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: 19

Answer

$$ y= f(x)=-\sqrt{3-2 x} , \quad x \ \leq \frac{3}{2} $$ we find that: $$ f^{-1}(x)=\frac{3-x^{2}}{2} , \quad x \leq 0. $$

Work Step by Step

we first write $$ y= f(x)=-\sqrt{3-2 x} , \quad x \ \leq \frac{3}{2} $$ Then we solve this equation for $ x$ as a function of $y$ $$ y^{2}=3-2 x $$ the later equation can be rewritten as the following $$ x= \frac{3-y^{2}}{2} $$ which tells us that $$ f^{-1}(y)= \frac{3-y^{2}}{2} . \quad \quad (i) $$ Since we want $x$ to be the independent variable, we reverse $x$ and $y $ in (i) to produce the formula $$ f^{-1}(x)=\frac{3-x^{2}}{2} . $$ We know the domain of $f^{ −1}$ is the range of $f$ whereas the range of $-\sqrt{3-2 x}$ is $ (- \infty , 0 ]$. Thus, if we want to make the domain of $f^{ −1}$ clear, we must express it explicitly by rewriting (i) as $$ f^{-1}(x)=\frac{3-x^{2}}{2} , \quad x \leq 0. $$
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