Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.4 Inverse Functions - Exercises Set 0.4 - Page 45: 17


$$ f^{-1}(x)=x^{\frac{1}{4}}-2, \quad \quad x \geq 16. $$

Work Step by Step

we first write $$ y= f(x)=(x+2)^{4}, \quad x \geq 0 $$ Then we solve this equation for $ x$ as a function of $y$ $$ y^{\frac{1}{4}}= x+2 $$ $$ x =y^{\frac{1}{4}}-2 $$ which tells us that $$ f^{-1}(y)=y^{\frac{1}{4}}-2. \quad \quad (i) $$ Since we want $x$ to be the independent variable, we reverse $x$ and $y $ in (i) to produce the formula $$ f^{-1}(x)=x^{\frac{1}{4}}-2. $$ We know the domain of $f^{ −1}$ is the range of $f$ whereas the range of $ f(x)=(x+2)^{4}$ is $ [16, \infty) $. Thus, if we want to make the domain of $f^{ −1}$ clear, we must express it explicitly by rewriting (i) as $$ f^{-1}(x)=x^{\frac{1}{4}}-2, \quad x \geq 16. $$
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