Question

Given `f(x)= x^5+x^3+x`, how do you find the inverse f(3) and f(f inverse (2))?

Answer 1

Examine the behaviour of `f(x)` and of `f^(-1)` to find:

`f^(-1)(3) = 1` and `f(f^(-1)(2)) = 2`

Explanation:

`f(x) = x^5+x^3+x` is a strictly monotonically increasing function with domain and range the whole of `RR`.

One way of seeing that is to look at its derivative:

`f'(x) = 5x^4 + 3x^2 + 1 > 0` for all `x in RR`

So its inverse `f^(-1)` is also strictly monotonic increasing with domain and range the whole of `RR`.

`f^(-1)(f(x)) = x` for all `x in RR`

Applying `f` to both sides we get:

`f(f^(-1)(f(x))) = f(x)` for all `x in RR`

Substitute `y = f(x)` to find:

`f(f^(-1)(y)) = y` for all `y in RR`

This last "for all `y in RR`" is due to the range of `f(x)` being the whole of `RR`.

So `f` is the inverse of `f^-1` too.

For example, `f(f^(-1)(2)) = 2`

Also, if `f(x) = 3`, then `f^(-1)(3) = x`

`f^(-1)(3)` is the root of `f(x) = 3`

That is:

`x^5+x^3+x = 3`

Notice that `x = 1` works, so this will be the only Real root and

`f^(-1)(3) = 1`