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

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

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critical f(x)=4x^3-3x

critical points

f (x) = 4 x^{3} - 3 x

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

Solution steps

4 x^{3} - 3 x

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

4 x^{3} - 3 x : - \infty < x < \infty

All critical points are in domain

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

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

x - 2

f (x) = x - \frac{4}{x ^{2}}

4 x + 3 y = - 2

f (x) = ∣ x ∣ - 2