MATH 121 - COLLEGE ALGEBRA Rational Functions Question Bank Set 1

MATH 121 - COLLEGE ALGEBRA -

Rational Functions

Question Bank - Set 1

Liberty University

Question 1

Question

Simplify the following rational function:

f(x) = 2x2−4x−6

x2−2x−3

Solution

Step 1: Factor the numerator and denominator. Step 2: Simplify by canceling

out common factors. Step 3: Write the simplified form of the rational function.

Step 1: Factor the numerator and denominator. The numerator 2x2−4x−6

can be factored as 2(x+ 1)(x−3). The denominator x2−2x−3can be factored

as (x−3)(x+ 1).

Step 2: Simplify by canceling out common factors.

f(x) = 2(x+ 1)(x−3)

(x−3)(x+ 1)

f(x) = 2(x+ 1)(x−3)

(x−3)(x+ 1)

Step 3: Write the simplified form of the rational function.

f(x) = 2

Therefore, the simplified form of the given rational function is f(x) = 2.

Question 2

Question

Let f(x) = 3x2−x−2

x2−5x+ 6 and g(x) = 2x2−x−1

x2−4x+ 3 be two rational functions.

Determine the domain of the composite function f(g(x)).

Solution

To determine the domain of the composite function f(g(x)), we need to consider

restrictions on both f(x)and g(x).

Step 1: Determine the domain of g(x)Since denominators cannot be

zero, we need to find the values of xthat would make the denominators of g(x)

equal to zero:

x2−4x+ 3 = 0

Factor the quadratic:

(x−3)(x−1) = 0

So, x= 3 and x= 1 are the values that would make the denominator of g(x)

equal to zero. Therefore, the domain of g(x)is all real numbers except x= 3

and x= 1.

Step 2: Determine the domain of f(g(x)) The domain of f(g(x)) is the

set of all xvalues for which g(x)is in the domain of f(x). Since g(x)cannot

be 1or 3, we need to exclude these values from the domain of f(x)as well. We

will find these values by substituting g(x)into f(x):

f(g(x)) = 3(g(x))2−g(x)−2

(g(x))2−5(g(x)) + 6

Substitute g(x):

f(g(x)) = 3(2x2−x−1

x2−4x+3 )2−2x2−x−1

x2−4x+3 −2

(2x2−x−1

x2−4x+3 )2−5(2x2−x−1

x2−4x+3 )+6

Simplify to find the domain. Since the domain of g(x)is not 1or 3, the domain

of f(g(x)) is all real numbers except 1or 3.

Question 3

Question

Simplify the following rational expression:

4x2−12x

x3−5x2+ 6x

Solution

Step 1: Factor out the greatest common factor in both the numerator and

denominator. 4x(x−3)

x(x−3)(x−2)

Step 2: Simplify the expression by canceling out common factors.

x−2

So, the simplified form of the rational expression is 4

x−2.

Question 4

Question

Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph

of the rational function:

f(x) = 2x3+ 3

x2−5x+ 6

Solution

Step 1: To find the vertical asymptotes, set the denominator equal to zero and

solve for x:

x2−5x+ 6 = 0

Factoring the quadratic equation gives:

(x−2)(x−3) = 0

So, x= 2 and x= 3 are vertical asymptotes.

Step 2: To find the horizontal asymptote, we compare the degrees of the

numerator and denominator. Since the degree of the numerator is greater than

the degree of the denominator, there is no horizontal asymptote.

Step 3: To find any holes in the graph, we need to simplify the function.

We can simplify the function by factoring out (x-3) from the numerator and

denominator:

f(x) = 2x3+ 3

(x−3)(x−2) =2(x2+ 3x+ 1)

(x−3)(x−2)

We can see that there is no common factor between the numerator and

denominator. Therefore, there are no holes in the graph.

In conclusion, the rational function has vertical asymptotes at x= 2 and

x= 3, no horizontal asymptote, and no holes in the graph.

Question 5

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 identify values of

xthat would make the denominator 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: Factor the quadratic equation:

(x−5)(x+ 1) = 0

Step 4: Solve for x:

x−5 = 0 or x+ 1 = 0

x= 5 or x=−1

Step 5: The domain of the function is all real numbers except where the

denominator is zero. Therefore, the domain of the function is:

(−∞,−1) ∪(−1,5) ∪(5,+∞)

Question 6

Question

Let f(x) = 2x2−x−3

x2−2x−3. Find the domain of the function f(x).

Solution

Step 1: To find the domain of f(x), we need to determine which values of xwill

make the denominator equal to zero, since division by zero is undefined.

