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

Jun 28, 2016

There is a minimum for $x = 3$.

#### Explanation:

First of all we search the zeros of the derivative.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{4} - 4 {x}^{3}\right) = 4 {x}^{3} - 12 {x}^{2}$

$f ' \left(x\right) = 0 \to 4 {x}^{3} - 12 {x}^{2} = 0$

One solution is of course $x = 0$ the second is obtained dividing by ${x}^{2}$

$4 x - 12 = 0 \to x = 3$.

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

$f ' ' \left(x\right) = 12 {x}^{2} - 24 x$

and we have

$f ' ' \left(0\right) = 0$, $f \left(3\right) = 36$.

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]}