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

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

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

critical points

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

y = - 2, y = 0, y = 2

Solution steps

3 y^{4} - 12 y^{2}

Find where

f^{'} (y)

is equal to zero or undefined

y = - 2, y = 0, y = 2

Identify critical points not in

f (y)

domain

Domain of

3 y^{4} - 12 y^{2} : - \infty < y < \infty

All critical points are in domain

y = - 2, y = 0, y = 2

f (x) = \frac{7 y + 637}{7} - 3

f (x) = (0.33)^{x}

f (x) = e^{2 x} + 1

f (x) = \frac{3 x + 7}{6 x}

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

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