How do you find the maximum, minimum and inflection points and concavity for the function $y=1/5(x^4-4x^3)$ ?

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Answer 1 · 2016-11-05T15:04:51

Inflection point at x=0, Minima at x=3
concave up in $(-oo,0)$ and $(3, oo)$
concave down in (0,3)

First get critical points by making $dy/dx=0$

$dy/dx= 1/5( 4x^3 -12x^2)$

= $4/5 x^2(x-3)$ =0 gives x=0, 3

Now using second derivative test, $(d^2y)/dx^2= 1/5(12x^2 -24x)$

At x=0, $(d^2y)/dx^2$ =0 hence it is a inflection point.

At x=3, $(d^2y)/dx^2= 1/5(108-72)=36/5$ which is Positive. Hence there is a minima at x=3

For concavity second derivative test can be used as shown below
enter image source here

How do you find the maximum, minimum and inflection points and concavity for the function y=1/5(x^4-4x^3)y=1/5(x^4-4x^3)?