we need to find the values of xwhere the
denominator x2−4is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x+ 2 = 0 or x−2 = 0
3
x=−2or x= 2
Step 3: Therefore, the values x=−2,2make the denominator equal to zero,
resulting in the function being undefined at these points. Thus, the domain
of the function is all real numbers except x=−2,2, which can be written in
interval notation as:
(−∞,−2) ∪(−2,2) ∪(2,∞)
Question 5
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: Determine the values of xthat make the denominator of the rational
function equal to zero. These values correspond to the vertical asymptotes of
the function.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
Step 2: Set each factor equal to zero to find the roots:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 3: The vertical asymptotes of the rational function are at x= 5 and
x=−1. Thus, the vertical asymptotes are x= 5 and x=−1.
Question 6
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to determine all
the values of xfor which the function is defined. Since division by zero is not
allowed, we need to find the values that make the denominator zero.
4
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 ⇒x= 5
x+ 1 = 0 ⇒x=−1
Step 5: Therefore, the domain of the function is all real numbers except
x= 5 and x=−1.
Domain: x∈R, x = 5,−1
Question 7
Question
Let f(x) = 2x3+3x2
−2x−3
x2−1and g(x) = x2
−1
x2+2x+1 . Find the domain of f◦g(x).
Solution
Step 1: Find the composition f◦g(x) = f(g(x)) by substituting g(x)into f(x).
f(g(x)) = f(x2−1
x2+ 2x+ 1)
Step 2: Simplify the expression inside fby replacing xwith x2
−1
x2+2x+1 .
f(x2−1
x2+ 2x+ 1)=
2(x2
−1
x2+2x+1 )3+ 3 (x2
−1
x2+2x+1 )2
−2(x2
−1
x2+2x+1 )−3
(x2−1
x2+2x+1 )2
−1
Step 3: Simplify the expression further.
f(x2−1
x2+ 2x+ 1)=
2(x2
−1)3
(x2+2x+1)3+3(x2
−1)2
(x2+2x+1)2−2(x2
−1)
x2+2x+1 −3
(x2−1)2
(x2+2x+1)2−1
Step 4: Determine the domain of f◦g(x)by considering restrictions on the
denominator. Since the denominator cannot be equal to 0, we need to find
where x2+ 2x+ 1 = 0.
x2+ 2x+ 1 = 0
5
(x+ 1)(x+ 1) = 0
(x+ 1)2= 0
x=−1
Therefore, the domain of f◦g(x)is all real numbers except x=−1.
Question 8
Question
Find the domain of the following rational function:
f(x) = 4x2−7x−3
x2−5x+ 6
Solution
To find the domain of a rational function, we need to identify the values of x
for which the function is defined. The function will be undefined at any value
of xthat makes the denominator equal to zero.
Step 1: Find the values of xthat make the denominator zero.
Set the denominator x2−5x+ 6 equal to zero and solve for x:
x2−5x+ 6 = 0
Step 2: Solve for x.
Factor the quadratic equation:
(x−2)(x−3) = 0
Set each factor equal to zero and solve for x:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 3: Determine the domain.
The values of xthat make the denominator zero are x= 2 and x= 3.
Therefore, the domain of the function f(x) = 4x2−7x−3
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Answer: The domain of the function is all real numbers except x= 2 and
x= 3.
Question 9
Question
Let f(x) = x3
−3x2
−4x+12
x2−2x−8. Determine the x-intercepts, y-intercepts, vertical
asymptotes, horizontal asymptotes, and the domain of the function f(x).
6
Solution
Step 1: To find the x-intercepts, set f(x) = 0 and solve for x.
0 = x3−3x2−4x+ 12
x2−2x−8
0 = x3−3x2−4x+ 12
Using synthetic division or polynomial long division, we can find that x= 3 is a
root, which means (x−3) is a factor. Dividing through gives (x−3)(x2+x−4) =
0. Factoring the quadratic factor further, we get (x−3)(x−1)(x+ 4) = 0, so
the x-intercepts are x= 3,x= 1, and x=−4.
Step 2: To find the y-intercept, set x= 0 and evaluate f(0).
f(0) = 03−3(0)2−4(0) + 12
02−2(0) −8=12
−8=−3
2
Thus, the y-intercept is at (0,−3
2).
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x.
x2−2x−8 = 0
(x−4)(x+ 2) = 0
So, the vertical asymptotes are x= 4 and x=−2.
Step 4: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator of the rational function. Since the degree of the numerator
is greater than the degree of the denominator, there is no horizontal asymptote.
Step 5: The domain of the function f(x)is all real numbers except where
the denominator is equal to zero. Thus, the domain is x= 4,−2.
Therefore, the x-intercepts are x= 3,x= 1, and x=−4, the y-intercept is
at (0,−3
2), the vertical asymptotes are at x= 4 and x=−2, the function has
no horizontal asymptote, and the domain of f(x)is x= 4,−2.
Question 10
Question
Let f(x) = 2x3
−x2
−9x
x2−x−6. Determine the domain of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify the values of xthat will
make the denominator of f(x)equal to zero, since division by zero is undefined.
Step 2: Set the denominator x2−x−6equal to zero and solve for x:
x2−x−6 = 0
7
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor equal to zero and solve for x: For x−3 = 0,x= 3
For x+ 2 = 0,x=−2
Step 5: Therefore, the values x= 3 and x=−2make the denominator
of f(x)zero. Thus, the domain of f(x)is all real numbers except x= 3 and
x=−2. In interval notation, this can be written as (−∞,−2)∪(−2,3)∪(3,∞).
Question 11
Question
Find the vertical asymptotes of the rational function given by f(x) = 3x2−x−2
x2−4x−5.
Solution
Step 1: First, factor the numerator and denominator of the rational function.
f(x) = 3x2−x−2
x2−4x−5=(3x+ 2)(x−1)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero. These values will give us the vertical asymptotes of the function. Setting
the denominator equal to zero:
(x−5)(x+ 1) = 0
This gives us two possible vertical asymptotes:
x= 5 and x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−3x−2
x2−4x+ 3 .
8
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero because division by zero
is undefined.
Step 2: Find the values of xthat make the denominator zero by solving the
equation x2−4x+ 3 = 0.
Step 3: Factor the quadratic equation x2−4x+3 = 0 to get (x−3)(x−1) = 0.
Step 4: Set each factor to zero and solve for x.
{x−3 = 0
x−1 = 0
Step 5: Solve the equations to find the values of x.
{x= 3
x= 1
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 3 and x= 1. In interval notation, the domain is (−∞,1) ∪
(1,3) ∪(3,∞).
Question 13
Question
Find the domain of the rational function f(x) = 2x2−x−3
x2−9.
Solution
Step 1: Remember that the domain of a rational function is all real numbers
except for the values of xthat make the denominator zero, since division by zero
is undefined. Hence, we need to find the values of xthat make the denominator
x2−9equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−9 = 0
Step 3: Factor the quadratic expression:
x2−9 = (x+ 3)(x−3) = 0
Step 4: Set each factor equal to zero and solve for x:
{x+ 3 = 0 =⇒x=−3
x−3 = 0 =⇒x= 3
9
Step 5: The values x=−3and x= 3 would make the denominator zero, so
they are not included in the domain of the function f(x).
