College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises - Page 468: 8

Answer

$f^{-1}(x)=\sqrt{25-x^2}, x\le25$.

Work Step by Step

Let us find the inverse function.Replace $f(x)$ with y. $y=\sqrt{25-x^2}$ Interchange x and y. $x=\sqrt{25-y^2}$ Solve for y and square each side. $x^2=25-y^2$ Subtract 25 from each side. $x^2+25=-y^2$ Divide each side by -1. $25=x^2-y^2$ Now take the square root of both sides. $y=\sqrt{25=x^2}$ Replace y with $f^{-1}(x)$ $f^{-1}(x)=\sqrt{25-x^2}$ The range of f is $y\le0$. This means that the domain of $f^{-1}$ has to be $x\le25$. The inverse function is $f^{-1}(x)=\sqrt{25-x^2}, x\le25$.
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