Discussion
The inverse of the function \(f(x) = x^{2} + 4x + 4, x \leq -2\) is \(f^{-1}(x) =\)
Answer: A
The inverse is \(f^{-1}(x) = -2 - \sqrt{x}, x \geq 0\).
Write \(f(x) = (x + 2)^{2}\). On \(x \leq -2\) we have \(x + 2 \leq 0\), so the function is one-to-one there and its range is \([0, \infty)\).
\[ \begin{aligned} y &= (x + 2)^{2} \\ x + 2 &= -\sqrt{y} \quad (x + 2 \leq 0) \\ x &= -2 - \sqrt{y} \end{aligned} \]
The negative square root is taken because \(x + 2\) is not positive. The domain of the inverse is the range of \(f\), namely \(x \geq 0\). The options restricting the domain to \(x \geq 1, 2, 4\) or 5 leave out values that \(f\) does take, and \(\sqrt{x - 1}\) does not undo \((x + 2)^{2}\).
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