Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 344: 64

Answer

$\frac{1}{5}$

Work Step by Step

$f(x)$ = $x^{3}+2x-1$, $a$ = $2$ $x^{3}+2x-1$ = $2$ $x^{3}+2x-3$ = $0$ $(x-1)$$(x^{2}+x+3)$ = $0$ $x$ = $1$ $f(1)$ = $2$ $so$ $f^{-1}(2)$ = $1$ $(f^{-1})$$'(2)$ = $\frac{1}{f'(f^{-1}(2))}$ = $\frac{1}{f'(1)}$ $f'(x)$ = $3x^{2}+2$ $f'(1)$ = $3(1)^{2}+2$ = $5$ $so$ $(f^{-1})$$'(2)$ = $\frac{1}{f'(1)}$ = $\frac{1}{5}$
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