Given the function \( f(x)=x^{2}-2, x \geq 0 \) (a) Find \( f^{-1}(x) \). (b) Graph \( f \) and \( f^{-1} \) in the same rectangular coordinate system. (c) Use interval notation to give the domain and the range of \( f \) and \( f^{-1} \).

(a) To find \( f^{-1}(x) \), we need to solve for \( x \) in the equation \( y=x^{2}-2 \). Rearranging, we get \( x^{2}=y+2 \), and taking the square root (noting that \( x \geq 0 \)), we have \( x=\sqrt{y+2} \). Therefore, \( f^{-1}(x)=\sqrt{x+2} \). (b) To graph \( f \) and \( f^{-1} \), we can plot some points and connect them. For \( f(x)=x^{2}-2 \), we can choose some values of \( x \) and compute the corresponding values of \( f(x) \): \begin{align*} f(0)&=-2 \\ f(1)&=-1 \\ f(2)&=2 \\ f(3)&=7 \\ f(4)&=14 \end{align*} Using these points, we can sketch the graph of \( f \): \begin{center} \begin{tikzpicture}[scale=0.8] \draw[<->] (-2,0) -- (4,0) node[right] {$x$}; \draw[<->] (0,-3) -- (0,15) node[above] {$y$}; \draw[domain=0:2.5,smooth,variable=\x,blue] plot ({\x},{\x*\x-2}); \draw[dashed] (-2,-2) -- (4,14); \foreach \x in {0,1,2,3,4} { \filldraw (\x,\x*\x-2) circle (2pt); } \node[below left] at (0,-2) {$-2$}; \node[below] at (1,-1) {$(-1,0)$}; \node[above] at (2,2) {$(2,2)$}; \node[above] at (3,7) {$(3,7)$}; \node[above] at (4,14) {$(4,14)$}; \end{tikzpicture} \end{center} To graph \( f^{-1}(x)=\sqrt{x+2} \), we can choose some values of \( x \) and compute the corresponding values of \( f^{-1}(x) \): \begin{align*} f^{-1}(0)&=\sqrt{2} \\ f^{-1}(1)&=\sqrt{3} \\ f^{-1}(2)&=2 \\ f^{-1}(3)&=\sqrt{5} \\ f^{-1}(4)&=\sqrt{6} \end{align*} Using these points, we can sketch the graph of \( f^{-1} \): \begin{center} \begin{tikzpicture}[scale=0.8] \draw[<->] (-2,0) -- (4,0) node[right] {$x$}; \draw[<->] (0,-3) -- (0,15) node[above] {$y$}; \draw[domain=-2:-1/2,smooth,variable=\x,red] plot ({\x},{(\x+2)^0.5}); \draw[domain=-2:14,smooth,variable=\x,red] plot ({\x},{(\x+2)^0.5}); \draw[dashed] (-2,-2) -- (4,14); \foreach \x in {0,1,2,3,4} { \filldraw (\x,{(\x+2)^0.5}) circle (2pt); } \node[above left] at (0,{(\x+2)^0.5}) {$\sqrt{2}$}; \node[above] at (1,{(\x+2)^0.5}) {$\sqrt{3}$}; \node[above] at (2,2) {$2$}; \node[above] at (3,{(\x+2)^0.5}) {$\sqrt{5}$}; \node[above] at (4,{(\x+2)^0.5}) {$\sqrt{6}$}; \end{tikzpicture} \end{center} (c) The domain of \( f \) is \( [0,\infty) \), since the function is defined only for non-negative values of \( x \). The range of \( f \) is \( [-2,\infty) \), since \( f(x) \geq -2 \) for all \( x \geq 0 \), and \( f(x) \to \infty \)