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
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