# Question #23f3c

Aug 20, 2016

Assuming that, by f-1(x) you mean ${f}^{- 1} \left(x\right)$ (the inverse of $f \left(x\right)$) then the answers are $1$ and $4$, respectively.

#### Explanation:

Let's first review the process of finding inverses. It consists of 4 steps:

1. Change $f \left(x\right)$ to $y$.
2. Switch $x$ and $y$.
3. Solve for $y$.
4. Change $y$ to ${f}^{- 1} \left(x\right) .$

As this method applies to $f \left(x\right) = 2 x + 2$, we have:
$y = 2 x + 2 \implies$ Changing $f \left(x\right)$ to $y$
$x = 2 y + 2 \implies$ Switching $x$ and $y$
$x - 2 = 2 y \to y = \frac{x - 2}{2} \implies$ Solving for $y$
${f}^{- 1} \left(x\right) = \frac{x - 2}{2} \implies$ Changing $y$ to ${f}^{- 1} \left(x\right)$

The question asks for ${f}^{- 1} \left(x\right)$ when $x = 4$, so:
${f}^{- 1} \left(x\right) = \frac{x - 2}{2}$
$\to {f}^{- 1} \left(x\right) = \frac{4 - 2}{2}$
$\to {f}^{- 1} \left(x\right) = 1$

The correct answer, then, is $1$.

We follow the same process for $f \left(x\right) = 2 x - 6$:
$y = 2 x - 6$
$x = 2 y - 6$
$2 y = x + 6$
$y = \frac{x + 6}{2} \to {f}^{- 1} \left(x\right) = \frac{x + 6}{2}$

Now we just plug in $2$ for $x$ and do the math:
${f}^{- 1} \left(x\right) = \frac{x + 6}{2}$
$\to {f}^{- 1} \left(x\right) = \frac{2 + 6}{2}$
$\to {f}^{- 1} \left(x\right) = 4$

The correct answer for this problem is $4$.