How do you find the local extrema for #f(x)= x^4-4x³#?

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Answer 1 · 2016-06-28T20:36:02

There is a minimum for #x=3#.

First of all we search the zeros of the derivative.

#f'(x)=d/dx(x^4-4x^3)=4x^3-12x^2#

#f'(x)=0->4x^3-12x^2=0#

One solution is of course #x=0# the second is obtained dividing by #x^2#

#4x-12=0->x=3#.

To see which kind of points are #x=0# and #x=3# we can study the second derivative

#f''(x)=12x^2-24x#

and we have

#f''(0)=0#, #f(3)=36#.

Then when #x=0# the point is an horizontal flex (otherwise known as an inflection point), while when #x=3# it is a minimum.

We can see this also from the plot.

graph{x^4-4x^3 [-52.9, 61.54, -29.3, 27.9]}