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

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Answer 1 · 2015-08-20T20:26:15

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

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

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

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 $f$ .

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