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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 18e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.3 - Problem 18e

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# Answer: Graphing rational functions Use the guidelines of

ISBN: 9780321570567 2

## Solution for problem 18E Chapter 4.3

Calculus: Early Transcendentals | 1st Edition

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Problem 18E

15-20. Graphing rational functions Use the guidelines of this section to make a complete graph of f.

$$f(x)=\frac{2 x-3}{2 x-8}$$

Step-by-Step Solution:
Step 1 of 3

Solution: Step1 Given function is f(x)= 2x8 The zero of the denominator is x=4 so, the domain is { x; x= / 4} 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)= 2 - 2(2x3) 2x8 (2x8) 2(2x8)2(2x3) = (2x8) = 4x162x+6 (2x8) = 102 (2x8) Again differentiate f’(x) to find f’’(x) we get, 20 f’’(x)= (2x8).2 40 = 3 (2x8) Step3 To get extreme values we have to use f’(x)=0 10 (2x8) =0 -10=0 There will be no critical points . Step4 To find inflection points we have to use f’’(x)=0 40 3 =0 (2x8) 40=0 There will be no inflection point. The function has no local maximum and minimum. Step5 Note that zeros of the denominator which in this case is x=4 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 2x-3= 0 or x=1.5 Therefore, (1.5,0)is x-intercept. Step6

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