What is the critical f(x)=(2-x^2)/(3x^2-1) ?

The critical f(x)=(2-x^2)/(3x^2-1) is x=0

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critical f(x)=(2-x^2)/(3x^2-1)

critical points

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

x = 0

Solution steps

\frac{2 - x ^{2}}{3 x ^{2} - 1}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - \frac{1}{3}, x = \frac{1}{3}

Identify critical points not in

f (x)

domain

Domain of

\frac{2 - x ^{2}}{3 x ^{2} - 1} : x < - \frac{1}{3} or - \frac{1}{3} < x < \frac{1}{3} or x > \frac{1}{3}

\frac{2 - x ^{2}}{3 x ^{2} - 1}

is not defined at

x = - \frac{1}{3}, x = \frac{1}{3}

, therefore:

x = 0

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

f (x, y) = x y + x^{4} + y^{4}

f (x, y) = e^{\frac{( x ^{8} + y ^{8} )}{8}}

\frac{x ^{3} - 4 x}{x ^{3} - x}

f (x) = 14 cos (θ) + 7 sin^{2} (θ)

What is the critical f(x)=(2-x^2)/(3x^2-1) ?
The critical f(x)=(2-x^2)/(3x^2-1) is x=0