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

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Answer 1 · 2015-08-06T11:13:21

You set the derivative to zero:

$f'(x)=4*x^3+3*4x^2+2*4x=0$

We can factorise:
$f'(x)=4x*(x^2+3x+2)=4x(x+1)(x+2)=0$
$->x=0orx=-1orx=-2$

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]}

How do you find the relative extrema of the function f (x)=x^4+4x^3+4x^2f(x)=x4+4x3+4x2f (x)=x^4+4x^3+4x^2?