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

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

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

critical points

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

x = - \frac{3}{2}, x = 0, x = \frac{3}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{3}{2}, x = 0, x = \frac{3}{2}

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - \frac{3}{2}, x = 0, x = \frac{3}{2}

\frac{x}{x + 5}

- 8 x + y - 8 = 0, (2, 5)

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

2 x^{2} - 12 x + 20

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

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