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

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

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

critical points

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

x = 0

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{1 + x ^{2}} - 3 x^{2} : - \infty < x < \infty

All critical points are in domain

x = 0

y = \frac{( x ^{2} )}{x ^{2} - 16}

f (x) = \frac{2 x + 3}{5 x + 4}

\frac{7}{x + 3}

y = \frac{4 x ^{2}}{x ^{2} - 2 x - 3}

x^{3} - 12 x + 3

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