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

Jul 6, 2015

Critical numbers for function$f$ are numbers in the domain of $f$ at which $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ does not exist. For this function the critical numbers are $0$ and $\frac{20}{7}$.

#### Explanation:

For $f \left(x\right) = {x}^{\frac{2}{5}} \left(x - 10\right)$, I prefer to rewrite the function before differentiating.

$f \left(x\right) = {x}^{\frac{7}{5}} - 10 {x}^{\frac{2}{5}}$

So,
$f ' \left(x\right) = \frac{7}{5} {x}^{\frac{2}{5}} - 10 \cdot \frac{2}{5} {x}^{- \frac{3}{5}}$

$= \frac{7 {x}^{\frac{2}{5}}}{5} - \frac{4}{x} ^ \left(\frac{3}{5}\right)$

$= \frac{7 x - 20}{5 {x}^{\frac{3}{5}}}$

$f ' \left(x\right) = 0$ at $x = \frac{20}{7}$ which is in the domain of $f$/

$f ' \left(x\right)$ does not exist at $5 {x}^{\frac{3}{5}} = 0$ which happens at $x = 0$. Note that $0$ is in the domain of the original function, $f$, so it is a critical number for $f$.

The critical numbers for $f$ are: $0$ and $\frac{20}{7}$.

f"(x) changes from positive to negative as we move past $0$, fo $f \left(0\right) = 0$ is a local maximum value of $f$
$f ' \left(x\right)$ changes from negative to positive as we move past $\frac{20}{7}$, so $f \left(\frac{20}{7}\right)$ is a local minimum value of $f$.
It was not asked, but here is the graph of the function: $f \left(x\right) = {x}^{\frac{2}{5}} \left(x - 10\right)$,