Answer
Answer:
$f^{-1}(x) = \frac{1}{2}x$ when $x \leq 0$ and
$f^{-1}(x) = \sqrt x$ when $x \gt 0$
Work Step by Step
When $x \leq 0$, we have $f(x) = y = 2x$.
Thus, $x = \frac{1}{2}y$.
To get the inverse from this, replace the $x$ on the left-hand side with $f^{-1}(x)$ and replace the $y$ on the right-hand side with $x$.
Thus, $f^{-1}(x) = \frac{1}{2}x$ when $x \leq 0$
When $x \gt 0$, we have $f(x) = y = x^{2}$.
Thus, $x = \sqrt y$.
To get the inverse from this, replace the $x$ on the left-hand side with $f^{-1}(x)$ and replace the $y$ on the right-hand side with $x$.
Thus, $f^{-1}(x) = \sqrt x$ when $x \gt 0$.
Note that the inverse of $x^{2}$, which is $\sqrt x$, is not defined for $x <0$. This is the reason why the function takes values $2x$ whenever $x < 0$.