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers excluding x=−3and x= 3, which can be expressed in interval notation
as (−∞,−3) ∪(−3,3) ∪(3,+∞).
Question 14
Question
Find the domain of the rational function:
f(x) = 3x2−2x−8
x2−5x+ 6
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xthat make the denominator equal to zero, since division by zero is
undefined. So, we need to solve the equation x2−5x+ 6 = 0 to find the values
of xthat would make the denominator zero.
Step 2: Factoring the quadratic equation x2−5x+ 6 = 0, we get (x−2)(x−
3) = 0.
Step 3: Setting each factor to zero, we find the roots: x−2 = 0 ⇒x= 2
and x−3 = 0 ⇒x= 3.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 2 and x= 3. This can be written in interval notation as
(−,2)(2,3)(3,).
Question 15
Question
Solve the following rational equation: 2x+3
x−1=4x−1
x+3 .
Solution
Step 1: Cross multiply to eliminate the fractions.
(x−1)(4x−1) = (x+ 3)(2x+ 3)
4x2−x−4x+ 1 = 2x2+ 3x+ 6x+ 9
4x2−5x+ 1 = 2x2+ 9x+ 9
10
Step 2: Move all terms to one side of the equation.
4x2−5x+ 1 −2x2−9x−9 = 0
2x2−14x−8 = 0
Step 3: Factor the quadratic equation.
2(x2−7x−4) = 0
2(x−8)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x.
x−8 = 0 or x+ 1 = 0
x= 8 or x=−1
Step 5: Check for extraneous solutions.
Since x= 1 would make the denominators zero in the original equation,
x= 1 is an extraneous solution.
Therefore, the solution to the rational equation is x= 8.
Question 16
Question
Simplify the following rational function:
3x2−8x−3
2x2−5x−3
Solution
Step 1: Factor the numerator and denominator:
3x2−8x−3 = (3x+ 1)(x−3)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the rational function with the factored forms:
(3x+ 1)(x−3)
(2x+ 1)(x−3)
Step 3: Cancel out the common factor of (x−3) in the numerator and
denominator: 3x+ 1
2x+ 1
Therefore, the simplified form of the given rational function is 3x+1
2x+1 .
11
Question 17
Question
Solve the rational equation: 3
x−1−1
x+3 =1
2.
Solution
Step 1: To solve the given rational equation, we will first find a common denomi-
nator for the fractions on the left side of the equation. The common denominator
will be (x−1)(x+ 3).
Step 2: Rewrite the equation with the common denominator:
(x+ 3) 3
x−1−(x−1) 1
x+ 3 =1
2(x−1)(x+ 3)
Step 3: Simplify the equation:
3(x+ 3) −(x−1) = 1
2(x2+ 2x−3)
Step 4: Expand and simplify the equation:
3x+ 9 −x+ 1 = 1
2x2+x−3
2
2x+ 10 = 1
2x2+x−3
2
Step 5: Multiply both sides of the equation by 2 to eliminate the fraction:
4x+ 20 = x2+ 2x−3
Step 6: Rearrange the equation into standard form:
0 = x2−2x−23
Step 7: Factor the quadratic equation:
0 = (x−5)(x+ 3)
Step 8: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 3 = 0 =⇒x=−3
Step 9: Check for extraneous solutions by plugging each potential solution
back into the original equation: - For x= 5:3
5−1−1
5+3 =1
2, which is true. -
For x=−3:3
−3−1−1
−3+3 =1
2, which is not valid as it results in division by 0.
Therefore, the only solution to the given rational equation is x= 5.
12
Question 18
Question
Solve for x:
3x
x−4−x
x−2=4
x−4
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation:
3x
x−4−x
x−2=4
x−4
3x(x−2)
(x−4)(x−2) −x(x−4)
(x−2)(x−4) =4
x−4
Step 2: Combine the fractions on the left side of the equation:
3x(x−2) −x(x−4)
(x−4)(x−2) =4
x−4
3x2−6x−x2+ 4x
(x−4)(x−2) =4
x−4
2x2−2x
(x−4)(x−2) =4
x−4
Step 3: Simplify the equation:
2x(x−1)
(x−4)(x−2) =4
x−4
Step 4: Clear the denominators by multiplying both sides by (x−4)(x−2):
2x(x−1) = 4(x−2)
Step 5: Expand and simplify:
2x2−2x= 4x−8
2x2−2x= 4x−8
Step 6: Rearrange the equation in standard form:
2x2−2x−4x+ 8 = 0
2x2−6x+ 8 = 0
Step 7: Factor the quadratic equation:
13
2(x2−3x+ 4) = 0
2(x−2)(x−2) = 0
Step 8: Find the solutions:
x= 2, x = 2
However, the repeating solution, x= 2, is extraneous as it makes the de-
nominator equal to zero. Thus, the solution to the equation is x= 2.
Question 19
Question
Find the domain of the rational function: f(x) = x2−4
x2−9.
Solution
To find the domain of a rational function, we need to identify all real numbers
that make the denominator equal to zero since division by zero is undefined.
Step 1: Identify the denominator In this case, the denominator is x2−9.
Step 2: Find the values that make the denominator zero Set the
denominator equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
The solutions are x= 3 and x=−3.
Step 3: Determine the domain The domain of the function is all real
numbers except for the values that make the denominator zero. Therefore, the
domain of the function is (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 20
Question
Solve the rational inequality: 3x2+7x+2
x2−2x−3>0.
14
Solution
Step 1: Find the critical points by setting the numerator and denominator equal
to zero and solving for x.
3x2+ 7x+ 2 = 0
(x+ 1)(3x+ 2) = 0
x=−1,−2
3
x2−2x−3 = 0
(x−3)(x+ 1) = 0
x= 3,−1
Step 2: Create a sign chart using the critical points −1,−2
3,3.
− −2
3−1 3
3x2+ 7x+ 2 −+ + +
x2−2x−3 + −+ +
Step 3: Analyze the sign chart to determine where the rational function is
positive. Since we are looking for when 3x2+7x+2
x2−2x−3>0, we want to find where this
expression is positive, meaning when the function is above the x-axis. Therefore,
the solution to the inequality is x∈(−∞,−1) ∪(−2
3,3).
Question 21
Question
Simplify the rational expression:
2x2−5x−3
x2+ 2x−3∇ · x2−4
x2+x−6
Solution
To simplify the given rational expression, we will first factor all the quadratic
expressions involved.
