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Instructor’s Solutions Manual, Section 2.5 Exercise 1

Solutions to Exercises, Section 2.5

For Exercises 1–4, write the domain of the given function r as a union ofintervals.

1. r(x) = 5x3 − 12x2 + 13x2 − 7

solution Because we have no other information about the domain ofr , we assume that the domain of r is the set of numbers where theexpression defining r makes sense, which means where thedenominator is not 0. The denominator of the expression defining r is0 if x = −√7 or x = √7. Thus the domain of r is the set of numbersother than −√7 and

√7. In other words, the domain of r is

(−∞,−√7)∪ (−√7,√

7)∪ (√7,∞).


2. r(x) = x5 + 3x4 − 62x2 − 5

solution Because we have no other information about the domain ofr , we assume that the domain of r is the set of numbers where theexpression defining r makes sense, which means where thedenominator is not 0. The denominator of the expression defining r is

0 if x = −√

52 or x =

√52 . Thus the domain of r is the set of numbers

other than −√

52 and

√52 . In other words, the domain of r is(−∞,−√5

2

)∪ (−√52 ,√

52

)∪ (√52 ,∞

).


3. r(x) = 4x7 + 8x2 − 1x2 − 2x − 6

solution To find where the expression defining r does not makesense, apply the quadratic formula to the equation x2 − 2x − 6 = 0,getting x = 1−√7 or x = 1+√7. Thus the domain of r is the set ofnumbers other than 1−√7 and 1+√7. In other words, the domain of ris (−∞,1−√7)∪ (1−√7,1+√7)∪ (1+√7,∞).


4. r(x) = 6x9 + x5 + 8x2 + 4x + 1

solution To find where the expression defining r does not makesense, apply the quadratic formula to the equation x2 + 4x + 1 = 0,getting x = −2−√3 or x = −2+√3. Thus the domain of r is the set ofnumbers other than −2−√3 and −2+√3. In other words, the domainof r is (−∞,−2−√3)∪ (−2−√3,−2+√3)∪ (−2+√3,∞).


Suppose

r(x) = 3x + 4

x2 + 1,

s(x) = x2 + 2

2x − 1,

t(x) = 5

4x3 + 3.

In Exercises 5–22, write the indicated expression as a ratio, with thenumerator and denominator each written as a sum of terms of theform cxm.

5. (r + s)(x)solution

(r + s)(x) = 3x + 4x2 + 1

+ x2 + 22x − 1

= (3x + 4)(2x − 1)(x2 + 1)(2x − 1)

+ (x2 + 2)(x2 + 1)(x2 + 1)(2x − 1)

= (3x + 4)(2x − 1)+ (x2 + 2)(x2 + 1)(x2 + 1)(2x − 1)

= 6x2 − 3x + 8x − 4+ x4 + x2 + 2x2 + 22x3 − x2 + 2x − 1


= x4 + 9x2 + 5x − 22x3 − x2 + 2x − 1


6. (r − s)(x)solution

(r − s)(x) = 3x + 4x2 + 1

− x2 + 22x − 1

= (3x + 4)(2x − 1)(x2 + 1)(2x − 1)

− (x2 + 2)(x2 + 1)(x2 + 1)(2x − 1)

= (3x + 4)(2x − 1)− (x2 + 2)(x2 + 1)(x2 + 1)(2x − 1)

= 6x2 − 3x + 8x − 4− x4 − x2 − 2x2 − 22x3 − x2 + 2x − 1

= −x4 + 3x2 + 5x − 62x3 − x2 + 2x − 1


7. (s − t)(x)solution

(s − t)(x) = x2 + 22x − 1

− 54x3 + 3

= (x2 + 2)(4x3 + 3)(2x − 1)(4x3 + 3)

− 5(2x − 1)(2x − 1)(4x3 + 3)

= (x2 + 2)(4x3 + 3)− 5(2x − 1)(2x − 1)(4x3 + 3)

= 4x5 + 8x3 + 3x2 − 10x + 118x4 − 4x3 + 6x − 3


8. (s + t)(x)solution

(s + t)(x) = x2 + 22x − 1

+ 54x3 + 3

= (x2 + 2)(4x3 + 3)(2x − 1)(4x3 + 3)

