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 of intervals. 1. r(x) 5x3 12x 2 + 13 x 2 7 Because we have no other information about the domain of r, we assume that the domain of r is the set of numbers where the expression defining r makes sense, which means where the denominator is not 0. The denominator of the expression defining r is 0ifx 7orx 7. Thus the domain of r is the set of numbers other than 7 and 7. In other words, the domain of r is (, 7) ( 7, 7) ( 7, ).
Instructor s Solutions Manual, Section 2.5 Exercise 2 2. r(x) x5 + 3x 4 6 2x 2 5 Because we have no other information about the domain of r, we assume that the domain of r is the set of numbers where the expression defining r makes sense, which means where the denominator is not 0. The denominator of the expression defining r is 5 0ifx 2 or x 5 2. Thus the domain of r is the set of numbers 5 other than 2 and 5 2. In other words, the domain of r is ( 5 ) ( 5, 2 2, 5 ) ( 5 2 2, ).
Instructor s Solutions Manual, Section 2.5 Exercise 3 3. r(x) 4x7 + 8x 2 1 x 2 2x 6 To find where the expression defining r does not make sense, apply the quadratic formula to the equation x 2 2x 6 0, getting x 1 7orx 1 + 7. Thus the domain of r is the set of numbers other than 1 7 and 1 + 7. In other words, the domain of r is (, 1 7) (1 7, 1 + 7) (1 + 7, ).
Instructor s Solutions Manual, Section 2.5 Exercise 4 4. r(x) 6x9 + x 5 + 8 x 2 + 4x + 1 To find where the expression defining r does not make sense, apply the quadratic formula to the equation x 2 + 4x + 1 0, getting x 2 3orx 2 + 3. Thus the domain of r is the set of numbers other than 2 3 and 2 + 3. In other words, the domain of r is (, 2 3) ( 2 3, 2 + 3) ( 2 + 3, ).
Instructor s Solutions Manual, Section 2.5 Exercise 5 Suppose r(x) 3x + 4 x 2 + 1, s(x) x2 + 2 2x 1, t(x) 5 4x 3 + 3. In Exercises 5 22, write the indicated expression as a ratio, with the numerator and denominator each written as a sum of terms of the form cx m. 5. (r + s)(x) (r + s)(x) 3x + 4 x 2 + 1 + x2 + 2 2x 1 (3x + 4)(2x 1) (x 2 + 1)(2x 1) + (x2 + 2)(x 2 + 1) (x 2 + 1)(2x 1) (3x + 4)(2x 1) + (x2 + 2)(x 2 + 1) (x 2 + 1)(2x 1) 6x2 3x + 8x 4 + x 4 + x 2 + 2x 2 + 2 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 5 x4 + 9x 2 + 5x 2 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 6 6. (r s)(x) (r s)(x) 3x + 4 x 2 + 1 x2 + 2 2x 1 (3x + 4)(2x 1) (x 2 + 1)(2x 1) (x2 + 2)(x 2 + 1) (x 2 + 1)(2x 1) (3x + 4)(2x 1) (x2 + 2)(x 2 + 1) (x 2 + 1)(2x 1) 6x2 3x + 8x 4 x 4 x 2 2x 2 2 2x 3 x 2 + 2x 1 x4 + 3x 2 + 5x 6 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 7 7. (s t)(x) (s t)(x) x2 + 2 2x 1 5 4x 3 + 3 (x2 + 2)(4x 3 + 3) (2x 1)(4x 3 + 3) 5(2x 1) (2x 1)(4x 3 + 3) (x2 + 2)(4x 3 + 3) 5(2x 1) (2x 1)(4x 3 + 3) 4x5 + 8x 3 + 3x 2 10x + 11 8x 4 4x 3 + 6x 3
