# What are the values and types of the critical points, if any, of f(x) = (x^4/5)(x-4)^2?

Dec 4, 2016

Turning points at $x = \frac{8}{3} \mathmr{and} x = 4$.
Flat, at x = Look at the two graphs.

#### Explanation:

It is easy to show that

$f ' = 0 , x = 0 , \frac{8}{3} \mathmr{and} 4$.

Further, f'' is 0 only at x = 0. Yet, f''' = 0 here but

the next ( even order ) f'''' is not.

So, origin is not a point of inflexion, and so, at this turning point,

the curve is flat, in the $\epsilon -$neighborhood $\left(- \epsilon , \epsilon\right)$

This flatness is zoomed in the second graph.

graph{5y-x^4(x-4)^2=0 [-40, 40, -20, 20]}

graph{5y-x^4(x-4)^2=0 [-5, 5, -2.5, 2.5]}