Step 2: Set the denominator x2−2x−3equal to zero and solve for x:

x2−2x−3 = 0

Step 3: Factoring the quadratic equation:

(x−3)(x+ 1) = 0

Step 4: Setting each factor to zero:

x−3 = 0 or x+ 1 = 0

Step 5: Solving for xin each case:

x= 3 or x=−1

Step 6: The values x= 3 and x=−1make the denominator of f(x)equal

to zero. Therefore, the domain of f(x)is all real numbers except x= 3 and

x=−1.

Step 7: Thus, the domain of the function f(x)is (−∞,−1) ∪(−1,3) ∪(3,∞).

Question 7

Question

Simplify the following rational function:

f(x) = 2x3−5x2+ 2x

x3−2x2−x+ 2

Solution

Step 1: Factor both the numerator and denominator.

The numerator is already factored, so we have:

f(x) = 2x(x2−5x+ 2)

x3−2x2−x+ 2

Now, let’s factor the denominator by grouping:

x3−2x2−x+ 2 = x2(x−2) −1(x−2)

= (x2−1)(x−2)

= (x+ 1)(x−1)(x−2)

So now we have:

f(x) = 2x(x2−5x+ 2)

(x+ 1)(x−1)(x−2)

Step 2: Simplify the rational function by canceling out common factors.

f(x) = 2x(x2−5x+ 2)

(x+ 1)(x−1)(x−2)

f(x) = 2x(x−1)(x−2)

(x+ 1)(x−1)(x−2)

f(x) = 2x

(x+ 1)

Therefore, the simplified form of the rational function is:

f(x) = 2x

x+ 1

Question 8

Question

Simplify the following rational expression:

3x2−7x−6

x2+ 2x−8

Solution

Step 1: Factor the numerator and denominator:

3x2−7x−6 = (3x+ 2)(x−3)

x2+ 2x−8 = (x+ 4)(x−2)

Step 2: Rewrite the expression with the factored terms:

(3x+ 2)(x−3)

(x+ 4)(x−2)

Step 3: Simplify the expression by canceling out common factors:

(3x+ 2)(x−3)

(x+ 4)(x−2) =3x+ 2

x+ 4

Therefore, the simplified form of the given rational expression is 3x+2

x+4 .

Question 9

Question

Find the domain of the rational function:

f(x) = x2+ 3x−4

x2−9

Solution

Step 1: We need to find the values of xfor which the denominator is not equal

to zero, since division by zero is undefined. Thus, we set the denominator x2−9

not equal to zero and solve for x:

x2−9= 0

(x+ 3)(x−3) = 0

This gives us two critical points, x=−3and x= 3.

Step 2: Next, we need to consider any restrictions on the function due to

the numerator. Since the numerator is a polynomial, it is defined for all real

numbers.

Step 3: Therefore, the domain of the rational function is all real numbers

except x=−3, x = 3.

Step 4: In interval notation, the domain of the function fis (−∞,−3) ∪

(−3,3) ∪(3,∞).

Question 10

Question

Find the vertical asymptotes, horizontal asymptotes, and any holes (points of

discontinuity) of the rational function:

f(x) = 3x2−2x−8

x2−4

Solution

Step 1: To find the vertical asymptotes, we need to determine where the de-

nominator of the rational function equals zero (excluding any points where the

numerator is also zero). Set the denominator x2−4 = 0 and solve for x.

x2−4 = 0

(x−2)(x+ 2) = 0

This gives x= 2 and x=−2. Therefore, the vertical asymptotes are at x= 2

and x=−2.

Step 2: To find the horizontal asymptote, we compare the degrees of the

numerator and denominator. Since the degree of the numerator is equal to the

degree of the denominator, the horizontal asymptote is the ratio of the leading

coeﬀicients.

lim

x→∞

3x2−2x−8

x2−4

The horizontal asymptote is y= 3.

Step 3: To find any holes in the graph (points of discontinuity), we factor

the numerator and denominator to see if there are any common factors that can

be canceled.

f(x) = (3x+ 4)(x−2)

(x+ 2)(x−2)

Since there are common factors of (x−2) in the numerator and denominator,

there is a hole at x= 2.

Therefore, the rational function has vertical asymptotes at x= 2 and x=

−2, a horizontal asymptote at y= 3, and a hole at x= 2.

Question 11

Question

Find the horizontal asymptotes of the rational function f(x) = 2x2−3x+1

x2+1 .

Solution

Step 1: To find the horizontal asymptotes, we need to compare the degrees of

the numerator and denominator of the rational function. If the degree of the

numerator is greater than or equal to the degree of the denominator, there is no

horizontal asymptote (HA) because the graph of the rational function will have

a slant asymptote. If the degree of the denominator is greater than the degree

of the numerator, there will be a horizontal asymptote at y= 0.