+ 5(2x − 1)(2x − 1)(4x3 + 3)

= (x2 + 2)(4x3 + 3)+ 5(2x − 1)(2x − 1)(4x3 + 3)

= 4x5 + 8x3 + 3x2 + 10x + 18x4 − 4x3 + 6x − 3


9. (3r − 2s)(x)

solution

(3r − 2s)(x) = 3(3x + 4x2 + 1

)− 2

(x2 + 22x − 1

)

= 9x + 12x2 + 1

− 2x2 + 42x − 1

= (9x + 12)(2x − 1)(x2 + 1)(2x − 1)

− (2x2 + 4)(x2 + 1)

(x2 + 1)(2x − 1)

= (9x + 12)(2x − 1)− (2x2 + 4)(x2 + 1)(x2 + 1)(2x − 1)

= 18x2 − 9x + 24x − 12− 2x4 − 6x2 − 42x3 − x2 + 2x − 1

= −2x4 + 12x2 + 15x − 162x3 − x2 + 2x − 1


10. (4r + 5s)(x)

solution

(4r + 5s)(x) = 4(3x + 4x2 + 1

)+ 5

(x2 + 22x − 1

)

= 12x + 16x2 + 1

+ 5x2 + 102x − 1

= (12x + 16)(2x − 1)(x2 + 1)(2x − 1)

+ (5x2 + 10)(x2 + 1)

(x2 + 1)(2x − 1)

= (12x + 16)(2x − 1)+ (5x2 + 10)(x2 + 1)(x2 + 1)(2x − 1)

= 24x2 − 12x + 32x − 16+ 5x4 + 15x2 + 102x3 − x2 + 2x − 1

= 5x4 + 39x2 + 20x − 62x3 − x2 + 2x − 1


11. (rs)(x)

solution

(rs)(x) = 3x + 4x2 + 1

· x2 + 2

2x − 1

= (3x + 4)(x2 + 2)(x2 + 1)(2x − 1)

= 3x3 + 4x2 + 6x + 82x3 − x2 + 2x − 1


12. (rt)(x)

solution

(rt)(x) = 3x + 4x2 + 1

· 54x3 + 3

= 15x + 20(x2 + 1)(4x3 + 3)

= 15x + 204x5 + 4x3 + 3x2 + 3


13.(r(x)

)2

solution

(r(x)

)2 =(3x + 4x2 + 1

)2

= (3x + 4)2

(x2 + 1)2

= 9x2 + 24x + 16x4 + 2x2 + 1


14.(s(x)

)2

solution

(s(x)

)2 =(x2 + 2

2x − 1

)2

= (x2 + 2)2

(2x − 1)2

= x4 + 4x2 + 44x2 − 4x + 1


15.(r(x)

)2t(x)

solution Using the expression that we computed for(r(x)

)2in the

solution to Exercise 13, we have

(r(x)

)2t(x) = 9x2 + 24x + 16x4 + 2x2 + 1

· 54x3 + 3

= 5(9x2 + 24x + 16)(x4 + 2x2 + 1)(4x3 + 3)

= 45x2 + 120x + 804x7 + 8x5 + 3x4 + 4x3 + 6x2 + 3

.


16.(s(x)

)2t(x)

solution Using the expression that we computed for(sr(x)

)2in the

solution to Exercise 14, we have

(s(x)

)2t(x) = x4 + 4x2 + 44x2 − 4x + 1

· 54x3 + 3

= 5(x4 + 4x2 + 4)(4x2 − 4x + 1)(4x3 + 3)

= 5x4 + 20x2 + 2016x5 − 16x4 + 4x3 + 12x2 − 12x + 3

.


17. (r ◦ s)(x)solution We have

(r ◦ s)(x) = r(s(x))= r

(x2 + 22x − 1

)

= 3(x2+2

2x−1

)+ 4(x2+22x−1

)2 + 1

= 3 (x2+2)

(2x−1) + 4(x2+2)2(2x−1)2 + 1

.

