# What are the critical points for f(x) = (x^2-10x)^4?

May 28, 2015

For critical numbers for $f$, we need the values of $x$ that are in the domain of $f$ and at which $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ does not exist.

For $f \left(x\right) = {\left({x}^{2} - 10 x\right)}^{4}$, we use the power rule and the chain rule to get:

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

This derivative always exists, so we only need the zeros.

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

$4 {\left({x}^{2} - 10 x\right)}^{3} \left(2 x - 10\right) = 4 {\left[x \left(x - 10\right)\right]}^{3} 2 \left(x - 5\right) = 8 {x}^{3} {\left(x - 10\right)}^{3} \left(x - 5\right)$

The zeros of $f '$ are: $0 , 5 , \mathmr{and} 10$

All three are in the domain of $f$, so the critical numbers are:

$0 , 5 , \mathmr{and} 10$

Alternative Terminology
I am used to calling these points on the line, "critical points". Alternative terminology may say that critical points, are points in the plane. In this terminology, we need to find the $y$ values. And the critical points will be:

$\left(0 , 0\right)$, $\left(5 , f \left(5\right)\right)$ and $\left(10 , f \left(10\right)\right)$