Step 2: In this case, the degree of the numerator is 2 and the degree of

the denominator is also 2. Since the degrees are equal, we need to find the

limit as xapproaches positive and negative infinity to determine the horizontal

asymptotes.

Step 3: As xapproaches positive infinity, the terms with the highest de-

gree in the numerator and denominator dominate the behavior of the function.

Therefore, the horizontal asymptote can be found by dividing the leading coef-

ficients of the highest degree terms.

Step 4: The horizontal asymptote for f(x) = 2x2−3x+1

x2+1 is y=2

1= 2 as x

approaches positive infinity.

Step 5: As xapproaches negative infinity, the horizontal asymptote will be

the same as the limit as xapproaches positive infinity in this case.

Step 6: Therefore, the horizontal asymptote for f(x) = 2x2−3x+1

x2+1 is y= 2.

So, the horizontal asymptote for the given rational function is y= 2.

Question 12

Question

Given the rational function f(x) = 2x2+5x−3

x2−4, find the domain of f(x).

Solution

Step 1: Determine where the denominator is equal to zero to find values that

make the function undefined. The denominator x2−4is equal to zero when

x=±2.

Step 2: So, the function f(x)is undefined at x=±2. Thus, the domain

of f(x)is all real numbers except x=±2. Therefore, the domain of f(x)is

(−∞,−2) ∪(−2,2) ∪(2,∞).

Question 13

Question

Let f(x) = x2+3x−4

2x−1. Find the vertical asymptote(s) of the function f(x).

Solution

Step 1: To find the vertical asymptote(s) of f(x), we need to determine where

the denominator of the rational function is equal to zero.

Step 2: Set the denominator 2x−1equal to zero and solve for x.

2x−1 = 0

Step 3: Add 1 to both sides.

2x= 1

Step 4: Divide both sides by 2.

x=1

Step 5: Therefore, the vertical asymptote of f(x)is x=1

Question 14

Question

Find the domain of the rational function:

f(x) = x2−1

x2−4x+ 3

Solution

Step 1: We start by determining the values of xthat will make the denominator

x2−4x+ 3 equal to zero. This will give us the values of xthat need to be

excluded from the domain.

x2−4x+ 3 = 0

(x−3)(x−1) = 0

x= 1,3

Step 2: Now, we need to consider the values of xthat will make the function

undefined, which are the values that make the denominator equal to zero.

x2−4x+ 3 = 0

(x−3)(x−1) = 0

x= 1,3

Step 3: Therefore, the domain of the function f(x)is all real numbers except

x= 1 and x= 3.

Domain = (−∞,1) ∪(1,3) ∪(3,∞)

Question 15

Question

Solve the rational inequality: 2

x−1−3

x+1 ≥1

x. Write your answer in interval

notation.

Solution

Step 1: Find a common denominator for the left side of the inequality. To add

or subtract fractions, we need a common denominator. In this case, we can

use the denominator of x(x−1)(x+ 1). So, rewriting the inequality with the

common denominator, we have:

2x(x+ 1)

x(x−1)(x+ 1) −3x(x−1)

x(x−1)(x+ 1) ≥(x−1)(x+ 1)

x(x−1)(x+ 1)

which simplifies to:

2x2+ 2x−3x2+ 3x

x(x−1)(x+ 1) ≥x2−1

x(x−1)(x+ 1)

Step 2: Combine like terms. Simplifying further, we get:

−x2+ 5x

x(x−1)(x+ 1) ≥x2−1

x(x−1)(x+ 1)

Step 3: Solve the rational inequality. To solve the inequality, we can bring

everything to one side to find a common denominator:

−x2+ 5x−(x2−1)

x(x−1)(x+ 1) ≥0

which simplifies to: 6x−1

x(x−1)(x+ 1) ≥0

Step 4: Determine the critical points. The critical points occur when the

numerator and denominator are equal to zero. So, x= 0,x= 1, and x=−1

are the critical points.

Step 5: Create a sign chart. We can create a sign chart to determine the

intervals where the inequality holds true.

x < −1−1< x < 0 0 < x < 1x > 1

6x−1− − + +

x− − + +

x−1− − − +

x+ 1 −+ + +

Inequality −+−+

Step 6: Write the solution in interval notation. From the sign chart, we

observe that the inequality is true for x∈(−1,0) ∪(1,∞). Therefore, the

solution to the rational inequality 2

x−1−3

x+1 ≥1

xin interval notation is (−1,0)∪

(1,∞).

Question 16

Question

Simplify the rational function:

4x2−8x+ 4

x2−4

Solution

Step 1: Factor the numerator and the denominator.

Numerator: 4x2−8x+ 4

= 4(x2−2x+ 1)

= 4(x−1)(x−1)

= 4(x−1)2

Denominator: x2−4

= (x+ 2)(x−2)

Step 2: Rewrite the rational function with factored terms.

