Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 227: 65

Answer

$f^{-1}(x)=\sqrt[3]{2-5x}$

Work Step by Step

$f(x)=\dfrac{2-x^{3}}{5}$ Rewrite this expression as $y=\dfrac{2-x^{3}}{5}$ and solve for $x$: $y=\dfrac{2-x^{3}}{5}$ Take the $5$ to multiply the left side: $5y=2-x^{3}$ Take $-x^{3}$ to the left side: $5y+x^{3}=2$ Take $5y$ to the right side: $x^{3}=2-5y$ Take the cubic root of both sides: $\sqrt[3]{x^{3}}=\sqrt[3]{2-5y}$ $x=\sqrt[3]{2-5y}$ Interchange $x$ and $y$: $y=\sqrt[3]{2-5x}$ The inverse of the initial function is $f^{-1}(x)=\sqrt[3]{2-5x}$
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