# How do use the first derivative test to determine the local extrema f(x)= -x^3 + 12x?

Refer to explanation

#### Explanation:

To find the local extrema we need the zeroes of the derivative of f

Hence we have that

$f ' \left(x\right) = 0 \implies - 3 {x}^{2} + 12 = 0 \implies - {x}^{2} + 4 = 0 \implies x = \pm 2$

Now we see how f'(x) behaves around the zeroes

graph{-x^2+4 [-10, 10, -5, 5]}

We see at x=-2 f'(x)>0 and at x=2 f'(x)<0

Hence at x=2 local maxima f(2)=16 and at x=-2 local minima f(-2)=-16

graph{-x^3+12x [-40, 40, -20, 20]}