Step 1: Factor all the quadratic expressions: For the numerator of the first
fraction, 2x2−5x−3, we have:
2x2−5x−3 = (2x+ 1)(x−3)
For the denominator of the first fraction, x2+ 2x−3, we have:
x2+ 2x−3 = (x+ 3)(x−1)
15
For the numerator of the second fraction, x2−4, we have:
x2−4 = (x+ 2)(x−2)
For the denominator of the second fraction, x2+x−6, we have:
x2+x−6 = (x+ 3)(x−2)
Step 2: Rewrite the expression with factored terms:
(2x+ 1)(x−3)
(x+ 3)(x−1) ∇ · (x+ 2)(x−2)
(x+ 3)(x−2)
Step 3: Simplify by multiplying by the reciprocal of the second fraction:
(2x+ 1)(x−3)
(x+ 3)(x−1) ×(x+ 3)(x−2)
(x+ 2)(x−2)
Step 4: Cancel out the common factors:
(2x+ 1)(x−3)(x+ 3)(x−2)
(x+ 3)(x−1)(x+ 2)
Step 5: Simplify the expression further:
2x+ 1
x+ 2
Therefore, the simplified form of the given rational expression is 2x+1
x+2 .
Question 22
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, we need to
consider the values of xthat would make the denominator equal to zero, since
division by zero is undefined.
Step 2: Set the denominator equal to zero and solve for xto find the values
that will make the function undefined:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
16
Step 4: Set each factor equal to zero:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 5: The values of xthat make the denominator zero are x= 2 and
x= 3. Therefore, the domain of the function excludes these values.
Step 6: So, the domain of the rational function f(x)is all real numbers
except x= 2 and x= 3. In interval notation, the domain can be written as
(−∞,2) ∪(2,3) ∪(3,∞).
Question 23
Question
Solve the inequality 2x−1
x+3 ≥2.
Solution
Step 1: Find the critical points by setting the expression equal to zero:
2x−1
x+ 3 = 2
2x−1 = 2(x+ 3)
2x−1 = 2x+ 6
0 = 7
Since the equation 0 = 7 is false, there are no critical points in this case.
Step 2: Determine the values not in the domain of the inequality: The
expression 2x−1
x+3 is not defined when x=−3. Therefore, x=−3.
Step 3: Choose test points: Select test points in each of the intervals deter-
mined by the critical points and points not in the domain. Let’s choose x=−4
and x= 0.
Step 4: Plug the test points into the inequality: For x=−4:
2(−4) −1
−4+3 = 3 >2
For x= 0:2(0) −1
0+3 =−1
3≥ 2
Step 5: Analyze the results: Since the expression is greater than 2 for x=−4
and not greater than 2 for x= 0, the solution to the inequality is:
x < −3
17
Question 24
Question
Solve the following rational inequality for x:
2x−3
x+ 1 ≥4
x−2.
Solution
To solve the rational inequality, we will first find the critical points where the
inequality may change and then test each interval defined by these critical points
to determine where the inequality holds.
Step 1: Find the critical points The critical points occur where the
numerator or denominator of one fraction is zero. Equating the numerator and
denominator of the left-hand side to zero gives us: 2x−3 = 0 ⇒x=3
2.
Equating the numerator and denominator of the right-hand side to zero gives
us: x−2 = 0 ⇒x= 2. So, we have critical points at x=3
2and x= 2.
Step 2: Test intervals and solution We will test three intervals: (−∞,3
2),
(3
2,2), and (2,∞)by choosing test points within each interval.
1. For x= 0, we have:
2(0) −3
0+1 ≥4
0−2⇒ −3≥ −2,
which is false. So, (−∞,3
2)is not part of the solution.
2. For x=5
2, we have:
2(5
2)−3
5
2+ 1 ≥4
5
2−2⇒4
5≥ −4,
which is true. So, (3
2,2) is part of the solution.
3. For x= 3, we have:
2(3) −3
3+1 ≥4
3−2⇒3≥4,
which is false. So, (2,∞)is not part of the solution.
Therefore, the solution to the inequality is x∈(3
2,2).
Question 25
Question
Solve the following rational equation for x:
2
x+ 3 −3
x−2=1
x
18
Solution
Step 1: Find a common denominator for all fractions on the left side of the
equation. In this case, the common denominator is (x+ 3)(x−2)x.
Step 2: Rewrite the equation with the common denominator:
2x(x−2)
x(x+ 3)(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 3: Combine the fractions on the left side of the equation:
2x(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 4: Simplify the numerator on the left side of the equation:
2x2−4x−3x2−9x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 5: Combine like terms in the numerator on the left side of the equation:
−x2−13x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 6: Cross multiply to eliminate the denominators:
(−x2−13x)(x) = (x+ 3)(x−2)(x)
Step 7: Expand both sides of the equation:
−x3−13x2=x3−2x2+ 3x2−6x
Step 8: Combine like terms and solve for x:
0 = 2x3+ 15x2−6x
Step 9: Factor out an xfrom the equation:
0 = x(2x2+ 15x−6)
Step 10: Solve for xby setting each factor to zero:
x= 0 or 2x2+ 15x−6 = 0
Step 11: Solve the quadratic equation 2x2+ 15x−6 = 0 using the quadratic
formula or factoring. The solutions are x=−3
2and x=2
3.
Thus, the solutions to the original rational equation are x= 0,x=−3
2, and
x=2
3.
19
Question 26
Question
Let f(x) = 5x2
−3x−2
2x2−7x+3 . Find the domain of the function f(x).
Solution
Step 1: In order to find the domain of a rational function, we need to determine
the values of xfor which the denominator is not equal to zero. This is because
division by zero is undefined. Step 2: Set the denominator of the function equal
to zero and solve for x:
2x2−7x+ 3 = 0
Step 3: We can factor the quadratic equation as:
(2x−1)(x−3) = 0
Step 4: Setting each factor to zero gives us the solutions:
2x−1 = 0 =⇒x=1
2
x−3 = 0 =⇒x= 3
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x=1
2and x= 3. In interval notation, the domain is (−∞,1
2)∪(1
2,3) ∪(3,∞).
Question 27
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Determine the values that make the denominator equal to zero, since
division by zero is undefined.
x2−5x+ 6 = 0
Step 2: Factor the quadratic expression.
(x−2)(x−3) = 0
Step 3: Find the roots of the quadratic equation.
x−2 = 0 ⇒x= 2
20
x−3 = 0 ⇒x= 3
Step 4: The domain is all real numbers except for the values that make the
denominator equal to zero. Therefore, the domain of the rational function is all
real numbers except x= 2 and x= 3.
D:x∈R, x = 2,3
Question 28
Question
Find the domain of the function f(x) = x2
−4
x2−9.
Solution
Step 1: The domain of a function is the set of all real numbers for which the
function is defined. In this case, the function f(x)is defined for all xsuch
that the denominator x2−9is not equal to zero, because division by zero is
undefined.
Step 2: We need to find the values of xthat make x2−9 = 0. This can be
factored as (x−3)(x+ 3) = 0.
Step 3: Setting each factor to zero gives us x−3 = 0 and x+ 3 = 0, which
lead to x= 3 and x=−3.
Step 4: Therefore, the domain of the function f(x) = x2
−4
x2−9is all real numbers
except x= 3 and x=−3. This can be written as (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 29
Question
Solve the rational inequality: x−2
x+3 ≤1.
