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]}

How do you find the local extrema for f(x)= x^4-4x³f(x)= x^4-4x³?