4(x−1)2

(x+ 2)(x−2)

Step 3: Simplify the expression by canceling out common factors between

the numerator and the denominator.

4(x−1)(x−1)

(x+ 2)(x−2)

4(x−1)

x+ 2

Therefore, the simplified form of the rational function is 4(x−1)

x+2 .

Question 17

Question

Find the horizontal asymptotes of the rational function:

f(x) = 3x2−2x+ 1

2x2−5x−3

Solution

To find the horizontal asymptotes of the rational function f(x), we need to

consider the limit of f(x)as xapproaches positive infinity and negative infinity.

Step 1: Find the degree of the numerator and denominator Identify

the leading terms of the numerator and denominator: - The leading term of the

numerator is 3x2. - The leading term of the denominator is 2x2.

Step 2: Find the limit as xapproaches infinity Considering the leading

terms, as xapproaches infinity:

lim

x→∞

3x2−2x+ 1

2x2−5x−3= lim

x→∞

3x2

2x2=3

Step 3: Find the limit as xapproaches negative infinity Considering

the leading terms, as xapproaches negative infinity:

lim

x→−∞

3x2−2x+ 1

2x2−5x−3= lim

x→−∞

3x2

2x2=3

Step 4: Conclusion Since both limits as xapproaches infinity and negative

infinity are 3

2, the horizontal asymptote of the rational function f(x)is y=3

Question 18

Question

Find the domain of the rational function:

f(x) = x2−5x+ 6

x2+ 3x−10

Solution

Step 1: The domain of a rational function is all real numbers except the values

that make the denominator equal to zero, since division by zero is undefined.

We need to find the values of xthat make the denominator x2+ 3x−10 equal

to zero.

Step 2: To find these values, we solve the quadratic equation x2+3x−10 = 0

by factoring or using the quadratic formula. Factoring, we get:

(x+ 5)(x−2) = 0

Step 3: Setting each factor to zero, we find the roots of the quadratic equa-

tion:

x+ 5 = 0 =⇒x=−5

x−2 = 0 =⇒x= 2

Step 4: Therefore, the values that make the denominator zero are x=−5

and x= 2. The domain of the function will be all real numbers except these

values.

Step 5: The domain of f(x)is (−∞,−5) ∪(−5,2) ∪(2,∞)

Question 19

Question

Simplify the rational function:

4x2−9

x2−1∇ · 3x2−3x

2x2−2

Solution

Step 1: First, let’s rewrite the division as multiplication by the reciprocal of the

second fraction. 4x2−9

x2−1·2x2−2

3x2−3x

Step 2: Factorize the numerator and denominator of each fraction.

(2x−3)(2x+ 3)

(x−1)(x+ 1) ·2(x2−1)

3x(x−1)

Step 3: Simplify the expression by canceling out common factors.

2(2x−3)(2x+ 3)(x+ 1)

Step 4: Multiply out the numerator to get the final simplified form.

8x2−9

Therefore, the simplified form of the given rational function is 8x2−9

3x.

Question 20

Question

Let f(x) = 2x2−5x+3

x2−4x−12 . Find the domain of the function f(x).

Solution

Step 1: To find the domain of f(x), we need to determine the values of xfor

which the function f(x)is defined. Since division by zero is undefined, we must

exclude any values of xthat would make the denominator equal to zero.

Step 2: The denominator x2−4x−12 is equal to zero when x2−4x−12 = 0.

We can factor this quadratic equation to find its roots:

x2−4x−12 = 0

(x−6)(x+ 2) = 0

Step 3: Setting each factor to zero gives us the roots:

x−6 = 0 =⇒x= 6

x+ 2 = 0 =⇒x=−2

Step 4: Therefore, the function f(x)is undefined at x= 6 and x=−2, so

these values must be excluded from the domain.

Step 5: The domain of f(x)is all real numbers except x= 6 and x=−2.

In interval notation, the domain is (−∞,−2) ∪(−2,6) ∪(6,∞).

Question 21

Question

Simplify the following rational expression:

(4x2−2x−6

2x2−8x−6)∇ · (2x2+ 4x+ 2

4x2−16 )

Solution

Step 1: First, rewrite the expression as a multiplication instead of division by

multiplying by the reciprocal of the second fraction.

(4x2−2x−6

2x2−8x−6)×(4x2−16

2x2+ 4x+ 2)

Step 2: Factor the numerators and denominators where possible. For the

first fraction:

4x2−2x−6 = 2(2x2−x−3) = 2(2x+ 1)(x−3)

2x2−8x−6 = 2(x2−4x−3) = 2(x−3)(x+ 1)

For the second fraction:

4x2−16 = 4(x2−4) = 4(x−2)(x+ 2)

2x2+ 4x+ 2 = 2(x2+ 2x+ 1) = 2(x+ 1)2

So the expression becomes:

2(2x+ 1)(x−3)

2(x−3)(x+ 1) ×4(x−2)(x+ 2)

2(x+ 1)2

Step 3: Simplify the expression by canceling out common factors.