Solution
To solve the rational inequality x−2
x+3 ≤1, we need to find the values of xthat
satisfy the inequality.
Step 1: Rewrite the inequality without the fraction:
x−2
x+ 3 ≤1⇒x−2≤x+ 3
Step 2: Solve the equation x−2 = x+ 3 to find the boundary points:
x−2 = x+ 3
x−x= 3 + 2
0 = 5
21
This equation has no solution, which means the boundary points of the inequal-
ity are undefined.
Step 3: Test a point in each interval to determine the solution set:
• Test x= 0:0−2
0+3 =−2
3≤1(true)
• Test x=−4:−4−2
−4+3 =−6
−1= 6 ≤1(false)
Step 4: Analyze the test points:
•x= 0 satisfies the inequality.
•x=−4does not satisfy the inequality.
Step 5: Express the solution set: The solution to the rational inequality
x−2
x+3 ≤1is x∈(−∞,0].
Therefore, the solution set is x∈(−∞,0] .
Question 30
Question
Simplify the following rational function:
f(x) = 4x3−27
x2−4
Solution
Step 1: Factor the numerator and denominator:
f(x) = 4x3−27
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Step 2: Simplify the expression by canceling out common factors:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
x2−4
Step 3: Notice that x2−4can be factored further as the difference of squares:
f(x) = (2x−3)(2x2+ 6x+ 9)
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Therefore, the simplified form of the rational function f(x)is:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
22
Question 31
Question
Solve the rational inequality: 4x−8
x+ 3 ≥1.
Solution
Step 1: Begin by rewriting the inequality:
4x−8
x+ 3 ≥1 =⇒4x−8≥x+ 3.
Step 2: Simplify the inequality:
4x−8≥x+ 3 =⇒3x≥11.
Step 3: Solve for x:
3x≥11 =⇒x≥11
3.
Therefore, the solution to the inequality is x≥11
3.
Question 32
Question
Simplify the rational function:
4x2−25
x2−9
Solution
Step 1: Factor the numerator and denominator:
4x2−25 = (2x+ 5)(2x−5)
x2−9 = (x+ 3)(x−3)
Step 2: Rewrite the rational function with factored numerator and denomi-
nator: 4x2−25
x2−9=(2x+ 5)(2x−5)
(x+ 3)(x−3)
Step 3: Simplify the rational function by canceling out common factors:
(2x+ 5)(2x−5)
(x+ 3)(x−3) =2x+ 5
x+ 3
Therefore, the simplified form of the rational function is 2x+5
x+3 .
23
Question 33
Question
Find the domain of the rational function:
f(x) = 3x2+ 5x−2
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to exclude any
x-values that make the denominator equal to zero, since division by zero is
undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor or use the quadratic
formula. Here, we will use factoring.
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1make the denominator zero, so they
are not in the domain of the rational function.
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1. In interval notation, the domain can be expressed
as:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 34
Question
Simplify the rational function 3x3
−5x2+2x
2x2−3x−2.
Solution
Step 1: Factor the numerator and the denominator.
Step 2: Simplify by canceling out common factors.
Step 1: First, let’s factor the numerator and the denominator: The numer-
ator 3x3−5x2+2xcan be expressed as x(3x2−5x+2), where 3x2−5x+2 factors
to (3x−2)(x−1). The denominator 2x2−3x−2factors to (2x+ 1)(x−2).
24
Step 2: Now, substitute the factored forms back into the expression and
simplify:
3x3−5x2+ 2x
2x2−3x−2=x(3x2−5x+ 2)
(2x+ 1)(x−2) =x(3x−2)(x−1)
(2x+ 1)(x−2) .
Notice that the terms (3x−2) and (x−2) can be canceled out, leaving us with:
x(x−1)
2x+ 1 .
Therefore, the simplified form of the rational function is x(x−1)
2x+1 .
Question 35
Question
Simplify the rational expression: 3x2+ 5x−2
2x2−3x−2.
Solution
To simplify the rational expression 3x2+ 5x−2
2x2−3x−2, we need to factor both the
numerator and denominator and then cancel out any common factors.
Step 1: Factor the numerator We need to find two numbers that multiply
to 3× −2 = −6and add up to 5. These numbers are 6and −1. So, we can
rewrite the numerator as: (3x+ 6)(x−1).
Step 2: Factor the denominator Similarly, we need to find two numbers
that multiply to 2×−2 = −4and add up to −3. These numbers are −4and 1.
So, we can rewrite the denominator as: (2x+ 1)(x−2).
Therefore, the rational expression becomes (3x+ 6)(x−1)
(2x+ 1)(x−2).
Step 3: Cancel out common factors We can cancel out the common
factor (x−1) in the numerator and denominator.
Therefore, the simplified form of the rational expression is 3x+ 6
2x+ 1 .
This is the final simplified form of the given rational expression.
25
Question 3
Question
Find the domain of the rational function: f(x) = 1
x2−4x.
Solution
Step 1: The domain of a rational function consists of all real numbers except
those that would make the denominator equal to zero. Therefore, we need to
find the values of xfor which x2−4x= 0.
Step 2: Let’s solve the equation x2−4x= 0 to find the values that make
the denominator zero.
x2−4x= 0
Step 3: Factoring out an xfrom the left side gives:
x(x−4) = 0
Step 4: Setting each factor to zero gives the roots:
x= 0 or x= 4
Step 5: So, the values x= 0 and x= 4 would make the denominator zero.
Therefore, the domain of the function is all real numbers except 0and 4, or in
interval notation: (−∞,0) ∪(0,4) ∪(4,∞).
Question 4
Question
Given the rational function f(x) = 3x2
−6x
x2−4, determine the domain of the func-
tion.
Solution
Step 1: The domain of a rational function is all real numbers except those
that make the denominator equal to zero. This is because division by zero is
undefined. So, to find the domain, we need to find the values of xwhere the
denominator x2−4is equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−4 = 0
(x+ 2)(x−2) = 0
x+ 2 = 0 or x−2 = 0
3
x=−2or x= 2
Step 3: Therefore, the values x=−2,2make the denominator equal to zero,
resulting in the function being undefined at these points. Thus, the domain
of the function is all real numbers except x=−2,2, which can be written in
interval notation as:
(−∞,−2) ∪(−2,2) ∪(2,∞)
Question 5
Question
Find the vertical asymptotes of the rational function:
f(x) = 3x2−5x−2
x2−4x−5
Solution
Step 1: Determine the values of xthat make the denominator of the rational
function equal to zero. These values correspond to the vertical asymptotes of
the function.
x2−4x−5 = 0
(x−5)(x+ 1) = 0
Step 2: Set each factor equal to zero to find the roots:
x−5 = 0 or x+ 1 = 0
x= 5 or x=−1
Step 3: The vertical asymptotes of the rational function are at x= 5 and
x=−1. Thus, the vertical asymptotes are x= 5 and x=−1.