Multiplying the numerator and denominator of the expression above by(2x − 1)2 gives

(r ◦ s)(x) = 3(x2 + 2)(2x − 1)+ 4(2x − 1)2

(x2 + 2)2 + (2x − 1)2

= 6x3 + 13x2 − 4x − 2x4 + 8x2 − 4x + 5

.


18. (s ◦ r)(x)solution We have

(s ◦ r)(x) = s(r(x))= s

(3x + 4x2 + 1

)

=(3x+4x2+1

)2 + 2

2(3x+4x2+1

)− 1

=(3x+4)2(x2+1)2 + 2

6x+8x2+1 − 1

.

Multiplying the numerator and denominator of the expression above by(x2 + 1)2 gives

(s ◦ r)(x) = (3x + 4)2 + 2(x2 + 1)2

(6x + 8)(x2 + 1)− (x2 + 1)2

= 2x4 + 13x2 + 24x + 18−x4 + 6x3 + 6x2 + 6x + 7

.


19. (r ◦ t)(x)solution We have

(r ◦ t)(x) = r(t(x))= r

( 54x3 + 3

)

= 3( 5

4x3+3

)+ 4( 54x3+3

)2 + 1

=15

4x3+3 + 425

(4x3+3)2 + 1.

Multiplying the numerator and denominator of the expression above by(4x3 + 3)2 gives

(r ◦ t)(x) = 15(4x3 + 3)+ 4(4x3 + 3)2

25+ (4x3 + 3)2

= 64x6 + 156x3 + 8116x6 + 24x3 + 34

.


20. (t ◦ r)(x)solution We have

(t ◦ r)(x) = t(r(x))= t

(3x + 4x2 + 1

)

= 5

4(3x+4x2+1

)3 + 3

= 54(3x+4)3(x2+1)3 + 3

.

Multiplying the numerator and denominator of the expression above by(x2 + 1)3 gives

(t ◦ r)(x) = 5(x2 + 1)3

4(3x + 4)3 + 3(x2 + 1)3

= 5x6 + 15x4 + 15x2 + 53x6 + 9x4 + 108x3 + 441x2 + 576x + 259

.


21. s(1+x)−s(1)x

solution Note that s(1) = 3. Thus

s(1+ x)− s(1)x

=(1+x)2+22(1+x)−1 − 3

x

=x2+2x+3

2x+1 − 3

x.

Multiplying the numerator and denominator of the expression above by2x + 1 gives

s(1+ x)− s(1)x

= x2 + 2x + 3− 6x − 3x(2x + 1)

= x2 − 4xx(2x + 1)

= x − 42x + 1

.


22. t(x−1)−t(−1)x

solution Note that t(−1) = −5. Thus

t(x − 1)− t(−1)x

=5

4(x−1)3+3 + 5

x.

Multiplying the numerator and denominator of the expression above by4(x − 1)3 + 3 gives

t(x − 1)− t(−1)x

= 5+ 5(4(x − 1)3 + 3

)x(4(x − 1)3 + 3

)= 20x2 − 60x + 60

4x3 − 12x2 + 12x − 1.


For Exercises 23–28, suppose

r(x) = x + 1

x2 + 3and s(x) = x + 2

x2 + 5.

23. What is the domain of r?

solution The denominator of the expression defining r is a nonzeronumber for every real number x, and thus the expression defining rmakes sense for every real number x. Because we have no otherindication of the domain of r , we thus assume that the domain of r isthe set of real numbers.


24. What is the domain of s?

solution The denominator of the expression defining s is a nonzeronumber for every real number x, and thus the expression defining smakes sense for every real number x. Because we have no otherindication of the domain of s, we thus assume that the domain of s isthe set of real numbers.


25. Find two distinct numbers x such that r(x) = 14 .

solution We need to solve the equation

x + 1x2 + 3

= 14

for x. Multiplying both sides by x2 + 3 and then multiplying both sidesby 4 and collecting all the terms on one side, we have

x2 − 4x − 1 = 0.

Using the quadratic formula, we get the solutions x = 2−√5 andx = 2+√5.


