# Question #9099e

Mar 9, 2017

$f {\left(x\right)}^{-} 1 = {x}^{2} / 8 + \frac{7}{2}$

This can also be written as one fraction:

$f {\left(x\right)}^{-} 1 = \frac{{x}^{2} + 28}{8}$

#### Explanation:

Rules
When finding the inverse, ${f}^{- 1} \left(x\right)$, the easiest method is to switch the variables $x$ and $y$. So $y = f \left(x\right)$ is now $x = f \left(y\right)$.
Afterwards, you need to isolate $y$ again to get the proper form of the inverse.

Application
$f \left(x\right) = y = 2 {\left(2 x - 7\right)}^{\frac{1}{2}} \text{ }$ Applying the "switch" ...

$f \left(y\right) = x = 2 {\left(2 y - 7\right)}^{\frac{1}{2}}$

Solving for $y$ ...

$x = 2 \setminus \sqrt{2 y - 7}$

$\frac{1}{2} x = \setminus \sqrt{2 y - 7}$

${\left(\frac{1}{2} x\right)}^{2} = 2 y - 7$

$\frac{1}{4} {x}^{2} + 7 = 2 y$

$\frac{1}{2} \left(\frac{1}{4} {x}^{2} + 7\right) = y$

$\frac{1}{8} {x}^{2} + \frac{7}{2} = y$

Confused? Feel free to drop a comment and tell me where you're still not understanding things! :)