Question

Let `f(x)=x^5+3x−2` and let `f^-1` denote the inverse of f, Then how do you find `(f^-1)'(2)`?

Answer 1

`[f^-1]'(2)=1/8`

Explanation:

For the sake of easier notation, we shall say that `f` is a function with inverse function `g`, that is, `f^-1(x)=g(x)`.

According to the definition of inverse functions,

`f(g(x))=x`

Differentiating through the chain rule gives

`f'(g(x))g'(x)=1`

Solving for the derivative of the inverse gives

`g'(x)=1/(f'(g(x)))`

So, we want to find `g'(2)`.

`g'(2)=1/(f'(g(2))`

We want to first find `g(2)`, however, we cannot write an expression for `g(x)`. What we have to remember is that the domain of the mother function is the range of its inverse function, and vice versa.

Note that if `f(x)=2`, then `x=g(2)`. This means we should let `f(x)=2` then solve for `x`, which is equal to `g(2)`.

`f(x)=2" "=>" "x^5+3x-2=2`

Continuing to solve yields

`x^5+3x-4=0`

This may look impossible, but notice that the sum of the coefficients of each term is `0`, that is, `1+3-4=0`. This means that `x=1` is a solution.

Dividing, we see that

`(x-1)(x^4+x^3+x^2+x+4)=0`

Note that `x^4+x^3+x^2+x+4>0` for all values of `x`, so it has no real roots. Thus, `f(x)=2` when `x=1`, i.e., `f(1)=2`.

This means that `g(2)=1`.

Returning to what we had earlier, we see that

`g'(2)=1/(f'(g(2)))=1/(f'(1))`

Find this by taking the derivative of `f`.

`f(x)=x^5+3x-2" "=>" "f'(x)=5x^4+3`

Thus `f'(1)=5+3=8`.

`g'(2)=1/8`

With your notation,

`[f^-1]'(2)=1/8`