26. Find two distinct numbers x such that s(x) = 18 .

solution We need to solve the equation

x + 2x2 + 5

= 18

for x. Multiplying both sides by x2 + 5 and then multiplying both sidesby 8 and collecting all the terms on one side, we have

x2 − 8x − 11 = 0.

Using the quadratic formula, we get the solutions x = 4− 3√

3 andx = 4+ 3

√3.


27. What is the range of r?

solution To find the range of r , we must find all numbers y such that

x + 1x2 + 3

= y

for at least one number x. Thus we will solve the equation above for xand then determine for which numbers y we get an expression for xthat makes sense. Multiplying both sides of the equation above byx2 + 3 and then collecting terms gives

yx2 − x + (3y − 1) = 0.

If y = 0, then this equation has the solution x = −1. If y �= 0, then usethe quadratic formula to solve the equation above for x, getting

x =1+

√1+ 4y − 12y2

2y

or

x =1−

√1+ 4y − 12y2

2y.

These expressions for x make sense precisely when 1+ 4y − 12y2 ≥ 0.Completing the square, we can rewrite this inequality as

−12((y − 1

6)2 − 1

9

) ≥ 0.


Thus we must have (y − 16)

2 ≤ 19 , which is equivalent to −1

3 ≤ y − 16 ≤ 1

3 .Adding 1

6 to each side of these inequalities gives −16 ≤ y ≤ 1

2 .

Thus the range of r is the interval [−16 ,

12].


28. What is the range of s?

solution To find the range of s, we must find all numbers y such that

x + 2x2 + 5

= y

for at least one number x. Thus we will solve the equation above for xand then determine for which numbers y we get an expression for xthat makes sense. Multiplying both sides of the equation above byx2 + 5 and then collecting terms gives

yx2 − x + (5y − 2) = 0.

If y = 0, then this equation has the solution x = −2. If y �= 0, then usethe quadratic formula to solve the equation above for x, getting

x =1+

√1+ 8y − 20y2

2y

or

x =1−

√1+ 8y − 20y2

2y.

These expressions for x make sense precisely when 1+ 8y − 20y2 ≥ 0.Completing the square, we can rewrite this inequality as

−20((y − 1

5)2 − 9

100

) ≥ 0.


Thus we must have (y − 15)

2 ≤ 9100 , which is equivalent to the

inequalities − 310 ≤ y − 1

5 ≤ 310 . Adding 1

5 to each side of theseinequalities gives − 1

10 ≤ y ≤ 12 .

Thus the range of s is the interval [− 110 ,

12].


In Exercises 29–34, write each expression in the form G(x) + R(x)q(x) , where

q is the denominator of the given expression and G and R arepolynomials with deg R < deg q.

29.2x + 1x − 3

solution 2x + 1x − 3

= 2(x − 3)+ 6+ 1x − 3

= 2+ 7x − 3


30.4x − 5x + 7

solution

4x − 5x + 7

= 4(x + 7)− 28− 5x + 7

= 4− 33x + 7


31.x2

3x − 1

solution x2

3x − 1=

x3 (3x − 1)+ x

3

3x − 1

= x3+

x3

3x − 1

= x3+

19(3x − 1)+ 1

9

3x − 1

= x3+ 1

9+ 1

9(3x − 1)


32.x2

4x + 3

solution

x2

4x + 3=

x4 (4x + 3)− 3x

4

4x + 3

= x4−

3x4

4x + 3

= x4−

316(4x + 3)− 9

16

4x + 3

= x4− 3

16+ 9

16(4x + 3)


