# Question #9a31c

Apr 8, 2017

${f}^{- 1} \left(x\right) = 4 \pm \sqrt{\frac{x + 2}{2}}$

#### Explanation:

$f \left(x\right) = 2 {\left(x - 4\right)}^{2} - 2$

let say $y = f \left(x\right)$, then $x = {f}^{- 1} \left(y\right)$

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

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

${\left(x - 4\right)}^{2} = \frac{y + 2}{2}$

$\left(x - 4\right) = \pm \sqrt{\frac{y + 2}{2}}$

$x = 4 \pm \sqrt{\frac{y + 2}{2}} = {f}^{- 1} \left(y\right)$

therefore,

${f}^{- 1} \left(x\right) = 4 \pm \sqrt{\frac{x + 2}{2}}$

Apr 8, 2017

Technically speaking, quadratic functions are not invertible, because the ONLY point where two x values do NOT map to a single y value is the vertex. But let's proceed.

#### Explanation:

Given: $f \left(x\right) = 2 {\left(x - 4\right)}^{2} - 2$

Substitute${f}^{-} 1 \left(x\right)$ for every x:

$f \left({f}^{-} 1 \left(x\right)\right) = 2 {\left({f}^{-} 1 \left(x\right) - 4\right)}^{2} - 2$

Use the property $f \left({f}^{-} 1 \left(x\right)\right) = x$ on the left side:

$x = 2 {\left({f}^{-} 1 \left(x\right) - 4\right)}^{2} - 2$

Subtract 2 from both sides:

$x - 2 = 2 {\left({f}^{-} 1 \left(x\right) - 4\right)}^{2}$

Divide both sides by 2:

$\frac{x}{2} - 1 = {\left({f}^{-} 1 \left(x\right) - 4\right)}^{2}$

Use the square root on both sides:

$\pm \sqrt{\frac{x}{2} - 1} = {f}^{-} 1 \left(x\right) - 4$

Please observe that, if you want to restrict the inverse to returning only real values, you should restrict x to be greater than or equal to 2, at this point.

${f}^{-} 1 \left(x\right) = 4 \pm \sqrt{\frac{x}{2} - 1}$
${f}^{-} 1 \left(x\right) = 4 + \sqrt{\frac{x}{2} - 1}$ and ${f}^{-} 1 \left(x\right) = 4 - \sqrt{\frac{x}{2} - 1}$
To assure that these are inverses of the original, you should verify that $f \left({f}^{-} 1 \left(x\right)\right) = x$ and ${f}^{-} 1 \left(f \left(x\right)\right) = x$