2(2x+ 1)(x−3)

2(x−3)(x+ 1) ×4(x−2)(x+ 2)

2(x+ 1)2=2(2x+ 1)

x+ 1 ×4(x+ 2)

2(x+ 1)

Step 4: Finally, simplify the expression further.

2(2x+ 1) ·4(x+ 2)

(x+ 1) ·2=8(2x+ 1)(x+ 2)

2(x+ 1) =8(2x2+ 5x+ 2)

2(x+ 1) = 4(2x2+5x+2) = 8x2+ 20x+ 8

Question 22

Question

Simplify the rational function: f(x) = 2x2−x−3

x2−4x−5.

Solution

Step 1: Factor the numerator and denominator of the rational function.

f(x) = 2x2−x−3

x2−4x−5

=(2x+ 3)(x−1)

(x−5)(x+ 1)

Step 2: Simplify the expression by canceling out common factors.

f(x) = (2x+ 3)(x−1)

(x−5)(x+ 1)

=2x+ 3

x−5

Therefore, the simplified form of the rational function f(x)is 2x+3

x−5.

Question 23

Question

Find the domain of the rational function:

f(x) = 5x2−4x−3

x2−9

Solution

Step 1: Identify the values of xthat would make the denominator x2−9equal

to zero. Since division by zero is undefined, these values must be excluded from

the domain.

x2−9 = 0

(x+ 3)(x−3) = 0

x=−3or x= 3

Step 2: Therefore, the domain of the function f(x)is all real numbers except

x=−3and x= 3. We express this as:

Domain: x∈R, x =−3,3

Question 24

Question

Simplify the rational expression: 3x2−5x−2

2x2−7x+3 .

Solution

To simplify the rational expression, we need to factor the numerator and de-

nominator.

Step 1: Factor the Numerator The numerator of the rational expression

is 3x2−5x−2. We need to find two numbers that multiply to 3·(−2) = −6

and add up to −5. Those numbers are −6and 1. So, the factored form of the

numerator is:

3x2−5x−2 = (3x+ 1)(x−2)

Step 2: Factor the Denominator The denominator of the rational ex-

pression is 2x2−7x+ 3. We need to find two numbers that multiply to 2·3 = 6

and add up to −7. Those numbers are −6and −1. So, the factored form of the

denominator is:

2x2−7x+ 3 = (2x−1)(x−3)

Step 3: Simplify the Rational Expression Now, we can rewrite the

rational expression in factored form and simplify:

3x2−5x−2

2x2−7x+ 3 =(3x+ 1)(x−2)

(2x−1)(x−3)

Therefore, the simplified form of the rational expression is (3x+1)(x−2)

(2x−1)(x−3) .

Question 25

Question

Simplify the following rational function:

f(x) = 3x2−x−10

x2−4x−5

Solution

Step 1: Factor the numerator and denominator. Step 2: Simplify the rational

function by canceling out common factors.

Step 1: To factor the numerator 3x2−x−10, we look for two numbers that

multiply to 3× −10 = −30 and add up to −1. The numbers are −6and 5. So

we rewrite the numerator as:

f(x) = (3x−10)(x+ 1)

x2−4x−5

To factor the denominator x2−4x−5, we look for two numbers that multiply

to −5and add up to −4. The numbers are −5and 1. So we rewrite the

denominator as:

f(x) = (3x−10)(x+ 1)

(x−5)(x+ 1)

Step 2: Now we can simplify the rational function by canceling out the

common factor (x+ 1) from the numerator and denominator:

f(x) = 3x−10

x−5

Therefore, the simplified form of the rational function f(x)is 3x−10

x−5.

Question 26

Question

Let f(x) = x2−2x−8

x2−3x−4. Find the vertical and horizontal asymptotes of f(x), if

they exist.

Solution

Step 1: First, let’s determine the vertical asymptotes of the function.

Step 2: To find the vertical asymptotes, we need to find the values of xthat

make the denominator zero, but not the numerator.

Step 3: Set the denominator x2−3x−4equal to zero and solve for x.

x2−3x−4 = 0

Step 4: Factoring the quadratic equation gives us:

(x−4)(x+ 1) = 0

Step 5: Therefore, the vertical asymptotes occur at x= 4 and x=−1.

Step 6: Next, let’s find the horizontal asymptote of the function.

Step 7: To find the horizontal asymptote, we examine the degrees of the

numerator and the denominator.

