What is the critical f(x)=2x^3-12x^2+8 ?

The critical f(x)=2x^3-12x^2+8 is x=0,x=4

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critical f(x)=2x^3-12x^2+8

critical points

f (x) = 2 x^{3} - 12 x^{2} + 8

x = 0, x = 4

Solution steps

2 x^{3} - 12 x^{2} + 8

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 4

Identify critical points not in

f (x)

domain

Domain of

2 x^{3} - 12 x^{2} + 8 : - \infty < x < \infty

All critical points are in domain

x = 0, x = 4

f (x) = 4^{x - 5}

y = \frac{1}{5} x + 2

14

f (x) = cot (3 x)

4 - x

What is the critical f(x)=2x^3-12x^2+8 ?
The critical f(x)=2x^3-12x^2+8 is x=0,x=4