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Answer 1 · 2017-12-12T00:26:36

Please see below.

Explanation:

.

You switch the $x$ $x$ and $y$ $y$ in each function and solve for $y$ $y$ :

$1 .$ $1.$ $y = - \frac{x}{12}$ $y=-x/12$

$x = - \frac{y}{12}$ $x=-y/12$ ,

$y = - 12 x$ $y=-12x$

$2 .$ $2.$ $y = \frac{x - 12}{4}$ $y=(x-12)/4$

$x = \frac{y - 12}{4}$ $x=(y-12)/4$

$4 x = y - 12$ $4x=y-12$

$y = 4 x + 12$ $y=4x+12$

$3 .$ $3.$ $y = \frac{3 x + 1}{6}$ $y=(3x+1)/6$

$x = \frac{3 y + 1}{6}$ $x=(3y+1)/6$

$6 x = 3 y + 1$ $6x=3y+1$

$3 y = 6 x - 1$ $3y=6x-1$

$y = \frac{6 x - 1}{3}$ $y=(6x-1)/3$

Answer 2 · 2017-12-12T00:38:37

Manipulate the equation so that $x$ $x$ is isolated on one side, getting:
1. $f^{- 1} (x) = - 12 x$ $f^-1(x) = -12x$
2. $f^{- 1} (x) = 4 x + 12$ $f^-1(x) = 4x + 12$
3. $f^{- 1} (x) = 2 x - \frac{1}{3}$ $f^-1(x) = 2x - 1/3$

Explanation:

The inverse of a function, say, $f (x)$ $f(x)$ , is a function $f^{- 1} (x)$ $f^-1(x)$ such that $f^{- 1} (f (x)) = x$ $f^-1(f(x)) = x$ . To obtain the inverse function, manipulate the equation so that $x$ $x$ is isolated on one side.

Function 1: $f (x) = - \frac{x}{12}$ $f(x) = -(x)/(12)$

We could multiply each side by $- 12$ $-12$ :

$- 12 f (x) = x$ $-12f(x) = x$

So, it must be that $f^{- 1} (x) = - 12 x$ $f^-1(x) = -12x$ . Let's test it to be sure:

$f^{- 1} (f (x)) = - 12 (f (x)) = - 12 (- \frac{x}{12}) = x$ $f^-1(f(x)) = -12(f(x)) = -12(-(x)/(12)) = x$ .

Alright!

Function 2: $f (x) = \frac{x - 12}{4}$ $f(x) = (x - 12)/4$

Multiply each side by $4$ $4$ :

$4 f (x) = x - 12$ $4f(x) = x - 12$

Add each side by $12$ $12$ :

$4 f (x) + 12 = x$ $4f(x) + 12 = x$

So $f^{- 1} (x) = 4 x + 12$ $f^-1(x) = 4x + 12$ .

$f^{- 1} (f (x)) = 4 (f (x)) + 12 = 4 (\frac{x - 12}{4}) + 12$ $f^-1(f(x)) = 4(f(x)) + 12 = 4((x - 12)/4) + 12$
$= x - 12 + 12 = x$ $= x - 12 + 12 = x$ .

Function 3: $f (x) = \frac{3 x + 1}{6}$ $f(x) = (3x + 1)/6$

Multiply each side by $6$ $6$ :

$6 f (x) = 3 x + 1$ $6f(x) = 3x + 1$

Subtract $1$ $1$ from each side:

$6 f (x) - 1 = 3 x$ $6f(x) - 1 = 3x$

Divide each side by $3$ $3$ :

$\frac{6}{3} f (x) - \frac{1}{3} = x$ $6/3 f(x) - 1/3 = x$

$2 f (x) - \frac{1}{3} = x$ $2f(x) - 1/3 = x$

So $f^{- 1} (x) = 2 x - \frac{1}{3}$ $f^-1(x) = 2x - 1/3$ .

$f^{- 1} (f (x)) = 2 (f (x)) - \frac{1}{3} = 2 (\frac{3 x + 1}{6}) - \frac{1}{3}$ $f^-1(f(x)) = 2(f(x)) - 1/3 = 2((3x + 1)/6) - 1/3$
$= \frac{3 x + 1}{3} - \frac{1}{3} = \frac{3 x}{3} = x$ $= (3x + 1)/3 - 1/3 = (3x)/3 = x$ .

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