Instructor s Solutions Manual, Section 2.5 Exercise 8 8. (s + t)(x) (s + t)(x) x2 + 2 2x 1 + 5 4x 3 + 3 (x2 + 2)(4x 3 + 3) (2x 1)(4x 3 + 3) + 5(2x 1) (2x 1)(4x 3 + 3) (x2 + 2)(4x 3 + 3) + 5(2x 1) (2x 1)(4x 3 + 3) 4x5 + 8x 3 + 3x 2 + 10x + 1 8x 4 4x 3 + 6x 3
Instructor s Solutions Manual, Section 2.5 Exercise 9 9. (3r 2s)(x) ( 3x + 4 ) ( x 2 + 2 ) (3r 2s)(x) 3 x 2 2 + 1 2x 1 9x + 12 x 2 + 1 2x2 + 4 2x 1 (9x + 12)(2x 1) (x 2 + 1)(2x 1) (2x2 + 4)(x 2 + 1) (x 2 + 1)(2x 1) (9x + 12)(2x 1) (2x2 + 4)(x 2 + 1) (x 2 + 1)(2x 1) 18x2 9x + 24x 12 2x 4 6x 2 4 2x 3 x 2 + 2x 1 2x4 + 12x 2 + 15x 16 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 10 10. (4r + 5s)(x) ( 3x + 4 ) ( x 2 + 2 ) (4r + 5s)(x) 4 x 2 + 5 + 1 2x 1 12x + 16 x 2 + 1 + 5x2 + 10 2x 1 (12x + 16)(2x 1) (x 2 + 1)(2x 1) + (5x2 + 10)(x 2 + 1) (x 2 + 1)(2x 1) (12x + 16)(2x 1) + (5x2 + 10)(x 2 + 1) (x 2 + 1)(2x 1) 24x2 12x + 32x 16 + 5x 4 + 15x 2 + 10 2x 3 x 2 + 2x 1 5x4 + 39x 2 + 20x 6 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 11 11. (r s)(x) (r s)(x) 3x + 4 x 2 + 1 x2 + 2 2x 1 (3x + 4)(x2 + 2) (x 2 + 1)(2x 1) 3x3 + 4x 2 + 6x + 8 2x 3 x 2 + 2x 1
Instructor s Solutions Manual, Section 2.5 Exercise 12 12. (r t)(x) (r t)(x) 3x + 4 x 2 + 1 5 4x 3 + 3 15x + 20 (x 2 + 1)(4x 3 + 3) 15x + 20 4x 5 + 4x 3 + 3x 2 + 3
Instructor s Solutions Manual, Section 2.5 Exercise 13 13. ( r(x) ) 2 ( ) ( 2 3x + 4 ) 2 r(x) x 2 + 1 (3x + 4)2 (x 2 + 1) 2 9x2 + 24x + 16 x 4 + 2x 2 + 1
Instructor s Solutions Manual, Section 2.5 Exercise 14 14. ( s(x) ) 2 ( ) ( 2 x 2 + 2 ) 2 s(x) 2x 1 (x2 + 2) 2 (2x 1) 2 x4 + 4x 2 + 4 4x 2 4x + 1
Instructor s Solutions Manual, Section 2.5 Exercise 15 15. ( r(x) ) 2t(x) Using the expression that we computed for ( r(x) ) 2 in the to Exercise 13, we have ( ) 2t(x) 9x 2 + 24x + 16 r(x) x 4 + 2x 2 + 1 5 4x 3 + 3 5(9x 2 + 24x + 16) (x 4 + 2x 2 + 1)(4x 3 + 3) 45x 2 + 120x + 80 4x 7 + 8x 5 + 3x 4 + 4x 3 + 6x 2 + 3.
Instructor s Solutions Manual, Section 2.5 Exercise 16 16. ( s(x) ) 2t(x) Using the expression that we computed for ( sr (x) ) 2 in the to Exercise 14, we have ( ) 2t(x) x 4 + 4x 2 + 4 s(x) 4x 2 4x + 1 5 4x 3 + 3 5(x 4 + 4x 2 + 4) (4x 2 4x + 1)(4x 3 + 3) 5x 4 + 20x 2 + 20 16x 5 16x 4 + 4x 3 + 12x 2 12x + 3.
