# How do you find the inverse of y = - |x-3| + 5?

Feb 9, 2017

${f}^{- 1} \left(x\right) = 3 \pm \left(5 - x\right)$

#### Explanation:

$y = - | x - 3 | + 5$

$| x - 3 | = 5 - y$

$x - 3 = \pm \left(5 - y\right)$

$x = 3 \pm \left(5 - y\right)$

The inverse function is therefore:

${f}^{- 1} \left(x\right) = 3 \pm \left(5 - x\right)$

There would be no harm in verifying this for $x = 2$ and $x = 4$, ie either side of $x = 3$ which is the point at which $\left(x - 3\right)$ changes sign.

For $x = \textcolor{red}{2}$, we get $y = 4$. Our inverse function suggests that ${f}^{- 1} \left(4\right) = 3 \pm 1 = 4 , \textcolor{red}{2}$.

And for $x = \textcolor{b l u e}{4}$, we also get $y = 4$, which we expect from the symmetry. Our inverse function suggests that ${f}^{- 1} \left(4\right) = 3 \pm 1 = \textcolor{b l u e}{4} , 2$.

graph{y = -|x-3| + 5 [-10, 10, -5, 5]}