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

1 Answer
Dec 4, 2016

Turning points at #x = 8/3 and x = 4#.
Flat, at x = Look at the two graphs.

Explanation:

It is easy to show that

#f' = 0, x = 0, 8/3 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 #(-epsilon, epsilon)#

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]}