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

Aug 20, 2015

The critical numbers are $0 , 1 , \text{and } \frac{4}{7}$

#### Explanation:

$f \left(x\right) = {x}^{4} {\left(x - 1\right)}^{3}$

$f ' \left(x\right) = 4 {x}^{3} {\left(x - 1\right)}^{3} + {x}^{4} 3 {\left(x - 1\right)}^{2}$

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

$4 {x}^{3} {\left(x - 1\right)}^{3} + 3 {x}^{4} {\left(x - 1\right)}^{2} = 0$

This is the sum of two terms:

$\underbrace{4 {x}^{3} {\left(x - 1\right)}^{3}} + \underbrace{3 {x}^{4} {\left(x - 1\right)}^{2}} = 0$

These terms have common factors of ${x}^{3}$ and ${\left(x - 1\right)}^{2}$, so remove those:

${x}^{3} {\left(x - 1\right)}^{2} \left[4 \left(x - 1\right) + 3 x\right] = 0$ $\text{ "" }$ (simplify in the brackets)

${x}^{3} {\left(x - 1\right)}^{2} \left[4 x - 4 + 3 x\right] = 0$

${x}^{3} {\left(x - 1\right)}^{2} \left(7 x - 4\right) = 0$

The roots of the equation and zeros of $f '$ are: $0 , 1 , \text{and } \frac{4}{7}$

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