Question

Help please?

Answer 1

Please see below.

Explanation:

.

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

`1.` `y=-x/12`

`x=-y/12`,

`y=-12x`

`2.` `y=(x-12)/4`

`x=(y-12)/4`

`4x=y-12`

`y=4x+12`

`3.` `y=(3x+1)/6`

`x=(3y+1)/6`

`6x=3y+1`

`3y=6x-1`

`y=(6x-1)/3`

Answer 2

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

Explanation:

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

Function 1: `f(x) = -(x)/(12)`

We could multiply each side by `-12`:

`-12f(x) = x`

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

`f^-1(f(x)) = -12(f(x)) = -12(-(x)/(12)) = x`.

Alright!

Function 2: `f(x) = (x - 12)/4`

Multiply each side by `4`:

`4f(x) = x - 12`

Add each side by `12`:

`4f(x) + 12 = x`

So `f^-1(x) = 4x + 12`.

`f^-1(f(x)) = 4(f(x)) + 12 = 4((x - 12)/4) + 12`
`= x - 12 + 12 = x`.

Function 3: `f(x) = (3x + 1)/6`

Multiply each side by `6`:

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

Subtract `1` from each side:

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

Divide each side by `3`:

`6/3 f(x) - 1/3 = x`

`2f(x) - 1/3 = x`

So `f^-1(x) = 2x - 1/3`.

`f^-1(f(x)) = 2(f(x)) - 1/3 = 2((3x + 1)/6) - 1/3`
`= (3x + 1)/3 - 1/3 = (3x)/3 = x`.