Question

Question #9a31c

Answer 1

`f^(-1) (x) = 4+-sqrt ((x + 2)/2)`

Explanation:

`f(x) = 2(x -4)^2 - 2`

let say `y =f(x)`, then `x = f^(-1) (y)`

`2(x -4)^2 - 2 = y`

`2(x -4)^2 = y + 2`

`(x -4)^2 = (y + 2)/2`

`(x -4) = +-sqrt ((y + 2)/2)`

`x = 4+-sqrt ((y + 2)/2) = f^(-1) (y)`

therefore,

`f^(-1) (x) = 4+-sqrt ((x + 2)/2)`

Answer 2

Technically speaking, quadratic functions are not invertible, because the ONLY point where two x values do NOT map to a single y value is the vertex. But let's proceed.

Explanation:

Given: `f(x) = 2(x-4)^2 -2`

Substitute`f^-1(x)` for every x:

`f(f^-1(x)) = 2(f^-1(x)-4)^2 -2`

Use the property `f(f^-1(x)) = x` on the left side:

`x = 2(f^-1(x)-4)^2 -2`

Subtract 2 from both sides:

`x-2 = 2(f^-1(x)-4)^2`

Divide both sides by 2:

`x/2-1 = (f^-1(x)-4)^2`

Use the square root on both sides:

`+-sqrt(x/2-1) = f^-1(x)-4`

Please observe that, if you want to restrict the inverse to returning only real values, you should restrict x to be greater than or equal to 2, at this point.

Add 4 to both sides:

`f^-1(x)= 4+-sqrt(x/2-1)`

This is two functions:

`f^-1(x)= 4+sqrt(x/2-1)` and `f^-1(x)= 4-sqrt(x/2-1)`

To assure that these are inverses of the original, you should verify that `f(f^-1(x)) = x` and `f^-1(f(x))=x`