Question 6
Question
Find the domain of the rational function:
f(x) = 2x2−5x−3
x2−4x−5
Solution
Step 1: To find the domain of a rational function, we need to determine all
the values of xfor which the function is defined. Since division by zero is not
allowed, we need to find the values that make the denominator zero.
4
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: Factor the quadratic equation:
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero:
x−5 = 0 ⇒x= 5
x+ 1 = 0 ⇒x=−1
Step 5: Therefore, the domain of the function is all real numbers except
x= 5 and x=−1.
Domain: x∈R, x = 5,−1
Question 7
Question
Let f(x) = 2x3+3x2
−2x−3
x2−1and g(x) = x2
−1
x2+2x+1 . Find the domain of f◦g(x).
Solution
Step 1: Find the composition f◦g(x) = f(g(x)) by substituting g(x)into f(x).
f(g(x)) = f(x2−1
x2+ 2x+ 1)
Step 2: Simplify the expression inside fby replacing xwith x2
−1
x2+2x+1 .
f(x2−1
x2+ 2x+ 1)=
2(x2
−1
x2+2x+1 )3+ 3 (x2
−1
x2+2x+1 )2
−2(x2
−1
x2+2x+1 )−3
(x2−1
x2+2x+1 )2
−1
Step 3: Simplify the expression further.
f(x2−1
x2+ 2x+ 1)=
2(x2
−1)3
(x2+2x+1)3+3(x2
−1)2
(x2+2x+1)2−2(x2
−1)
x2+2x+1 −3
(x2−1)2
(x2+2x+1)2−1
Step 4: Determine the domain of f◦g(x)by considering restrictions on the
denominator. Since the denominator cannot be equal to 0, we need to find
where x2+ 2x+ 1 = 0.
x2+ 2x+ 1 = 0
5
(x+ 1)(x+ 1) = 0
(x+ 1)2= 0
x=−1
Therefore, the domain of f◦g(x)is all real numbers except x=−1.
Question 8
Question
Find the domain of the following rational function:
f(x) = 4x2−7x−3
x2−5x+ 6
Solution
To find the domain of a rational function, we need to identify the values of x
for which the function is defined. The function will be undefined at any value
of xthat makes the denominator equal to zero.
Step 1: Find the values of xthat make the denominator zero.
Set the denominator x2−5x+ 6 equal to zero and solve for x:
x2−5x+ 6 = 0
Step 2: Solve for x.
Factor the quadratic equation:
(x−2)(x−3) = 0
Set each factor equal to zero and solve for x:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 3: Determine the domain.
The values of xthat make the denominator zero are x= 2 and x= 3.
Therefore, the domain of the function f(x) = 4x2−7x−3
x2−5x+ 6 is all real numbers
except x= 2 and x= 3.
Answer: The domain of the function is all real numbers except x= 2 and
x= 3.
Question 9
Question
Let f(x) = x3
−3x2
−4x+12
x2−2x−8. Determine the x-intercepts, y-intercepts, vertical
asymptotes, horizontal asymptotes, and the domain of the function f(x).
6
Solution
Step 1: To find the x-intercepts, set f(x) = 0 and solve for x.
0 = x3−3x2−4x+ 12
x2−2x−8
0 = x3−3x2−4x+ 12
Using synthetic division or polynomial long division, we can find that x= 3 is a
root, which means (x−3) is a factor. Dividing through gives (x−3)(x2+x−4) =
0. Factoring the quadratic factor further, we get (x−3)(x−1)(x+ 4) = 0, so
the x-intercepts are x= 3,x= 1, and x=−4.
Step 2: To find the y-intercept, set x= 0 and evaluate f(0).
f(0) = 03−3(0)2−4(0) + 12
02−2(0) −8=12
−8=−3
2
Thus, the y-intercept is at (0,−3
2).
Step 3: To find the vertical asymptotes, set the denominator equal to zero
and solve for x.
x2−2x−8 = 0
(x−4)(x+ 2) = 0
So, the vertical asymptotes are x= 4 and x=−2.
Step 4: To find the horizontal asymptote, compare the degrees of the numer-
ator and denominator of the rational function. Since the degree of the numerator
is greater than the degree of the denominator, there is no horizontal asymptote.
Step 5: The domain of the function f(x)is all real numbers except where
the denominator is equal to zero. Thus, the domain is x= 4,−2.
Therefore, the x-intercepts are x= 3,x= 1, and x=−4, the y-intercept is
at (0,−3
2), the vertical asymptotes are at x= 4 and x=−2, the function has
no horizontal asymptote, and the domain of f(x)is x= 4,−2.
Question 10
Question
Let f(x) = 2x3
−x2
−9x
x2−x−6. Determine the domain of f(x).
Solution
Step 1: To find the domain of f(x), we need to identify the values of xthat will
make the denominator of f(x)equal to zero, since division by zero is undefined.
Step 2: Set the denominator x2−x−6equal to zero and solve for x:
x2−x−6 = 0
7
Step 3: Factor the quadratic equation:
(x−3)(x+ 2) = 0
Step 4: Set each factor equal to zero and solve for x: For x−3 = 0,x= 3
For x+ 2 = 0,x=−2
Step 5: Therefore, the values x= 3 and x=−2make the denominator
of f(x)zero. Thus, the domain of f(x)is all real numbers except x= 3 and
x=−2. In interval notation, this can be written as (−∞,−2)∪(−2,3)∪(3,∞).
Question 11
Question
Find the vertical asymptotes of the rational function given by f(x) = 3x2−x−2
x2−4x−5.
Solution
Step 1: First, factor the numerator and denominator of the rational function.
f(x) = 3x2−x−2
x2−4x−5=(3x+ 2)(x−1)
(x−5)(x+ 1)
Step 2: Next, identify the values of xthat make the denominator equal to
zero. These values will give us the vertical asymptotes of the function. Setting
the denominator equal to zero:
(x−5)(x+ 1) = 0
This gives us two possible vertical asymptotes:
x= 5 and x=−1
Step 3: Therefore, the vertical asymptotes of the rational function f(x)are
x= 5 and x=−1.
Question 12
Question
Determine the domain of the rational function:
f(x) = 2x2−3x−2
x2−4x+ 3 .
8
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xfor which the function is defined. The function is defined for all x
values except where the denominator is equal to zero because division by zero
is undefined.
Step 2: Find the values of xthat make the denominator zero by solving the
equation x2−4x+ 3 = 0.
Step 3: Factor the quadratic equation x2−4x+3 = 0 to get (x−3)(x−1) = 0.
Step 4: Set each factor to zero and solve for x.
{x−3 = 0
x−1 = 0
Step 5: Solve the equations to find the values of x.
{x= 3
x= 1
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 3 and x= 1. In interval notation, the domain is (−∞,1) ∪
(1,3) ∪(3,∞).
Question 13
Question
Find the domain of the rational function f(x) = 2x2−x−3
x2−9.
Solution
Step 1: Remember that the domain of a rational function is all real numbers
except for the values of xthat make the denominator zero, since division by zero
is undefined. Hence, we need to find the values of xthat make the denominator
x2−9equal to zero.
