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Graphing rational functions Use the guidelines of this

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 16E Chapter 4.3

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 16E

Graphing rational functions ?Use the guidelines of this section to make a complete graph of f.

Step-by-Step Solution:
Step 1 of 3

Solution: Step1 Given function is f(x)= x +12 2x+1 The zero of the denominator is x=-0.5 so, the domain is { x; x/ -0.5} This function consists of an even function divided by an odd function. The product of even and odd function is odd. Therefore, the graph is symmetric about the origin. Step2 Differentiate the given equation to find f’(x) we get, f’(x)= (2x)/(2x+1) - (2(x^2+12))/(2x+1)^2 = (2x(2x+1)-2(x^2+12))/(2x+1)^2 = (4x^2+2x-2x^2-24)/(2x+1)^2 = (2x^2+2x-24)/(2x+1)^2 = (2(x^2+x-12))/(2x+1)^2 Again differentiate f’(x) to find f’’(x) we get, f’’(x)=(4x+2)/(2x+1)^2 - (4(2x^2+2x-24))/(2x+1)^3 = ((4x+2)(2x+1)-8x^2-8x+96)/(2x+1)^3 = (8x^2+4x+4x+2-8x^2-8x+96)/(2x+1)^3 = 98/(2x+1)^3 Step3 To get extreme values we have to use f’(x)=0 => (2(x^2+x-12))/(2x+1)^2 =0 =>2(x^2+x-12)=0 => x^2+x-12=0 =>x^2+4x-3x-12=0 => x(x+4)-3(x+4)=0 =>(x-3)(x+4)=0 =>x=3,-4 Critical points are 3,-4 Step4 To find inflection points we have to use f’’(x)=0 =>98/(2x+1)^3=0 Evaluate f’’(x) at the critical points we get, f”(3 )=98/(2x+1)^3 = 98/(2*3+1)^3 = 98/(7)^3 =98/343 =0.286>0 has a local minimum at x=3 And f”(-4)= 98/(2x+1)^3 = 98/(2*-4+1)^3 =98/(-8+1)^3 =98/(-7)^3 = 98/-343 =-0.286<0 Has a local Maximum at x=-4 Step5 The corresponding function values are f(3 )=(x^2+12)/(2x+1) = ((3)^2+12)/(2*3+1) =(9+12)/(6+1) =21/7=3 , f(4 )= (x^2+12)/(2x+1) =((-4)^2+12)/(2*-4+1) =(16+12)/(-8+1) = 28/-7 =-4 Therefore, there will be no inflection points.And the local maximum at (-4,-4) , Local minimum at (3,3) Note that zeros of the denominator which in this case is x=-0.5 is the candidate for vertical asymptotes. The zeros of the rational fraction coincide with the zeros of the numerator.In this case the zeros of f(x) satisfy x^2+12= 0 Therefore, (0,12)is y-intercept. Step6

Step 2 of 3

Chapter 4.3, Problem 16E is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Graphing rational functions Use the guidelines of this