What are the local extrema, if any, of $f (x) = x^{2} (x + 2)$ $f(x) =x^2(x+2)$ ?

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Answer 1 · 2017-11-07T16:58:55

$x = 0, - \frac{4}{3}$ $x=0, -4/3$

Find the derivative of $f (x) = x^{2} (x + 2)$ $f(x)=x^2(x+2)$ .
You will have to use the product rule.

$f'(x)=x^2+(x+2)2x=x^2+2x^2+4x=3x^2+4x$
$f'(x)=x(3x+4)$

Set $f'(x)$ equal to zero to find the critical points.

$x=0$
$3x+4=0 rarr x=-4/3$

$f(x)$ has local extrema at $x=0, -4/3$ .
OR
$f(x)$ has local extrema at the points (0, 0) and ( $-4/3$ , $32/27$ ).

What are the local extrema, if any, of f(x) =x^2(x+2) f(x)=x2(x+2)f(x) =x^2(x+2) ?