Step 2: Set the denominator equal to zero and solve for x:
x2−9 = 0
Step 3: Factor the quadratic expression:
x2−9 = (x+ 3)(x−3) = 0
Step 4: Set each factor equal to zero and solve for x:
{x+ 3 = 0 =⇒x=−3
x−3 = 0 =⇒x= 3
9
Step 5: The values x=−3and x= 3 would make the denominator zero, so
they are not included in the domain of the function f(x).
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers excluding x=−3and x= 3, which can be expressed in interval notation
as (−∞,−3) ∪(−3,3) ∪(3,+∞).
Question 14
Question
Find the domain of the rational function:
f(x) = 3x2−2x−8
x2−5x+ 6
Solution
Step 1: To find the domain of a rational function, we need to determine all the
values of xthat make the denominator equal to zero, since division by zero is
undefined. So, we need to solve the equation x2−5x+ 6 = 0 to find the values
of xthat would make the denominator zero.
Step 2: Factoring the quadratic equation x2−5x+ 6 = 0, we get (x−2)(x−
3) = 0.
Step 3: Setting each factor to zero, we find the roots: x−2 = 0 ⇒x= 2
and x−3 = 0 ⇒x= 3.
Step 4: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 2 and x= 3. This can be written in interval notation as
(−,2)(2,3)(3,).
Question 15
Question
Solve the following rational equation: 2x+3
x−1=4x−1
x+3 .
Solution
Step 1: Cross multiply to eliminate the fractions.
(x−1)(4x−1) = (x+ 3)(2x+ 3)
4x2−x−4x+ 1 = 2x2+ 3x+ 6x+ 9
4x2−5x+ 1 = 2x2+ 9x+ 9
10
Step 2: Move all terms to one side of the equation.
4x2−5x+ 1 −2x2−9x−9 = 0
2x2−14x−8 = 0
Step 3: Factor the quadratic equation.
2(x2−7x−4) = 0
2(x−8)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x.
x−8 = 0 or x+ 1 = 0
x= 8 or x=−1
Step 5: Check for extraneous solutions.
Since x= 1 would make the denominators zero in the original equation,
x= 1 is an extraneous solution.
Therefore, the solution to the rational equation is x= 8.
Question 16
Question
Simplify the following rational function:
3x2−8x−3
2x2−5x−3
Solution
Step 1: Factor the numerator and denominator:
3x2−8x−3 = (3x+ 1)(x−3)
2x2−5x−3 = (2x+ 1)(x−3)
Step 2: Rewrite the rational function with the factored forms:
(3x+ 1)(x−3)
(2x+ 1)(x−3)
Step 3: Cancel out the common factor of (x−3) in the numerator and
denominator: 3x+ 1
2x+ 1
Therefore, the simplified form of the given rational function is 3x+1
2x+1 .
11
Question 17
Question
Solve the rational equation: 3
x−1−1
x+3 =1
2.
Solution
Step 1: To solve the given rational equation, we will first find a common denomi-
nator for the fractions on the left side of the equation. The common denominator
will be (x−1)(x+ 3).
Step 2: Rewrite the equation with the common denominator:
(x+ 3) 3
x−1−(x−1) 1
x+ 3 =1
2(x−1)(x+ 3)
Step 3: Simplify the equation:
3(x+ 3) −(x−1) = 1
2(x2+ 2x−3)
Step 4: Expand and simplify the equation:
3x+ 9 −x+ 1 = 1
2x2+x−3
2
2x+ 10 = 1
2x2+x−3
2
Step 5: Multiply both sides of the equation by 2 to eliminate the fraction:
4x+ 20 = x2+ 2x−3
Step 6: Rearrange the equation into standard form:
0 = x2−2x−23
Step 7: Factor the quadratic equation:
0 = (x−5)(x+ 3)
Step 8: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 3 = 0 =⇒x=−3
Step 9: Check for extraneous solutions by plugging each potential solution
back into the original equation: - For x= 5:3
5−1−1
5+3 =1
2, which is true. -
For x=−3:3
−3−1−1
−3+3 =1
2, which is not valid as it results in division by 0.
Therefore, the only solution to the given rational equation is x= 5.
12
Question 18
Question
Solve for x:
3x
x−4−x
x−2=4
x−4
Solution
Step 1: Find a common denominator for the fractions on the left side of the
equation:
3x
x−4−x
x−2=4
x−4
3x(x−2)
(x−4)(x−2) −x(x−4)
(x−2)(x−4) =4
x−4
Step 2: Combine the fractions on the left side of the equation:
3x(x−2) −x(x−4)
(x−4)(x−2) =4
x−4
3x2−6x−x2+ 4x
(x−4)(x−2) =4
x−4
2x2−2x
(x−4)(x−2) =4
x−4
Step 3: Simplify the equation:
2x(x−1)
(x−4)(x−2) =4
x−4
Step 4: Clear the denominators by multiplying both sides by (x−4)(x−2):
2x(x−1) = 4(x−2)
Step 5: Expand and simplify:
2x2−2x= 4x−8
2x2−2x= 4x−8
Step 6: Rearrange the equation in standard form:
2x2−2x−4x+ 8 = 0
2x2−6x+ 8 = 0
Step 7: Factor the quadratic equation:
13
2(x2−3x+ 4) = 0
2(x−2)(x−2) = 0
Step 8: Find the solutions:
x= 2, x = 2
However, the repeating solution, x= 2, is extraneous as it makes the de-
nominator equal to zero. Thus, the solution to the equation is x= 2.
Question 19
Question
Find the domain of the rational function: f(x) = x2−4
x2−9.
Solution
To find the domain of a rational function, we need to identify all real numbers
that make the denominator equal to zero since division by zero is undefined.
Step 1: Identify the denominator In this case, the denominator is x2−9.
Step 2: Find the values that make the denominator zero Set the
denominator equal to zero and solve for x:
x2−9 = 0
(x+ 3)(x−3) = 0
The solutions are x= 3 and x=−3.
Step 3: Determine the domain The domain of the function is all real
numbers except for the values that make the denominator zero. Therefore, the
domain of the function is (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 20
Question
Solve the rational inequality: 3x2+7x+2
x2−2x−3>0.
14
Solution
Step 1: Find the critical points by setting the numerator and denominator equal
to zero and solving for x.
3x2+ 7x+ 2 = 0
(x+ 1)(3x+ 2) = 0
x=−1,−2
3
x2−2x−3 = 0
(x−3)(x+ 1) = 0
x= 3,−1
Step 2: Create a sign chart using the critical points −1,−2
3,3.
− −2
3−1 3
3x2+ 7x+ 2 −+ + +
x2−2x−3 + −+ +
Step 3: Analyze the sign chart to determine where the rational function is
positive. Since we are looking for when 3x2+7x+2
x2−2x−3>0, we want to find where this
expression is positive, meaning when the function is above the x-axis. Therefore,
the solution to the inequality is x∈(−∞,−1) ∪(−2
3,3).