Instructor s Solutions Manual, Section 2.5 Exercise 17 17. (r s)(x) We have (r s)(x) r ( s(x) ) r ( x 2 + 2 ) 2x 1 3( x2 +2 2x 1 ) + 4 ( x 2 +2 2x 1 ) 2 + 1 3 (x2 +2) (2x 1) + 4 (x 2 +2) 2 (2x 1) + 1. 2 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 (x 2 + 2) 2 + (2x 1) 2 6x3 + 13x 2 4x 2 x 4 + 8x 2 4x + 5.
Instructor s Solutions Manual, Section 2.5 Exercise 18 18. (s r )(x) We have (s r )(x) s ( r(x) ) s ( 3x + 4 ) x 2 + 1 ( 3x+4 x 2 +1 ) 2 + 2 2 ( 3x+4 x 2 +1 ) 1 (3x+4) 2 (x 2 +1) + 2 2 6x+8 x 2 +1 1. Multiplying the numerator and denominator of the expression above by (x 2 + 1) 2 gives (s r )(x) (3x + 4) 2 + 2(x 2 + 1) 2 (6x + 8)(x 2 + 1) (x 2 + 1) 2 2x4 + 13x 2 + 24x + 18 x 4 + 6x 3 + 6x 2 + 6x + 7.
Instructor s Solutions Manual, Section 2.5 Exercise 19 19. (r t)(x) We have (r t)(x) r ( t(x) ) ( 5 ) r 4x 3 + 3 3( 5 4x 3 +3 ) + 4 ( 5 4x 3 +3 ) 2 + 1 15 4x 3 +3 + 4 25 (4x 3 +3) + 1. 2 Multiplying the numerator and denominator of the expression above by (4x 3 + 3) 2 gives (r t)(x) 15(4x3 + 3) + 4(4x 3 + 3) 2 25 + (4x 3 + 3) 2 64x6 + 156x 3 + 81 16x 6 + 24x 3 + 34.
Instructor s Solutions Manual, Section 2.5 Exercise 20 20. (t r )(x) We have (t r )(x) t ( r(x) ) ( 3x + 4 ) t x 2 + 1 5 4 ( 3x+4 x 2 +1 ) 3 + 3 5 4(3x+4) 3 (x 2 +1) + 3. 3 Multiplying the numerator and denominator of the expression above by (x 2 + 1) 3 gives (t r )(x) 5(x 2 + 1) 3 4(3x + 4) 3 + 3(x 2 + 1) 3 5x 6 + 15x 4 + 15x 2 + 5 3x 6 + 9x 4 + 108x 3 + 441x 2 + 576x + 259.
Instructor s Solutions Manual, Section 2.5 Exercise 21 21. s(1+x) s(1) x Note that s(1) 3. Thus s(1 + x) s(1) x (1+x) 2 +2 2(1+x) 1 3 x x 2 +2x+3 2x+1 3 x. Multiplying the numerator and denominator of the expression above by 2x + 1 gives s(1 + x) s(1) x x2 + 2x + 3 6x 3 x(2x + 1) x2 4x x(2x + 1) x 4 2x + 1.
Instructor s Solutions Manual, Section 2.5 Exercise 22 22. t(x 1) t( 1) x 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 by 4(x 1) 3 + 3 gives t(x 1) t( 1) x 5 + 5( 4(x 1) 3 + 3 ) x ( 4(x 1) 3 + 3 ) 20x 2 60x + 60 4x 3 12x 2 + 12x 1.
Instructor s Solutions Manual, Section 2.5 Exercise 23 For Exercises 23 28, suppose r(x) x + 1 x 2 + 3 and s(x) x + 2 x 2 + 5. 23. What is the domain of r? The denominator of the expression defining r is a nonzero number for every real number x, and thus the expression defining r makes sense for every real number x. Because we have no other indication of the domain of r, we thus assume that the domain of r is the set of real numbers.
Instructor s Solutions Manual, Section 2.5 Exercise 24 24. What is the domain of s? The denominator of the expression defining s is a nonzero number for every real number x, and thus the expression defining s makes sense for every real number x. Because we have no other indication of the domain of s, we thus assume that the domain of s is the set of real numbers.
