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

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Answer 1 · 2018-01-23T15:27:12

The critical points are $=(-1,0)$ and $=(-1/4, -27/256)$

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

Here,

$f(x)=x(x+1)^3$

Apply the product rule for differentiation

$(uv)'=u'v+uv'$

$u=x$ , $=>$ , $u'=1$

$v=(x+1)^3$ , $=>$ , $v'=3(x+1)^2$

$f'(x)=1*(x+1)^3+3x(x+1)^2$

$=(x+1)^2(x+1+3x)$

$=(x+1)^2(4x+1)$

$f'(x)=0$ , $=>$

The critical points are when $x=-1$ and $x=-1/4$

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

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