Question 21
Question
Simplify the rational expression:
2x2−5x−3
x2+ 2x−3∇ · x2−4
x2+x−6
Solution
To simplify the given rational expression, we will first factor all the quadratic
expressions involved.
Step 1: Factor all the quadratic expressions: For the numerator of the first
fraction, 2x2−5x−3, we have:
2x2−5x−3 = (2x+ 1)(x−3)
For the denominator of the first fraction, x2+ 2x−3, we have:
x2+ 2x−3 = (x+ 3)(x−1)
15
For the numerator of the second fraction, x2−4, we have:
x2−4 = (x+ 2)(x−2)
For the denominator of the second fraction, x2+x−6, we have:
x2+x−6 = (x+ 3)(x−2)
Step 2: Rewrite the expression with factored terms:
(2x+ 1)(x−3)
(x+ 3)(x−1) ∇ · (x+ 2)(x−2)
(x+ 3)(x−2)
Step 3: Simplify by multiplying by the reciprocal of the second fraction:
(2x+ 1)(x−3)
(x+ 3)(x−1) ×(x+ 3)(x−2)
(x+ 2)(x−2)
Step 4: Cancel out the common factors:
(2x+ 1)(x−3)(x+ 3)(x−2)
(x+ 3)(x−1)(x+ 2)
Step 5: Simplify the expression further:
2x+ 1
x+ 2
Therefore, the simplified form of the given rational expression is 2x+1
x+2 .
Question 22
Question
Find the domain of the rational function: f(x) = x2−4
x2−5x+ 6.
Solution
Step 1: Recall that the domain of a function is the set of all real numbers for
which the function is defined. In the case of rational functions, we need to
consider the values of xthat would make the denominator equal to zero, since
division by zero is undefined.
Step 2: Set the denominator equal to zero and solve for xto find the values
that will make the function undefined:
x2−5x+ 6 = 0
Step 3: Factor the quadratic equation:
(x−2)(x−3) = 0
16
Step 4: Set each factor equal to zero:
x−2 = 0 =⇒x= 2
x−3 = 0 =⇒x= 3
Step 5: The values of xthat make the denominator zero are x= 2 and
x= 3. Therefore, the domain of the function excludes these values.
Step 6: So, the domain of the rational function f(x)is all real numbers
except x= 2 and x= 3. In interval notation, the domain can be written as
(−∞,2) ∪(2,3) ∪(3,∞).
Question 23
Question
Solve the inequality 2x−1
x+3 ≥2.
Solution
Step 1: Find the critical points by setting the expression equal to zero:
2x−1
x+ 3 = 2
2x−1 = 2(x+ 3)
2x−1 = 2x+ 6
0 = 7
Since the equation 0 = 7 is false, there are no critical points in this case.
Step 2: Determine the values not in the domain of the inequality: The
expression 2x−1
x+3 is not defined when x=−3. Therefore, x=−3.
Step 3: Choose test points: Select test points in each of the intervals deter-
mined by the critical points and points not in the domain. Let’s choose x=−4
and x= 0.
Step 4: Plug the test points into the inequality: For x=−4:
2(−4) −1
−4+3 = 3 >2
For x= 0:2(0) −1
0+3 =−1
3≥ 2
Step 5: Analyze the results: Since the expression is greater than 2 for x=−4
and not greater than 2 for x= 0, the solution to the inequality is:
x < −3
17
Question 24
Question
Solve the following rational inequality for x:
2x−3
x+ 1 ≥4
x−2.
Solution
To solve the rational inequality, we will first find the critical points where the
inequality may change and then test each interval defined by these critical points
to determine where the inequality holds.
Step 1: Find the critical points The critical points occur where the
numerator or denominator of one fraction is zero. Equating the numerator and
denominator of the left-hand side to zero gives us: 2x−3 = 0 ⇒x=3
2.
Equating the numerator and denominator of the right-hand side to zero gives
us: x−2 = 0 ⇒x= 2. So, we have critical points at x=3
2and x= 2.
Step 2: Test intervals and solution We will test three intervals: (−∞,3
2),
(3
2,2), and (2,∞)by choosing test points within each interval.
1. For x= 0, we have:
2(0) −3
0+1 ≥4
0−2⇒ −3≥ −2,
which is false. So, (−∞,3
2)is not part of the solution.
2. For x=5
2, we have:
2(5
2)−3
5
2+ 1 ≥4
5
2−2⇒4
5≥ −4,
which is true. So, (3
2,2) is part of the solution.
3. For x= 3, we have:
2(3) −3
3+1 ≥4
3−2⇒3≥4,
which is false. So, (2,∞)is not part of the solution.
Therefore, the solution to the inequality is x∈(3
2,2).
Question 25
Question
Solve the following rational equation for x:
2
x+ 3 −3
x−2=1
x
18
Solution
Step 1: Find a common denominator for all fractions on the left side of the
equation. In this case, the common denominator is (x+ 3)(x−2)x.
Step 2: Rewrite the equation with the common denominator:
2x(x−2)
x(x+ 3)(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 3: Combine the fractions on the left side of the equation:
2x(x−2) −3x(x+ 3)
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 4: Simplify the numerator on the left side of the equation:
2x2−4x−3x2−9x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 5: Combine like terms in the numerator on the left side of the equation:
−x2−13x
x(x+ 3)(x−2) =(x+ 3)(x−2)
x(x+ 3)(x−2)
Step 6: Cross multiply to eliminate the denominators:
(−x2−13x)(x) = (x+ 3)(x−2)(x)
Step 7: Expand both sides of the equation:
−x3−13x2=x3−2x2+ 3x2−6x
Step 8: Combine like terms and solve for x:
0 = 2x3+ 15x2−6x
Step 9: Factor out an xfrom the equation:
0 = x(2x2+ 15x−6)
Step 10: Solve for xby setting each factor to zero:
x= 0 or 2x2+ 15x−6 = 0
Step 11: Solve the quadratic equation 2x2+ 15x−6 = 0 using the quadratic
formula or factoring. The solutions are x=−3
2and x=2
3.
Thus, the solutions to the original rational equation are x= 0,x=−3
2, and
x=2
3.
19
Question 26
Question
Let f(x) = 5x2
−3x−2
2x2−7x+3 . Find the domain of the function f(x).
Solution
Step 1: In order to find the domain of a rational function, we need to determine
the values of xfor which the denominator is not equal to zero. This is because
division by zero is undefined. Step 2: Set the denominator of the function equal
to zero and solve for x:
2x2−7x+ 3 = 0
Step 3: We can factor the quadratic equation as:
(2x−1)(x−3) = 0
Step 4: Setting each factor to zero gives us the solutions:
2x−1 = 0 =⇒x=1
2
x−3 = 0 =⇒x= 3
Step 5: Therefore, the domain of the function f(x)is all real numbers except
x=1
2and x= 3. In interval notation, the domain is (−∞,1
2)∪(1
2,3) ∪(3,∞).
