# What are the critical points of f(x) = x(x + 1)^3?

Jan 23, 2018

The critical points are $= \left(- 1 , 0\right)$ and $= \left(- \frac{1}{4} , - \frac{27}{256}\right)$

#### Explanation:

Calculate the first derivative and determine the critical points $f ' \left(x\right) = 0$

Here,

$f \left(x\right) = x {\left(x + 1\right)}^{3}$

Apply the product rule for differentiation

$\left(u v\right) ' = u ' v + u v '$

$u = x$, $\implies$, $u ' = 1$

$v = {\left(x + 1\right)}^{3}$, $\implies$, $v ' = 3 {\left(x + 1\right)}^{2}$

$f ' \left(x\right) = 1 \cdot {\left(x + 1\right)}^{3} + 3 x {\left(x + 1\right)}^{2}$

$= {\left(x + 1\right)}^{2} \left(x + 1 + 3 x\right)$

$= {\left(x + 1\right)}^{2} \left(4 x + 1\right)$

$f ' \left(x\right) = 0$, $\implies$

The critical points are when $x = - 1$ and $x = - \frac{1}{4}$

graph{x(x+1)^3 [-1.994, 1.044, -0.595, 0.924]}