Question

Given `f(x)=1/8x-3` and `g(x)=x^3`, how do you find `(f^-1of^-1)(6)`?

Answer 1

`(f^(-1) @ f^(-1))(6) = 600`

Explanation:

We have:

`f(x) = 1/8x-3`

And so we can construct the inverse, `f^(-1)(x)` by writing:

`y = 1/8x-3`

So that:

`1/8x = y+3 => x = 8y+24`

Thus we have:

`f^(-1)(x) = 8x+24`

So that:

`(f^(-1) @ f^(-1))(x) = f^(-1)( 8x+24 )`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8( 8x+24 ) + 24`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 64x + 194 + 24`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 64x + 216`

Hence:

`(f^(-1) @ f^(-1))(6) = 64*6+216 = 600`

Noting that `g(x)` is not required.