What is the critical f(x)=-(x^3)/3+16x ?

The critical f(x)=-(x^3)/3+16x is x=-4,x=4

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critical f(x)=-(x^3)/3+16x

critical points

f (x) = - \frac{x ^{3}}{3} + 16 x

x = - 4, x = 4

Solution steps

- \frac{x ^{3}}{3} + 16 x

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

- \frac{x ^{3}}{3} + 16 x : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 4

f (x) = 4 - 2 2 - 4 x

f (x) = \frac{1}{5} cos (3 x)

x^{2} + x

2 x + 6

f (x) = lo g_{5} (x + 1)

What is the critical f(x)=-(x^3)/3+16x ?
The critical f(x)=-(x^3)/3+16x is x=-4,x=4