Instructor s Solutions Manual, Section 2.5 Exercise 25 25. Find two distinct numbers x such that r(x) 1 4. We need to solve the equation x + 1 x 2 + 3 1 4 for x. Multiplying both sides by x 2 + 3 and then multiplying both sides by 4 and collecting all the terms on one side, we have x 2 4x 1 0. Using the quadratic formula, we get the s x 2 5 and x 2 + 5.
Instructor s Solutions Manual, Section 2.5 Exercise 26 26. Find two distinct numbers x such that s(x) 1 8. We need to solve the equation x + 2 x 2 + 5 1 8 for x. Multiplying both sides by x 2 + 5 and then multiplying both sides by 8 and collecting all the terms on one side, we have x 2 8x 11 0. Using the quadratic formula, we get the s x 4 3 3 and x 4 + 3 3.
Instructor s Solutions Manual, Section 2.5 Exercise 27 27. What is the range of r? To find the range of r, we must find all numbers y such that x + 1 x 2 + 3 y for at least one number x. Thus we will solve the equation above for x and then determine for which numbers y we get an expression for x that makes sense. Multiplying both sides of the equation above by x 2 + 3 and then collecting terms gives yx 2 x + (3y 1) 0. If y 0, then this equation has the x 1. If y 0, then use the quadratic formula to solve the equation above for x, getting x 1 + 1 + 4y 12y 2 2y or x 1 1 + 4y 12y 2. 2y These expressions for x make sense precisely when 1 + 4y 12y 2 0. Completing the square, we can rewrite this inequality as 12 ( (y 1 6 )2 1 9 ) 0.
Instructor s Solutions Manual, Section 2.5 Exercise 27 Thus we must have (y 1 6 )2 1 9, which is equivalent to 1 3 y 1 6 1 3. Adding 1 6 to each side of these inequalities gives 1 6 y 1 2. Thus the range of r is the interval [ 1 6, 1 2 ].
Instructor s Solutions Manual, Section 2.5 Exercise 28 28. What is the range of s? To find the range of s, we must find all numbers y such that x + 2 x 2 + 5 y for at least one number x. Thus we will solve the equation above for x and then determine for which numbers y we get an expression for x that makes sense. Multiplying both sides of the equation above by x 2 + 5 and then collecting terms gives yx 2 x + (5y 2) 0. If y 0, then this equation has the x 2. If y 0, then use the quadratic formula to solve the equation above for x, getting x 1 + 1 + 8y 20y 2 2y or x 1 1 + 8y 20y 2. 2y These expressions for x make sense precisely when 1 + 8y 20y 2 0. Completing the square, we can rewrite this inequality as 20 ( (y 1 5 )2 9 100 ) 0.
Instructor s Solutions Manual, Section 2.5 Exercise 28 Thus we must have (y 1 5 )2 9 100, which is equivalent to the inequalities 3 10 y 1 5 3 10. Adding 1 5 to each side of these inequalities gives 1 10 y 1 2. Thus the range of s is the interval [ 1 10, 1 2 ].
