What is the critical f(x)=3x^4-4x^3-12x^2+6 ?

The critical f(x)=3x^4-4x^3-12x^2+6 is x=-1,x=0,x=2

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critical f(x)=3x^4-4x^3-12x^2+6

critical points

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 6

x = - 1, x = 0, x = 2

Solution steps

3 x^{4} - 4 x^{3} - 12 x^{2} + 6

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 0, x = 2

Identify critical points not in

f (x)

domain

Domain of

3 x^{4} - 4 x^{3} - 12 x^{2} + 6 : - \infty < x < \infty

All critical points are in domain

x = - 1, x = 0, x = 2

f (x) = \frac{1}{2} x^{2} - 2 x^{2} + 3 x + 5

f (x) = \frac{4}{x ^{2} - 16}

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

f (x) = x (16 - x)^{3}

f (x, y) = x^{2} - 3 x y + 6 y^{2} + 5 x - 2 y + 8

What is the critical f(x)=3x^4-4x^3-12x^2+6 ?
The critical f(x)=3x^4-4x^3-12x^2+6 is x=-1,x=0,x=2