# How do you find the inverse of g(x) = (x -5)^2?

May 30, 2018

${g}^{-} 1 \left(x\right) = 5 + \sqrt{x}$

or

${g}^{-} 1 \left(x\right) = 5 - \sqrt{x}$

#### Explanation:

$g \left(x\right) = {\left(x - 5\right)}^{2}$

$y = {\left(x - 5\right)}^{2}$

Switch the $x$ and $y$:

$x = {\left(y - 5\right)}^{2}$

solve for $x$:

$\pm \sqrt{x} = \sqrt{{\left(y - 5\right)}^{2}}$

$\pm \sqrt{x} = y - 5$

$y = 5 \pm \sqrt{x}$

Now you have the problem that this inverse is not a function unless you restrict its range.

${g}^{-} 1 \left(x\right) = 5 + \sqrt{x}$

or

${g}^{-} 1 \left(x\right) = 5 - \sqrt{x}$