Instructor s Solutions Manual, Section 2.5 Exercise 29 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 are polynomials with deg R<deg q. 29. 2x + 1 x 3 2x + 1 x 3 2(x 3) + 6 + 1 x 3 2 + 7 x 3
Instructor s Solutions Manual, Section 2.5 Exercise 30 30. 4x 5 x + 7 4x 5 x + 7 4(x + 7) 28 5 x + 7 4 33 x + 7
Instructor s Solutions Manual, Section 2.5 Exercise 31 31. x 2 3x 1 x 2 3x 1 x 3 (3x 1) + x 3 3x 1 x 3 + x 3 3x 1 x 3 + 1 9 (3x 1) + 1 9 3x 1 x 3 + 1 9 + 1 9(3x 1)
Instructor s Solutions Manual, Section 2.5 Exercise 32 32. x 2 4x + 3 x 2 4x + 3 x 4 (4x + 3) 3x 4 4x + 3 x 3x 4 4 4x + 3 x 3 9 4 16 (4x + 3) 16 4x + 3 x 4 3 16 + 9 16(4x + 3)
Instructor s Solutions Manual, Section 2.5 Exercise 33 33. x 6 + 3x 3 + 1 x 2 + 2x + 5 x 6 + 3x 3 + 1 x 2 + 2x + 5 x4 (x 2 + 2x + 5) 2x 5 5x 4 + 3x 3 + 1 x 2 + 2x + 5 x 4 + 2x5 5x 4 + 3x 3 + 1 x 2 + 2x + 5 x 4 + ( 2x3 )(x 2 + 2x + 5) x 2 + 2x + 5 + 4x4 + 10x 3 5x 4 + 3x 3 + 1 x 2 + 2x + 5 x 4 2x 3 + x4 + 13x 3 + 1 x 2 + 2x + 5 x 4 2x 3 + ( x2 )(x 2 + 2x + 5) x 2 + 2x + 5 + 2x3 + 5x 2 + 13x 3 + 1 x 2 + 2x + 5 x 4 2x 3 x 2 + 15x3 + 5x 2 + 1 x 2 + 2x + 5
Instructor s Solutions Manual, Section 2.5 Exercise 33 x 4 2x 3 x 2 + 15x(x2 + 2x + 5) 30x 2 75x + 5x 2 + 1 x 2 + 2x + 5 x 4 2x 3 x 2 + 15x + 25x2 75x + 1 x 2 + 2x + 5 x 4 2x 3 x 2 + 15x + 25(x2 + 2x + 5) + 50x + 125 75x + 1 x 2 + 2x + 5 x 4 2x 3 x 2 + 15x 25 + 25x + 126 x 2 + 2x + 5
Instructor s Solutions Manual, Section 2.5 Exercise 34 34. x 6 4x 2 + 5 x 2 3x + 1 x 6 4x 2 + 5 x 2 3x + 1 x4 (x 2 3x + 1) + 3x 5 x 4 4x 2 + 5 x 2 3x + 1 x 4 + 3x5 x 4 4x 2 + 5 x 2 3x + 1 x 4 + (3x3 )(x 2 3x + 1) x 2 3x + 1 + 9x4 3x 3 x 4 4x 2 + 5 x 2 3x + 1 x 4 + 3x 3 + 8x4 3x 3 4x 2 + 5 x 2 3x + 1 x 4 + 3x 3 + (8x2 )(x 2 3x + 1) x 2 3x + 1 + 24x3 8x 2 3x 3 4x 2 + 5 x 2 3x + 1 x 4 + 3x 3 + 8x 2 + 21x3 12x 2 + 5 x 2 3x + 1
Instructor s Solutions Manual, Section 2.5 Exercise 34 x 4 + 3x 3 + 8x 2 + 21x(x2 3x + 1) + 63x 2 21x 12x 2 + 5 x 2 3x + 1 x 4 + 3x 3 + 8x 2 + 21x + 51x2 21x + 5 x 2 3x + 1 x 4 + 3x 3 + 8x 2 + 21x + 51(x2 3x + 1) + 153x 51 21x + 5 x 2 3x + 1 x 4 + 3x 3 + 8x 2 + 21x + 51 + 132x 46 x 2 3x + 1
Instructor s Solutions Manual, Section 2.5 Exercise 35 35. Find a constant c such that r(10 100 ) 6, where r(x) cx3 + 20x 2 15x + 17 5x 3 + 4x 2 + 18x + 7. Because 10 100 is a very large number, we need to estimate the value of r(x) for very large values of x. The highest-degree term in the numerator of r is cx 3 (unless we choose c 0); the highest-degree term in the denominator of r is 5x 3. Factoring out these terms and considering only very large values of x, we have r(x) cx3( 1 + 20 cx 15 cx + 17 ) 2 cx 3 5x 3( 1 + 4 5x + 18 5x + 7 ) 2 5x 3 c ( 20 1 + 5 cx 15 cx + 17 ) 2 cx ( 3 4 1 + 5x + 18 5x + 7 ) 2 5x 3 c 5. For x very large, ( 1 + 20 cx 15 cx + 17 ) ( 4 2 cx and 1 + 3 5x + 18 5x + 7 ) 2 5x are both 3 very close to 1, which explains how we got the approximation above. The approximation above shows that r(10 100 ) c 5. Hence we want to choose c so that c 5 6. Thus we take c 30.
