Invers Function

\(f(x) = x^2, \: x < - 3 \)

Domain of \(f\)

\(D_f : x < - 3 \)

Range of \(f\)

For \(x \rightarrow -3, \: y \rightarrow 9\)

For \(x \rightarrow -\sim, \: y \rightarrow +\sim\)

\(R_f : y > 9\)

Determine invers function

Given \(y = x^2\), we need to find \(x = f(y)\).

\begin{equation*}
\begin{split}
y & =x^2 \\
x & = \pm \sqrt{y}
\end{split}
\end{equation*}

Since \(x < - 3\), then we must choose \(x = - \sqrt{y}\)

Hence, the invers function is:

\begin{equation*}
\begin{split}
f^{-1} (x) & = - \sqrt{x}
\end{split}
\end{equation*}

Domain of \(f^{-1} (x)\) is \(x > 9\)

Range of \(f^{-1} (x)\) is \(y < -3\)

Note:

Domain of \(f^{-1}\) is same with range of \(f\)

Range of \(f^{-1}\) is same with domain of \(f\)