University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 49: 28


$$f^{-1}(x)=2x+7$$ - Domain: $(-\infty,\infty)$ - Range: $(-\infty,\infty)$

Work Step by Step

$$y=f(x)=\frac{1}{2}x-\frac{7}{2}\hspace{1cm}$$ To find its inverse: 1) Solve for $x$ in terms of $y$: $$y=\frac{1}{2}x-\frac{7}{2}$$ $$\frac{1}{2}x=y+\frac{7}{2}$$ $$x=2y+7$$ 2) Interchange $x$ and $y$: $$y=2x+7$$ Therefore, $$f^{-1}(x)=2x+7$$ - Domain: $x$ is defined in $R$. So the domain of $f^{-1}$ is $(-\infty,\infty)$. - Range: $x$ ranges from $-\infty$ to $\infty$. All the same, $2x+7$ also ranges from $-\infty$ to $\infty$. So the range of $f^{-1}$ is $(-\infty,\infty)$ *Check: $f(f^{-1}(x))=\frac{1}{2}(2x+7)-\frac{7}{2}=x+\frac{7}{2}-\frac{7}{2}=x$ $f^{-1}(f(x))=2\Big(\frac{1}{2}x-\frac{7}{2}\Big)+7=x-7+7=x$ Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
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