# How do you find the relative extrema of the function f (x)=x^4+4x^3+4x^2?

Aug 6, 2015

You set the derivative to zero:

#### Explanation:

$f ' \left(x\right) = 4 \cdot {x}^{3} + 3 \cdot 4 {x}^{2} + 2 \cdot 4 x = 0$

We can factorise:
$f ' \left(x\right) = 4 x \cdot \left({x}^{2} + 3 x + 2\right) = 4 x \left(x + 1\right) \left(x + 2\right) = 0$
$\to x = 0 \mathmr{and} x = - 1 \mathmr{and} x = - 2$

Fill these in into the original function, and you'll get the coordinates:
graph{x^4+4x^3+4x^2 [-3.442, 2.717, -0.573, 2.503]}