33.x6 + 3x3 + 1x2 + 2x + 5

solution

x6 + 3x3 + 1x2 + 2x + 5

= x4(x2 + 2x + 5)− 2x5 − 5x4 + 3x3 + 1x2 + 2x + 5

= x4 + −2x5 − 5x4 + 3x3 + 1x2 + 2x + 5

= x4 + (−2x3)(x2 + 2x + 5)x2 + 2x + 5

+ 4x4 + 10x3 − 5x4 + 3x3 + 1x2 + 2x + 5

= x4 − 2x3 + −x4 + 13x3 + 1x2 + 2x + 5

= x4 − 2x3 + (−x2)(x2 + 2x + 5)x2 + 2x + 5

+ 2x3 + 5x2 + 13x3 + 1x2 + 2x + 5

= x4 − 2x3 − x2 + 15x3 + 5x2 + 1x2 + 2x + 5


= x4 − 2x3 − x2

+ 15x(x2 + 2x + 5)− 30x2 − 75x + 5x2 + 1x2 + 2x + 5

= x4 − 2x3 − x2 + 15x + −25x2 − 75x + 1x2 + 2x + 5

= x4 − 2x3 − x2 + 15x

+ −25(x2 + 2x + 5)+ 50x + 125− 75x + 1x2 + 2x + 5

= x4 − 2x3 − x2 + 15x − 25+ −25x + 126x2 + 2x + 5


34.x6 − 4x2 + 5x2 − 3x + 1

solution

x6 − 4x2 + 5x2 − 3x + 1

= x4(x2 − 3x + 1)+ 3x5 − x4 − 4x2 + 5x2 − 3x + 1

= x4 + 3x5 − x4 − 4x2 + 5x2 − 3x + 1

= x4 + (3x3)(x2 − 3x + 1)x2 − 3x + 1

+ 9x4 − 3x3 − x4 − 4x2 + 5x2 − 3x + 1

= x4 + 3x3 + 8x4 − 3x3 − 4x2 + 5x2 − 3x + 1

= x4 + 3x3 + (8x2)(x2 − 3x + 1)x2 − 3x + 1

+ 24x3 − 8x2 − 3x3 − 4x2 + 5x2 − 3x + 1

= x4 + 3x3 + 8x2 + 21x3 − 12x2 + 5x2 − 3x + 1


= x4 + 3x3 + 8x2

+ 21x(x2 − 3x + 1)+ 63x2 − 21x − 12x2 + 5x2 − 3x + 1

= x4 + 3x3 + 8x2 + 21x + 51x2 − 21x + 5x2 − 3x + 1

= x4 + 3x3 + 8x2 + 21x

+ 51(x2 − 3x + 1)+ 153x − 51− 21x + 5x2 − 3x + 1

= x4 + 3x3 + 8x2 + 21x + 51+ 132x − 46x2 − 3x + 1


35. Find a constant c such that r(10100) ≈ 6, where

r(x) = cx3 + 20x2 − 15x + 175x3 + 4x2 + 18x + 7

.

solution Because 10100 is a very large number, we need to estimatethe value of r(x) for very large values of x. The highest-degree term inthe numerator of r is cx3 (unless we choose c = 0); the highest-degreeterm in the denominator of r is 5x3. Factoring out these terms andconsidering only very large values of x, we have

r(x) = cx3(1+ 20

cx − 15cx2 + 17

cx3

)5x3

(1+ 4

5x + 185x2 + 7

5x3

)= c

5·(1+ 20

cx − 15cx2 + 17

cx3

)(1+ 4

5x + 185x2 + 7

5x3

)≈ c

5.

For x very large,(1+ 20

cx − 15cx2 + 17

cx3

)and

(1+ 4

5x + 185x2 + 7

5x3

)are both

very close to 1, which explains how we got the approximation above.

The approximation above shows that r(10100) ≈ c5 . Hence we want to

choose c so that c5 = 6. Thus we take c = 30.


36. Find a constant c such that r(21000) ≈ 5, where

r(x) = 3x4 − 2x3 + 8x + 7cx4 − 9x + 2

.

solution Because 21000 is a very large number, we need to estimatethe value of r(x) for very large values of x. The highest-degree term inthe numerator of r is 3x4; the highest-degree term in the denominatorof r is cx4 (unless we choose c = 0). Factoring out these terms andconsidering only very large values of x, we have

r(x) = 3x4(1− 2

3x + 83x3 + 7

3x4

)cx4

(1− 9

cx3 + 2cx4

)= 3c·(1− 2

3x + 83x3 + 7

3x4

)(1− 9

cx3 + 2cx4

)≈ 3c.