Step 8: Since the degrees of the numerator and the denominator are the

same, the horizontal asymptote is the ratio of the leading coeﬀicients.

Step 9: The horizontal asymptote is y=1

1= 1.

Step 10: Therefore, the vertical asymptotes are x= 4 and x=−1, and the

horizontal asymptote is y= 1.

Question 27

Question

Find the domain of the rational function:

f(x) = 5x2−11x+ 2

x2−4x−12

Solution

Step 1: Determine the values for which the denominator is equal to zero, since

division by zero is undefined. Set the denominator equal to zero and solve for

x2−4x−12 = 0

(x−6)(x+ 2) = 0

The solutions are x= 6 and x=−2.

Step 2: The domain of the rational function is all real numbers except for

the values of xthat make the denominator zero. Thus, the domain of f(x)is

all real numbers xsuch that x= 6 and x=−2.

Therefore, the domain of f(x)is (−∞,−2) ∪(−2,6) ∪(6,∞).

Question 28

Question

Find the domain of the rational function:

f(x) = x2+ 4x+ 4

x2−4

Solution

Step 1: We start by finding the values of xthat make the denominator equal to

zero, since division by zero is undefined.

x2−4 = 0

Factoring the denominator gives:

(x+ 2)(x−2) = 0

So, x=−2and x= 2 are the values that make the denominator zero.

Step 2: Our domain will be all real numbers except those which make the

denominator zero. Therefore, the domain of the function f(x)is given by:

(−∞,−2) ∪(−2,2) ∪(2,∞)

Question 29

Question

Let f(x) = 3x2−1

x+2 and g(x) = x+2

x−1. Find (f◦g)(x)and simplify the resulting

rational function.

Solution

Step 1: First, we need to find (f◦g)(x), which is equal to f(g(x)).

Step 2: Substitute g(x)into the function f(x):

f(g(x)) = f(x+ 2

x−1)=

3(x+2

x−1)2

−1

x+2

x−1+ 2

Step 3: Simplify the expression inside f:

f(g(x)) =

3(x2+4x+4

(x−1)2)−1

x+2

x−1+ 2

Step 4: Simplify the numerator:

f(g(x)) =

3(x2+4x+4

(x−1)2)−1

x+2

x−1+ 2 =

3x2+12x+12

(x−1)2−1

x+2

x−1+ 2

Step 5: Multiply the terms:

f(g(x)) =

3x2+12x+12−(x−1)2

(x−1)2

x+2

x−1+ 2

Step 6: Expand the numerator:

f(g(x)) =

3x2+12x+12−(x2−2x+1)

(x−1)2

x+2

x−1+ 2

Step 7: Simplify the numerator further:

f(g(x)) =

3x2+12x+12−x2+2x−1

(x−1)2

x+2

x−1+ 2 =

2x2+14x+11

(x−1)2

x+2

x−1+ 2

Step 8: Multiply the fractions and simplify:

f(g(x)) = (2x2+ 14x+ 11)(x−1)

(x−1)2×(x−1) + 2(x−1)

x+ 2 =2x2+ 14x+ 11

x+ 2

Therefore, (f◦g)(x) = 2x2+14x+11

x+2 .

Question 30

Question

Simplify the rational function:

f(x) = 2x2−5x−3

x2−16

Solution

Step 1: Factor the numerator and denominator:

f(x) = 2x2−5x−3

x2−16

=(2x+ 1)(x−3)

(x+ 4)(x−4)

Step 2: Simplify the expression by canceling out common factors:

f(x) = (2x+ 1)(x−3)

(x+ 4)(x−4)

=2x+ 1

x+ 4

Question 31

Question

Let f(x) = x2−5x+ 6

x2−4x. Find the domain of the function f(x).

Solution

Step 1: The domain of a rational function consists of all real numbers for which

the denominator is not equal to zero. In this case, the denominator x2−4x

cannot equal zero. So, solve the equation x2−4x= 0 to find the values that

make the denominator zero.

Step 2: Factor out xfrom x2−4xto get x(x−4) = 0.

Step 3: Set each factor to zero: x= 0 or x−4 = 0

Step 4: Solve for x: For x= 0, we have 0. For x−4 = 0, we have x= 4.

Step 5: Therefore, the domain of the function f(x)is all real numbers except

x= 0 and x= 4.

Step 6: In interval notation, the domain can be written as (−∞,0) ∪(0,4) ∪

(4,∞).

Question 32

Question

Let f(x) = x3−3x2+2x

x2−5x+6 . Determine the vertical asymptotes of the function f(x).

Solution

Step 1: To find the vertical asymptotes of the function f(x), we need to identify

the values of xthat make the denominator equal to zero, but not the numerator.

