Answer
$$f^{-1}(x)=2x+7$$
- 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, $f(f^{-1}(x))=f^{-1}(f(x))=x$