For x very large,(1− 2

3x + 83x3 + 7

3x4

)and

(1− 9

cx3 + 2cx4

)are both very

close to 1, which explains how we got the approximation above.

The approximation above shows that r(21000) ≈ 3c . Hence we want to

choose c so that 3c = 5. Thus we take c = 3

5 .


For Exercises 37–40, find the asymptotes of the graph of the givenfunction r.

37. r(x) = 6x4 + 4x3 − 72x4 + 3x2 + 5

solution The denominator of this rational function is never 0, so weonly need to worry about the behavior of r near ±∞. For |x| very large,we have

r(x) = 6x4 + 4x3 − 72x4 + 3x2 + 5

= 6x4(1+ 2

3x − 76x4

)2x4

(1+ 3

2x2 + 52x4

)≈ 3.

Thus the line y = 3 is an asymptote of the graph of r , as shown below:


�15 �5 5 15x

3

y

The graph of6x4 + 4x3 − 72x4 + 3x2 + 5

on the interval [−15,15].


38. r(x) = 6x6 − 7x3 + 33x6 + 5x4 + x2 + 1

solution The denominator of this rational function is never 0, so weonly need to worry about the behavior of r near ±∞. For |x| very large,we have

r(x) = 6x6 − 7x3 + 33x6 + 5x4 + x2 + 1

= 6x6(1− 7

6x3 + 12x6

)3x6

(1+ 5

3x2 + 13x4 + 1

3x6

)≈ 2.

Thus the line y = 2 is an asymptote of the graph of r , as shown below:

�10 �5 5 10x

2

y

The graph of6x6 − 7x3 + 3

3x6 + 5x4 + x2 + 1on the interval [−10,10].


39. r(x) = 3x + 1x2 + x − 2

solution The denominator of this rational function is 0 when

x2 + x − 2 = 0.

Solving this equation either by factoring or using the quadraticformula, we get x = −2 or x = 1. Because the degree of the numeratoris less than the degree of the denominator, the value of this function isclose to 0 when |x| is large. Thus the asymptotes of the graph of r arethe lines x = −2, x = 1, and y = 0, as shown below:

�6 �2 1 6x

�20

20

y

The graph of3x + 1

x2 + x − 2on the interval

[−6,6], truncated on the vertical axisto the interval [−20,20].


40. r(x) = 9x + 5x2 − x − 6

solution The denominator of this rational function is 0 when

x2 − x − 6 = 0.

Solving this equation either by factoring or using the quadraticformula, we get x = −2 or x = 3. Because the degree of the numeratoris less than the degree of the denominator, the value of this function isclose to 0 when |x| is large. Thus the asymptotes of the graph of r arethe lines x = −2, x = 3, and y = 0, as shown below:

�8 �2 3 8x

�20

20

y

The graph of9x + 5

x2 − x − 6on the interval

[−6,6], truncated on the vertical axisto the interval [−20,20].

Instructor’s Solutions Manual, Section 2.5 Problem 41

Solutions to Problems, Section 2.5

41. Suppose s(x) = x2 + 22x − 1

.

(a) Show that the point (1,3) is on the graph of s.

(b) Show that the slope of a line containing (1,3) and a point on thegraph of s very close to (1,3) is approximately −4.

[Hint: Use the result of Exercise 21.]

solution

(a) Note that

s(1) = 12 + 22 · 1− 1

= 3.

Thus the point (1,3) is on the graph of s.

(b) Suppose x is a very small nonzero number. Thus((1+ x, s(1+ x)) is a

point on the graph of s that is very close to (1,3). The slope of the linecontaining (1,3) and

((1+ x, s(1+ x)) is

s(1+ x)− 3(1+ x)− 1

= s(1+ x)− s(1)x

= x − 42x + 1

,

where the last equality comes from Exercise 21. Because x is verysmall, x−4

2x+1 is close to −4, and thus the last equation shows that theslope of this line is approximately −4.


42. Suppose t(x) = 54x3 + 3

.

(a) Show that the point (−1,−5) is on the graph of t.

