Question #23f3c

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Question #23f3c

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Answer 1 · 2016-08-20T01:29:35

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

Explanation:

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

Change $f (x)$ $f(x)$ to $y$ $y$ .
Switch $x$ $x$ and $y$ $y$ .
Solve for $y$ $y$ .
Change $y$ $y$ to $f^{- 1} (x) .$ $f^(-1)(x).$

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

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

The correct answer, then, is $1$ $1$ .

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

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

The correct answer for this problem is $4$ $4$ .

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Question #23f3c

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