Instructor s Solutions Manual, Section 2.5 Exercise 36 36. Find a constant c such that r(2 1000 ) 5, where r(x) 3x4 2x 3 + 8x + 7 cx 4. 9x + 2 Because 2 1000 is a very large number, we need to estimate the value of r(x) for very large values of x. The highest-degree term in the numerator of r is 3x 4 ; the highest-degree term in the denominator of r is cx 4 (unless we choose c 0). Factoring out these terms and considering only very large values of x, we have r(x) 3x4( 1 2 3x + 8 3x + 7 ) 3 3x 4 cx 4( 1 9 cx + 2 ) 3 cx 4 3 ( 2 1 c 3x + 8 3x + 7 ) 3 3x ( 4 9 1 cx + 2 ) 3 cx 4 3 c. For x very large, ( 1 2 3x + 8 3x 3 + 7 3x 4 ) and ( 1 9 cx 3 + 2 cx 4 ) are both very close to 1, which explains how we got the approximation above. The approximation above shows that r(2 1000 ) 3 c. Hence we want to choose c so that 3 c 5. Thus we take c 3 5.
Instructor s Solutions Manual, Section 2.5 Exercise 37 For Exercises 37 40, find the asymptotes of the graph of the given function r. 37. r(x) 6x4 + 4x 3 7 2x 4 + 3x 2 + 5 The denominator of this rational function is never 0, so we only need to worry about the behavior of r near ±. For x very large, we have r(x) 6x4 + 4x 3 7 2x 4 + 3x 2 + 5 6x4( 1 + 2 3x 7 ) 6x 4 2x ( 4 1 + 3 2x + 5 ) 2 2x 4 3. Thus the line y 3 is an asymptote of the graph of r, as shown below:
Instructor s Solutions Manual, Section 2.5 Exercise 37 3 y 15 5 5 15 x The graph of 6x4 + 4x 3 7 2x 4 + 3x 2 + 5 on the interval [ 15, 15].
Instructor s Solutions Manual, Section 2.5 Exercise 38 6x 6 7x 3 + 3 38. r(x) 3x 6 + 5x 4 + x 2 + 1 The denominator of this rational function is never 0, so we only need to worry about the behavior of r near ±. For x very large, we have 6x 6 7x 3 + 3 r(x) 3x 6 + 5x 4 + x 2 + 1 6x 6( 1 7 6x + 1 ) 3 2x 6 3x ( 6 1 + 5 3x + 1 2 3x + 1 ) 4 3x 6 2. Thus the line y 2 is an asymptote of the graph of r, as shown below: y 2 10 5 5 10 6x 6 7x 3 + 3 The graph of 3x 6 + 5x 4 + x 2 + 1 on the interval [ 10, 10]. x
Instructor s Solutions Manual, Section 2.5 Exercise 39 39. r(x) 3x + 1 x 2 + x 2 The denominator of this rational function is 0 when x 2 + x 2 0. Solving this equation either by factoring or using the quadratic formula, we get x 2orx 1. Because the degree of the numerator is less than the degree of the denominator, the value of this function is close to 0 when x is large. Thus the asymptotes of the graph of r are the lines x 2, x 1, and y 0, as shown below: 20 y 6 2 1 6 x 20 3x + 1 The graph of x 2 on the interval + x 2 [ 6, 6], truncated on the vertical axis to the interval [ 20, 20].
Instructor s Solutions Manual, Section 2.5 Exercise 40 40. r(x) 9x + 5 x 2 x 6 The denominator of this rational function is 0 when x 2 x 6 0. Solving this equation either by factoring or using the quadratic formula, we get x 2orx 3. Because the degree of the numerator is less than the degree of the denominator, the value of this function is close to 0 when x is large. Thus the asymptotes of the graph of r are the lines x 2, x 3, and y 0, as shown below: 20 y 8 2 3 8 x 20 9x + 5 The graph of x 2 on the interval x 6 [ 6, 6], truncated on the vertical axis to the interval [ 20, 20].
