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

The critical f(x)=4x^3-3x^2+8 is x=0,x= 1/2

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critical f(x)=4x^3-3x^2+8

critical points

f (x) = 4 x^{3} - 3 x^{2} + 8

x = 0, x = \frac{1}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{1}{2}

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 0, x = \frac{1}{2}

f (x) = \frac{( x - 9 )}{( x - 2 )}

\frac{2}{x} + 4

5 x + 3 y = 2

\frac{x + 5}{x ^{2} - 25}

f (x) = x^{2} - 2 x + 2

What is the critical f(x)=4x^3-3x^2+8 ?
The critical f(x)=4x^3-3x^2+8 is x=0,x= 1/2