## Thomas' Calculus 13th Edition

$f^{-1}(x)=2x+7$ Domain of $f^{-1}$ is $(-\infty,\infty)$ Range of $f^{-1}$ is $(-\infty,\infty)$
Given $$f(x) = \frac{1}{2}x-\frac{7}{2}$$ Find the inverse: \begin{aligned} y &= \frac{1}{2}x-\frac{7}{2} \\2y &=x-7 \\ x&=2y+7 \\ \text {Switch }& x \ and\ y:\\ y=&2x+7 =f^{-1}(x) \end{aligned} We see that: Domain of $f^{-1}$ is $(-\infty,\infty)$ and Range of $f^{-1}$ is $(-\infty,\infty)$ Confirm the inverse: \begin{aligned} f\left(f^{-1}(x)\right) &=f(2 x+7) \\ &=\frac{1}{2}(2 x+7)-\frac{7}{2} \\ &=x+\frac{7}{2}-\frac{7}{2} \\ &=x \\ f^{-1}(f(x)) &=f^{-1}\left(\frac{1}{2} x-\frac{7}{2}\right) \\ &=2\left(\frac{1}{2} x-\frac{7}{2}\right)+7 \\ &=x-7+7 \\ &=x \end{aligned}