What is the critical f(x)=x^4-6x^2+9 ?

The critical f(x)=x^4-6x^2+9 is x=-sqrt(3),x=0,x=sqrt(3)

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critical f(x)=x^4-6x^2+9

critical points

f (x) = x^{4} - 6 x^{2} + 9

x = - 3, x = 0, x = 3

Solution steps

x^{4} - 6 x^{2} + 9

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0, x = 3

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 6 x^{2} + 9 : - \infty < x < \infty

All critical points are in domain

x = - 3, x = 0, x = 3

t^{\frac{3}{4}} - 2 t^{\frac{1}{4}}

f (x, y) = - 14 + 5 x^{2} + x y + y^{2}

y = x^{4} - 4 x^{3}

4 x y - x^{4} - 2 y^{2} + 2

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

What is the critical f(x)=x^4-6x^2+9 ?
The critical f(x)=x^4-6x^2+9 is x=-sqrt(3),x=0,x=sqrt(3)