Notice that the left side of the graph corresponds to y=-x, whereas the right side corresponds to another line. We can find the former by simply using two points. In this case (0,0) and (5,-3).
Then the equation of the line on the right side is:
[tex]y=-\frac{3}{5}x[/tex]
So, the function is:
[tex]y=f(x)=\begin{cases}-x;-7\le x\le0 \\ -\frac{3}{5}x;0\le x\le5\end{cases}[/tex]
Since function f(x) has two parts, so does the inverse function.
Remember the definition of an inverse function, let g(x) be a function:
[tex]\begin{gathered} g^{-1}\circ g(x)=x, \\ g\circ g^{-1}(y)=y \end{gathered}[/tex]
As for the left side of the function:
[tex]\begin{gathered} f^{-1}\circ f(x)=x \\ andf^{-1}\circ f(x)=f^{-1}(-x) \\ \Rightarrow f^{-1}(-x)=x \\ \Rightarrow f^{-1}(x)=-x \end{gathered}[/tex]
Now, the right side:
[tex]\begin{gathered} f^{-1}\circ f(x)=x \\ \Rightarrow f^{-1}\circ f(x)=f^{-1}(-\frac{3}{5}x) \\ \Rightarrow f^{-1}(-\frac{3}{5}x)=x \\ \Rightarrow f^{-1}(x)=-\frac{5}{3}x \end{gathered}[/tex]
Thus, the inverse function is:
[tex]f^{-1}(x)=\begin{cases}-x,-7\le x\le0 \\ -\frac{5}{3}x,0\le x\le5\end{cases}[/tex]
The range of the inverse function is:
For the first part of the function:
[tex]range(f^{-1}(x))=\begin{cases}f(x)\in\lbrack0,7\rbrack,x\in\lbrack-7,0\rbrack \\ f(x)\in\lbrack-\frac{25}{3},0\rbrack,x\in\lbrack0,5\rbrack\end{cases}[/tex]
Merging both results:
[tex]\begin{gathered} range(f^{-1}(x))=\lbrack-\frac{25}{3},7\rbrack \\ domain(f^{-1}(x))=\lbrack-7,5\rbrack \end{gathered}[/tex]
