# What are the critical points of g(x)=x^(2/5)(x-10)?

Oct 29, 2015

$0$ and $\frac{20}{7}$

#### Explanation:

$g \left(x\right) = {x}^{\frac{2}{5}} \left(x - 10\right)$

Note first that "Dom(g) = (-oo,oo) so every point at which $f ' \left(x\right)$ is undefined or $0$ is a critical number.

Let's find $g ' \left(x\right)$ using the product rule:

$g ' \left(x\right) = \frac{2}{5} {x}^{- \frac{3}{5}} \left(x - 10\right) + {x}^{\frac{2}{5}} \left(1\right)$.

Let's now factor the denominators out of the expression:

$g ' \left(x\right) = \frac{1}{5} {x}^{- \frac{3}{5}} \left[2 \left(x - 10\right) + 5 x\right]$.

We'll simplify the brackets and write the denominator as a denominator:

g'(x) = (7x-20)/(5 x^(3/5).

$g ' \left(x\right)$ is not defined for $x = 0$ and is $0$ at $x = \frac{20}{7}$.

Both are in $\text{Dom} \left(g\right)$, so both are critical numbers for $g$.