These values will give us vertical asymptotes. So, we solve x2−5x+ 6 = 0.

Step 2: The equation x2−5x+ 6 = 0 can be factored as (x−2)(x−3) = 0.

Setting each factor to zero, we have x−2 = 0 and x−3 = 0.

Step 3: Solving x−2 = 0, we get x= 2. Solving x−3 = 0, we get x= 3.

Therefore, the vertical asymptotes of the function f(x)are x= 2 and x= 3.

Question 33

Question

Simplify the following rational expression:

4x2−4x−3

x2−9

Solution

Step 1: Factor both the numerator and denominator:

•Numerator: We can factor the numerator as 4x2−4x−3 = (2x−3)(2x+

1).

•Denominator: The denominator can be factored as x2−9 = (x+3)(x−3).

Step 2: Now, rewrite the expression with factored numerator and denomi-

nator: (2x−3)(2x+ 1)

(x+ 3)(x−3)

Step 3: Simplify the expression by cancelling out common factors:

(2x−3)(2x+ 1)

(x+ 3)(x−3) =2x−3

x+ 3

Therefore, the simplified form of 4x2−4x−3

x2−9is 2x−3

x+3 .

Question 34

Question

Simplify the following rational expression:

2x2+ 7x−15

4x2−9

Solution

Step 1: Factor both the numerator and denominator. Step 2: Simplify the

expression by canceling out common factors.

Step 1: Factor the numerator and denominator. The numerator, 2x2+

7x−15, factors to (2x−3)(x+ 5). The denominator, 4x2−9, is a difference of

squares and factors to (2x+ 3)(2x−3).

Step 2: Simplify the expression by canceling out common factors.

2x2+ 7x−15

4x2−9=(2x−3)(x+ 5)

(2x+ 3)(2x−3)

Cancel out the common factor (2x−3) in the numerator and denominator.

=x+ 5

2x+ 3

Therefore, the simplified form of the given rational expression is x+5

2x+3 .

Question 2

Question

Let f(x) = 3x2−x−2

x2−5x+ 6 and g(x) = 2x2−x−1

x2−4x+ 3 be two rational functions.

Determine the domain of the composite function f(g(x)).

Solution

To determine the domain of the composite function f(g(x)), we need to consider

restrictions on both f(x)and g(x).

Step 1: Determine the domain of g(x)Since denominators cannot be

zero, we need to find the values of xthat would make the denominators of g(x)

equal to zero:

x2−4x+ 3 = 0

Factor the quadratic:

(x−3)(x−1) = 0

So, x= 3 and x= 1 are the values that would make the denominator of g(x)

equal to zero. Therefore, the domain of g(x)is all real numbers except x= 3

and x= 1.

Step 2: Determine the domain of f(g(x)) The domain of f(g(x)) is the

set of all xvalues for which g(x)is in the domain of f(x). Since g(x)cannot

be 1or 3, we need to exclude these values from the domain of f(x)as well. We

will find these values by substituting g(x)into f(x):

f(g(x)) = 3(g(x))2−g(x)−2

(g(x))2−5(g(x)) + 6

Substitute g(x):

f(g(x)) = 3(2x2−x−1

x2−4x+3 )2−2x2−x−1

x2−4x+3 −2

(2x2−x−1

x2−4x+3 )2−5(2x2−x−1

x2−4x+3 )+6

Simplify to find the domain. Since the domain of g(x)is not 1or 3, the domain

of f(g(x)) is all real numbers except 1or 3.

Question 3

Question

Simplify the following rational expression:

4x2−12x

x3−5x2+ 6x

Solution

Step 1: Factor out the greatest common factor in both the numerator and

denominator. 4x(x−3)

x(x−3)(x−2)

Step 2: Simplify the expression by canceling out common factors.

x−2

So, the simplified form of the rational expression is 4

x−2.

Question 4

Question

Find the vertical asymptotes, horizontal asymptotes, and any holes in the graph

of the rational function:

f(x) = 2x3+ 3

x2−5x+ 6

Solution

Step 1: To find the vertical asymptotes, set the denominator equal to zero and

solve for x:

x2−5x+ 6 = 0

Factoring the quadratic equation gives:

(x−2)(x−3) = 0

So, x= 2 and x= 3 are vertical asymptotes.

Step 2: To find the horizontal asymptote, we compare the degrees of the

numerator and denominator. Since the degree of the numerator is greater than

the degree of the denominator, there is no horizontal asymptote.

Step 3: To find any holes in the graph, we need to simplify the function.

We can simplify the function by factoring out (x-3) from the numerator and

denominator:

f(x) = 2x3+ 3

(x−3)(x−2) =2(x2+ 3x+ 1)

(x−3)(x−2)

We can see that there is no common factor between the numerator and

denominator. Therefore, there are no holes in the graph.

