University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.8 - Derivatives of Inverse Functions and Logarithms - Exercises - Page 174: 8



Work Step by Step

$$f(x)=x^2-4x-5\hspace{1cm}x\gt2$$ To find $df^{-1}/dx$ at $x=0=f(5)$, or $\left.\frac{df^{-1}}{dx}\right|_{x=0=f(5)}$, we need to apply Theorem 3. In detail, 1) Find $\left.\frac{df}{dx}\right|_{x=5}$: We have $$\frac{df}{dx}=2x-4$$ So, $$\left.\frac{df}{dx}\right|_{x=5}=2\times5-4=6$$ 2) According to Theorem 3: $$\left.\frac{df^{-1}}{dx}\right|_{x=f(5)}=\frac{1}{\left.\frac{df}{dx}\right|_{x=5}}=\frac{1}{6}$$
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