# How do you find the critical points for f(x) = (x^2-10x)^4 and the local max and min?

Jul 3, 2016

three critical points: $x = 0 , 5 , 10$
Relative minimum at $x = 0 , 10$
Relative maximum at $x = 5$

#### Explanation:

First, we need to find $f ' \left(x\right)$.
Using rules for taking derivative, we get following:
$f ' \left(x\right) = 4 \cdot {\left({x}^{2} - 10 x\right)}^{3} \cdot \left(2 x - 10\right)$

Critical points occur when the slope is zero meaning $f ' \left(x\right) = 0$.

$f ' \left(x\right) = 4 \cdot {\left({x}^{2} - 10 x\right)}^{3} \cdot \left(2 x - 10\right) = 0$

$\left(2 x - 10\right) = 0$ => $x = 5$
$\left({x}^{2} - 10 x\right) = 0$ => $x = 0 , 10$

So, three critical points would be $x = 0 , 5 , 10$.

To find relative maximum and minimum, we can do a sign test.

For $x \in$(-∞, 0) => $f ' \left(x\right)$ is negative.
For $x \in$(0, 5) => $f ' \left(x\right)$ is positive.
For $x \in$(5, 10) => $f ' \left(x\right)$ is negative.
For $x \in$(10, ∞) => $f ' \left(x\right)$ is positive.

At $x = 0$, function has relative minimum because $f ' \left(x\right)$ changes signs from negative to positive.

At $x = 5$, function has relative maximum because $f ' \left(x\right)$ changes signs from positive to negative.

At $x = 10$, function has relative minimum because $f ' \left(x\right)$ changes signs from negative to positive.