In conclusion, the rational function has vertical asymptotes at x= 2 and

x= 3, no horizontal asymptote, and no holes in the graph.

Question 5

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 identify values of

xthat would make the denominator 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: Factor the quadratic equation:

(x−5)(x+ 1) = 0

Step 4: Solve for x:

x−5 = 0 or x+ 1 = 0

x= 5 or x=−1

Step 5: The domain of the function is all real numbers except where the

denominator is zero. Therefore, the domain of the function is:

(−∞,−1) ∪(−1,5) ∪(5,+∞)

Question 6

Question

Let f(x) = 2x2−x−3

x2−2x−3. Find the domain of the function f(x).

Solution

Step 1: To find the domain of f(x), we need to determine which values of xwill

make the denominator equal to zero, since division by zero is undefined.

Step 2: Set the denominator x2−2x−3equal to zero and solve for x:

x2−2x−3 = 0

Step 3: Factoring the quadratic equation:

(x−3)(x+ 1) = 0

Step 4: Setting each factor to zero:

x−3 = 0 or x+ 1 = 0

Step 5: Solving for xin each case:

x= 3 or x=−1

Step 6: The values x= 3 and x=−1make the denominator of f(x)equal

to zero. Therefore, the domain of f(x)is all real numbers except x= 3 and

x=−1.

Step 7: Thus, the domain of the function f(x)is (−∞,−1) ∪(−1,3) ∪(3,∞).

Question 7

Question

Simplify the following rational function:

f(x) = 2x3−5x2+ 2x

x3−2x2−x+ 2

Solution

Step 1: Factor both the numerator and denominator.

The numerator is already factored, so we have:

f(x) = 2x(x2−5x+ 2)

x3−2x2−x+ 2

Now, let’s factor the denominator by grouping:

x3−2x2−x+ 2 = x2(x−2) −1(x−2)

= (x2−1)(x−2)

= (x+ 1)(x−1)(x−2)

So now we have:

f(x) = 2x(x2−5x+ 2)

(x+ 1)(x−1)(x−2)

Step 2: Simplify the rational function by canceling out common factors.

f(x) = 2x(x2−5x+ 2)

(x+ 1)(x−1)(x−2)

f(x) = 2x(x−1)(x−2)

(x+ 1)(x−1)(x−2)

f(x) = 2x

(x+ 1)

Therefore, the simplified form of the rational function is:

f(x) = 2x

x+ 1

Question 8

Question

Simplify the following rational expression:

3x2−7x−6

x2+ 2x−8

Solution

Step 1: Factor the numerator and denominator:

3x2−7x−6 = (3x+ 2)(x−3)

x2+ 2x−8 = (x+ 4)(x−2)

Step 2: Rewrite the expression with the factored terms:

(3x+ 2)(x−3)

(x+ 4)(x−2)

Step 3: Simplify the expression by canceling out common factors:

(3x+ 2)(x−3)

(x+ 4)(x−2) =3x+ 2

x+ 4

Therefore, the simplified form of the given rational expression is 3x+2

x+4 .

Question 9

Question

Find the domain of the rational function:

f(x) = x2+ 3x−4

x2−9

Solution

Step 1: We need to find the values of xfor which the denominator is not equal

to zero, since division by zero is undefined. Thus, we set the denominator x2−9

not equal to zero and solve for x:

x2−9= 0

(x+ 3)(x−3) = 0

This gives us two critical points, x=−3and x= 3.

Step 2: Next, we need to consider any restrictions on the function due to

the numerator. Since the numerator is a polynomial, it is defined for all real

numbers.

Step 3: Therefore, the domain of the rational function is all real numbers

except x=−3, x = 3.

Step 4: In interval notation, the domain of the function fis (−∞,−3) ∪

(−3,3) ∪(3,∞).

Question 10

Question

Find the vertical asymptotes, horizontal asymptotes, and any holes (points of

discontinuity) of the rational function:

f(x) = 3x2−2x−8

x2−4

Solution

Step 1: To find the vertical asymptotes, we need to determine where the de-

nominator of the rational function equals zero (excluding any points where the

numerator is also zero). Set the denominator x2−4 = 0 and solve for x.

x2−4 = 0

(x−2)(x+ 2) = 0

This gives x= 2 and x=−2. Therefore, the vertical asymptotes are at x= 2

and x=−2.

Step 2: To find the horizontal asymptote, we compare the degrees of the

numerator and denominator. Since the degree of the numerator is equal to the

degree of the denominator, the horizontal asymptote is the ratio of the leading

coeﬀicients.

lim

x→∞

3x2−2x−8

x2−4

The horizontal asymptote is y= 3.