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

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

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

critical points

y = \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

x 7 - x

f (x) = 2 x^{3} - 9 x^{2} + 2

f (x) = 3 y^{3} - x^{2} \cdot y^{2} + 8 y^{2} + 4 x^{2} - 20 y

f (x) = 2 x + \frac{300000}{x}

f (x) = - 2 x^{2} - y^{2} + 8 x + 10 y - 5 x y

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