# Question #312d6

Dec 14, 2016

${h}^{- 1} \left(x\right) = - \sqrt{4 - x}$

#### Explanation:

The inverse of a function $f$ is a function ${f}^{- 1}$ such that ${f}^{- 1} \left(f \left(x\right)\right) = x$ for every $x$ in the domain of $f$.

One trick for finding the inverse of a function $f$ is to set $x = f \left(y\right)$ and then solve for $y$. The resulting equation will be of the form $y = {f}^{- 1} \left(x\right)$.

If we apply that trick to the given function, though, we run into an issue:

Let $x = h \left(y\right) = 4 - {y}^{2}$

$\implies {y}^{2} = 4 - x$

$\implies y = \pm \sqrt{4 - x}$

It seems, then, that we would have ${h}^{- 1} \left(x\right) = \pm \sqrt{4 - x}$. Unfortunately, this is not a function, as by definition, a function can only produce one value for any given input. Here we rely on the additional information given in the problem.

We are given that $x \le 0$, meaning that ${h}^{- 1} \left(h \left(x\right)\right) \le 0$. This tells us that the inverse function should be producing negative values, meaning we want $- \sqrt{4 - x}$, rather than $\sqrt{4 - x}$. Thus, our final answer is

${h}^{- 1} \left(x\right) = - \sqrt{4 - x}$

(As a side note, we do not have to worry about taking the square root of a negative number as the domain of ${h}^{- 1}$ is the same as the range of $h$, which is $\left(- \infty , 4\right]$. Thus we will not apply ${h}^{- 1}$ to any value greater than $4$.)