Question 27
Question
Find the domain of the rational function:
f(x) = x2−4
x2−5x+ 6
Solution
Step 1: Determine the values that make the denominator equal to zero, since
division by zero is undefined.
x2−5x+ 6 = 0
Step 2: Factor the quadratic expression.
(x−2)(x−3) = 0
Step 3: Find the roots of the quadratic equation.
x−2 = 0 ⇒x= 2
20
x−3 = 0 ⇒x= 3
Step 4: The domain is all real numbers except for the values that make the
denominator equal to zero. Therefore, the domain of the rational function is all
real numbers except x= 2 and x= 3.
D:x∈R, x = 2,3
Question 28
Question
Find the domain of the function f(x) = x2
−4
x2−9.
Solution
Step 1: The domain of a function is the set of all real numbers for which the
function is defined. In this case, the function f(x)is defined for all xsuch
that the denominator x2−9is not equal to zero, because division by zero is
undefined.
Step 2: We need to find the values of xthat make x2−9 = 0. This can be
factored as (x−3)(x+ 3) = 0.
Step 3: Setting each factor to zero gives us x−3 = 0 and x+ 3 = 0, which
lead to x= 3 and x=−3.
Step 4: Therefore, the domain of the function f(x) = x2
−4
x2−9is all real numbers
except x= 3 and x=−3. This can be written as (−∞,−3) ∪(−3,3) ∪(3,∞).
Question 29
Question
Solve the rational inequality: x−2
x+3 ≤1.
Solution
To solve the rational inequality x−2
x+3 ≤1, we need to find the values of xthat
satisfy the inequality.
Step 1: Rewrite the inequality without the fraction:
x−2
x+ 3 ≤1⇒x−2≤x+ 3
Step 2: Solve the equation x−2 = x+ 3 to find the boundary points:
x−2 = x+ 3
x−x= 3 + 2
0 = 5
21
This equation has no solution, which means the boundary points of the inequal-
ity are undefined.
Step 3: Test a point in each interval to determine the solution set:
• Test x= 0:0−2
0+3 =−2
3≤1(true)
• Test x=−4:−4−2
−4+3 =−6
−1= 6 ≤1(false)
Step 4: Analyze the test points:
•x= 0 satisfies the inequality.
•x=−4does not satisfy the inequality.
Step 5: Express the solution set: The solution to the rational inequality
x−2
x+3 ≤1is x∈(−∞,0].
Therefore, the solution set is x∈(−∞,0] .
Question 30
Question
Simplify the following rational function:
f(x) = 4x3−27
x2−4
Solution
Step 1: Factor the numerator and denominator:
f(x) = 4x3−27
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Step 2: Simplify the expression by canceling out common factors:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
x2−4
Step 3: Notice that x2−4can be factored further as the difference of squares:
f(x) = (2x−3)(2x2+ 6x+ 9)
x2−4=(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2) =(2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
Therefore, the simplified form of the rational function f(x)is:
f(x) = (2x−3)(2x2+ 6x+ 9)
(x−2)(x+ 2)
22
Question 31
Question
Solve the rational inequality: 4x−8
x+ 3 ≥1.
Solution
Step 1: Begin by rewriting the inequality:
4x−8
x+ 3 ≥1 =⇒4x−8≥x+ 3.
Step 2: Simplify the inequality:
4x−8≥x+ 3 =⇒3x≥11.
Step 3: Solve for x:
3x≥11 =⇒x≥11
3.
Therefore, the solution to the inequality is x≥11
3.
Question 32
Question
Simplify the rational function:
4x2−25
x2−9
Solution
Step 1: Factor the numerator and denominator:
4x2−25 = (2x+ 5)(2x−5)
x2−9 = (x+ 3)(x−3)
Step 2: Rewrite the rational function with factored numerator and denomi-
nator: 4x2−25
x2−9=(2x+ 5)(2x−5)
(x+ 3)(x−3)
Step 3: Simplify the rational function by canceling out common factors:
(2x+ 5)(2x−5)
(x+ 3)(x−3) =2x+ 5
x+ 3
Therefore, the simplified form of the rational function is 2x+5
x+3 .
23
Question 33
Question
Find the domain of the rational function:
f(x) = 3x2+ 5x−2
x2−4x−5
Solution
Step 1: To find the domain of the rational function, we need to exclude any
x-values that make the denominator equal to zero, since division by zero is
undefined.
Step 2: Set the denominator equal to zero and solve for x:
x2−4x−5 = 0
Step 3: To solve the quadratic equation, we can factor or use the quadratic
formula. Here, we will use factoring.
(x−5)(x+ 1) = 0
Step 4: Set each factor to zero and solve for x:
x−5 = 0 =⇒x= 5
x+ 1 = 0 =⇒x=−1
Step 5: The values x= 5 and x=−1make the denominator zero, so they
are not in the domain of the rational function.
Step 6: Therefore, the domain of the rational function f(x)is all real num-
bers except x= 5 and x=−1. In interval notation, the domain can be expressed
as:
(−∞,−1) ∪(−1,5) ∪(5,∞)
Question 34
Question
Simplify the rational function 3x3
−5x2+2x
2x2−3x−2.
Solution
Step 1: Factor the numerator and the denominator.
Step 2: Simplify by canceling out common factors.
Step 1: First, let’s factor the numerator and the denominator: The numer-
ator 3x3−5x2+2xcan be expressed as x(3x2−5x+2), where 3x2−5x+2 factors
to (3x−2)(x−1). The denominator 2x2−3x−2factors to (2x+ 1)(x−2).
24
Step 2: Now, substitute the factored forms back into the expression and
simplify:
3x3−5x2+ 2x
2x2−3x−2=x(3x2−5x+ 2)
(2x+ 1)(x−2) =x(3x−2)(x−1)
(2x+ 1)(x−2) .
Notice that the terms (3x−2) and (x−2) can be canceled out, leaving us with:
x(x−1)
2x+ 1 .
Therefore, the simplified form of the rational function is x(x−1)
2x+1 .
Question 35
Question
Simplify the rational expression: 3x2+ 5x−2
2x2−3x−2.
Solution
To simplify the rational expression 3x2+ 5x−2
2x2−3x−2, we need to factor both the
numerator and denominator and then cancel out any common factors.
Step 1: Factor the numerator We need to find two numbers that multiply
to 3× −2 = −6and add up to 5. These numbers are 6and −1. So, we can
rewrite the numerator as: (3x+ 6)(x−1).
Step 2: Factor the denominator Similarly, we need to find two numbers
that multiply to 2×−2 = −4and add up to −3. These numbers are −4and 1.
So, we can rewrite the denominator as: (2x+ 1)(x−2).
Therefore, the rational expression becomes (3x+ 6)(x−1)
(2x+ 1)(x−2).
Step 3: Cancel out common factors We can cancel out the common
factor (x−1) in the numerator and denominator.
Therefore, the simplified form of the rational expression is 3x+ 6
2x+ 1 .
This is the final simplified form of the given rational expression.
25
