\documentclass[12pt]{article}
\usepackage{amsmath,amssymb}
\usepackage{tkz-euclide,tkz-fct}
\usetkzobj{all}

\begin{document}
The graph of $f(x)=-x^2+7$ is a parabola that opens downward, and is shifted 7 units upward. Because the domain is restricted to $x\le 0$, we sketch only that part of the parabola that lies to the left of $x=0$ (see the figure on the left). Note that this piece satisfies the horizontal line test, so $f$ is a one-to-one function and its inverse exists. The inverse is found by reflecting the graph of $f$ across the line $y=x$, which produces the graph shown in the figure on the right. 

\medbreak
\begin{minipage}{0.45\linewidth}
\centering
\begin{tikzpicture}[scale=0.25]
\tkzInit[xmin=-10,xmax=10,ymin=-10,ymax=10]
\tkzGrid
\tkzLabelX[step=20]
\tkzLabelY[step=20]
\tkzDrawX[right=3pt]
\tkzDrawY[above=3pt]
\tkzFct[thick,color=blue,domain=-10:0]{-\x**2+7}
\tkzText[below=3pt,text=blue](-4.12310562561766,-10){$f$}
\tkzFct[thick,color=green,domain=-10:10]{\x}
\tkzText[above,text=green](10,10){$y=x$}
\end{tikzpicture}
\end{minipage}\hfil
\begin{minipage}{0.45\linewidth}
\centering
\begin{tikzpicture}[scale=0.25]
\tkzInit[xmin=-10,xmax=10,ymin=-10,ymax=10]
\tkzGrid
\tkzLabelX[step=20]
\tkzLabelY[step=20]
\tkzDrawX[right=3pt]
\tkzDrawY[above=3pt]
\tkzFct[thick,color=blue,domain=-10:7]{-1*sqrt(7-\x)}
\tkzText[left=3pt,text=blue](-10,-4.12310562561766){$f^{-1}$}
\end{tikzpicture}
\end{minipage}

\medbreak

One might surmise from the figure on the right that the equation of the inverse function is $f^{-1}(x)=-\sqrt{-(x-7)}$, but let's use an algebraic approach to verify the result. 
\end{document}