Instructor s Solutions Manual, Section 2.5 Problem 41 Solutions to Problems, Section 2.5 41. Suppose s(x) x2 + 2 2x 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 the graph of s very close to (1, 3) is approximately 4. [Hint: Use the result of Exercise 21.] (a) Note that s(1) 12 + 2 2 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 line containing (1, 3) and ( (1 + x,s(1 + x) ) is s(1 + x) 3 (1 + x) 1 s(1 + x) s(1) x x 4 2x + 1, where the last equality comes from Exercise 21. Because x is very x 4 small, 2x+1 is close to 4, and thus the last equation shows that the slope of this line is approximately 4.
Instructor s Solutions Manual, Section 2.5 Problem 42 5 42. Suppose t(x) 4x 3 + 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 a point on the graph of t very close to ( 1, 5). [Hint: Use the result of Exercise 22.] (a) Note that t( 1) 5 4( 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 the line containing ( 1, 5) and ( (x 1,t(x 1) ) is t(x 1) + 5 (x 1) + 1 t(x 1) t( 1) x 20x 2 60x + 60 4x 3 12x 2 + 12x 1, where the last equality comes from Exercise 22. Because x is very 20x small, 2 60x+60 4x 3 12x 2 +12x 1 is close to 60, and thus the last equation shows that the slope of this line is approximately 60.
Instructor s Solutions Manual, Section 2.5 Problem 43 43. Explain how the result in the previous section for the maximum number of peaks and valleys in the graph of a polynomial is a special case of the result in this section for the maximum number of peaks and valleys in the graph of a rational function. Suppose p is a polynomial. We can think of p as the rational function p q, where q is the degree 0 polynomial defined by q(x) 1. By the result in this section concerning the maximum number of peaks and valley in the graph of a rational function, p q can have at most deg p + deg q 1 peaks and valleys. Because p q p and deg q 0, we can restate this result as saying that p can have at most deg p 1 peaks and valleys, which is the result stated in the previous section for polynomials.
Instructor s Solutions Manual, Section 2.5 Problem 44 44. Explain why the composition of a polynomial and a rational function (in either order) is a rational function. 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 ( (s p)(x) ) p(x) (t p)(x). Problem 55 in Section 2.4 tells us that both s p and t p are polynomials. Thus r p, which by the equation above equals s p t p,isthe ratio of two polynomials and hence is a rational function. Now we consider the composition p r in the other order. There exist numbers a 0,a 1,...,a n such that p(x) a 0 + a 1 x + +a n x n. Thus (p r )(x) p ( r(x) ) a 0 + a 1 r(x)+ +a n r(x) n s(x) a 0 + a 1 t(x) + +a s(x) n n t(x) n, which implies that
Instructor s Solutions Manual, Section 2.5 Problem 44 s n s p r a 0 + a 1 t + +a n t n. If k is a positive integer, then s k and t k are polynomials and hence sk is t k a ratio of polynomials and thus is a rational function. Thus each term a k s k t k above is a rational function. Because the sum of rational functions is a rational function, the equation above implies that p r is a rational function.
Instructor s Solutions Manual, Section 2.5 Problem 45 45. Explain why the composition of two rational functions is a rational function. Suppose r and q are rational functions. Thus there exist polynomials s and t such that r s t. We have (r q)(x) r ( q(x) ) s( q(x) ) t ( (s q)(x) ) q(x) (t q)(x). The previous problem tells us that both s p and t p are rational functions. Thus r p, which by the equation above equals s p t p,isthe ratio of two rational functions and hence is a rational function.
Instructor s Solutions Manual, Section 2.5 Problem 46 46. Suppose p is a polynomial and r is a number. Explain why there is a polynomial G such that for every number x r. p(x) p(r) x r G(x) 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 factorization due 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 equation above and then divide both sides by x r to conclude that for every number x r. p(x) p(r) x r G(x)