Question

How do you find the critical numbers of `f(x) = x^4(x-1)^3`?

Answer 1

The critical numbers are `0, 1, "and "4/7`

Explanation:

`f(x) = x^4(x-1)^3`

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

Since `f'` is a polynomial, it is never undefined, so we need only solve:

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

This is the sum of two terms:

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

These terms have common factors of `x^3` and `(x-1)^2`, so remove those:

`x^3(x-1)^2[4(x-1) + 3x] = 0` `" "" "` (simplify in the brackets)

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

`x^3(x-1)^2(7x-4) = 0`

The roots of the equation and zeros of `f'` are: `0, 1, "and "4/7`

All three of those numbers are in the domain of `f` so they are all critical numbers for `f`.