What is the critical f(x)=-(x^3)/3+64x ?

The critical f(x)=-(x^3)/3+64x is x=-8,x=8

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critical f(x)=-(x^3)/3+64x

critical points

f (x) = - \frac{x ^{3}}{3} + 64 x

x = - 8, x = 8

Solution steps

- \frac{x ^{3}}{3} + 64 x

Find where

f^{'} (x)

is equal to zero or undefined

x = - 8, x = 8

Identify critical points not in

f (x)

domain

Domain of

- \frac{x ^{3}}{3} + 64 x : - \infty < x < \infty

All critical points are in domain

x = - 8, x = 8

cos (cos (x))

f (x) = x^{2} (x^{2} + 9) (x^{3} + 2 x)

f (x) = (\frac{2}{3})^{x}

57 + 30 cos (\frac{π}{6} t)

- (7)^{x}

What is the critical f(x)=-(x^3)/3+64x ?
The critical f(x)=-(x^3)/3+64x is x=-8,x=8