(b) Give an estimate for the slope of a line containing (−1,−5) and apoint on the graph of t very close to (−1,−5).

[Hint: Use the result of Exercise 22.]

solution

(a) Note that

t(−1) = 54(−1)3 + 3

= −5.

Thus the point (−1,−5) is on the graph of t.

(b) Suppose x is a very small nonzero number. Thus((x − 1, t(x − 1)

)is a

point on the graph of t that is very close to (−1,−5). The slope of theline containing (−1,−5) and

((x − 1, t(x − 1)

)is

t(x − 1)+ 5(x − 1)+ 1

= t(x − 1)− t(−1)x

= 20x2 − 60x + 604x3 − 12x2 + 12x − 1

,

where the last equality comes from Exercise 22. Because x is verysmall, 20x2−60x+60

4x3−12x2+12x−1 is close to −60, and thus the last equation showsthat the slope of this line is approximately −60.


43. Explain how the result in the previous section for the maximumnumber of peaks and valleys in the graph of a polynomial is a specialcase of the result in this section for the maximum number of peaks andvalleys in the graph of a rational function.

solution Suppose p is a polynomial. We can think of p as the rationalfunction p

q , where q is the degree 0 polynomial defined by q(x) = 1. Bythe result in this section concerning the maximum number of peaksand valley in the graph of a rational function, pq can have at most

degp + degq − 1

peaks and valleys. Because pq = p and degq = 0, we can restate this

result as saying that p can have at most

degp − 1

peaks and valleys, which is the result stated in the previous section forpolynomials.


44. Explain why the composition of a polynomial and a rational function (ineither order) is a rational function.

solution Suppose p is a polynomial and r is a rational function.Thus there exist polynomials s and t such that r = s

t .

First consider the composition r ◦ p. We have

(r ◦ p)(x) = r(p(x)) = s(p(x)

)t(p(x)

) = (s ◦ p)(x)(t ◦ p)(x) .

Problem 55 in Section 2.4 tells us that both s ◦ p and t ◦ p arepolynomials. Thus r ◦ p, which by the equation above equals s◦p

t◦p , is theratio of two polynomials and hence is a rational function.

Now we consider the composition p ◦ r in the other order. There existnumbers a0, a1, . . . , an such that

p(x) = a0 + a1x + · · · + anxn.

Thus

(p ◦ r)(x) = p(r(x))= a0 + a1r(x)+ · · · + anr(x)n

= a0 + a1s(x)t(x)

+ · · · + an s(x)n

t(x)n,

which implies that


p ◦ r = a0 + a1st+ · · · + an s

n

tn.

If k is a positive integer, then sk and tk are polynomials and hence sktk is

a ratio of polynomials and thus is a rational function. Thus each term

ak sk

tk above is a rational function. Because the sum of rational functionsis a rational function, the equation above implies that p ◦ r is a rationalfunction.


45. Explain why the composition of two rational functions is a rationalfunction.

solution Suppose r and q are rational functions. Thus there existpolynomials s and t such that r = s

t .

We have

(r ◦ q)(x) = r(q(x)) = s(q(x)

)t(q(x)

) = (s ◦ q)(x)(t ◦ q)(x) .

The previous problem tells us that both s ◦ p and t ◦ p are rationalfunctions. Thus r ◦ p, which by the equation above equals s◦p

t◦p , is theratio of two rational functions and hence is a rational function.


46. Suppose p is a polynomial and r is a number. Explain why there is apolynomial G such that

p(x)− p(r)x − r = G(x)

for every number x �= r .

solution Define a polynomial P by

P(x) = p(x)− p(r).

Note that P(r) = p(r)− p(r) = 0. Thus by our result on factorizationdue to a zero, there is a polynomial G such that

P(x) = (x − r)G(x)

for every real number x. Replace P(x) by p(x)− p(r) in the equationabove and then divide both sides by x − r to conclude that

p(x)− p(r)x − r = G(x)

for every number x �= r .

solutions to exercises, section 2 - math.wisc.edupark/fall2013/precalculus/2.5sol.pdf ·...

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