Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises: 28

Answer

$$f'(x)=\frac{2x^2-6x+2}{(2x-3)^2}$$ The domain of both $f(x)$ and $f'(x)$ is $(-\infty,\frac{3}{2})U(\frac{3}{2},\infty)$

Work Step by Step

$$f(x)=\frac{x^2-1}{2x-3}$$ According to definition, $$f'(x)=\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{\frac{(x+h)^2-1}{2(x+h)-3}-\frac{x^2-1}{2x-3}}{h}$$ $$f'(x)=\lim\limits_{h\to0}\frac{[(x+h)^2-1](2x-3)-[2(x+h)-3](x^2-1)}{h[2(x+h)-3](2x-3)}$$ $$f'(x)=\lim\limits_{h\to0}\frac{(x+h)^2(2x-3)-(2x-3)-2(x+h)(x^2-1)+3(x^2-1)}{h[2(x+h)-3](2x-3)}$$ $$f'(x)=\lim\limits_{h\to0}\frac{(x^2+h^2+2xh)(2x-3)-2x+3-2x^3+2x-2x^2h+2h+3x^2-3}{h[2(x+h)-3](2x-3)}$$ $$f'(x)=\lim\limits_{h\to0}\frac{2x^3-3x^2+2xh^2-3h^2+4x^2h-6xh-2x^3-2x^2h+2h+3x^2}{h[2(x+h)-3](2x-3)}$$ $$f'(x)=\lim\limits_{h\to0}\frac{2xh^2-3h^2+2x^2h-6xh+2h}{h[2(x+h)-3](2x-3)}$$ $$f'(x)=\lim\limits_{h\to0}\frac{(2xh-3h+2x^2-6x+2)}{[2(x+h)-3](2x-3)}$$ $$f'(x)=\frac{2x\times0-3\times0+2x^2-6x+2}{[2(x+0)-3](2x-3)}$$ $$f'(x)=\frac{2x^2-6x+2}{(2x-3)^2}$$ *Domain of $f(x)$ $f(x)$ is defined for $R$ except where $2x-3=0$, or $x=\frac{3}{2}$ Therefore, the domain of $f(x)$ is $(-\infty,\frac{3}{2})U(\frac{3}{2},\infty)$ *Domain of $f'(x)$ $f'(x)$ is defined for $R$ except where $(2x-3)^2=0$, or $x=\frac{3}{2}$ Therefore, the domain of $f'(x)$ is $(-\infty,\frac{3}{2})U(\frac{3}{2},\infty)$
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