What is the critical f(x)= 1/(sqrt(4-x^2)) ?

The critical f(x)= 1/(sqrt(4-x^2)) is x=0

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critical f(x)= 1/(sqrt(4-x^2))

critical points

f (x) = \frac{1}{4 - x ^{2}}

x = 0

Solution steps

\frac{1}{4 - x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{4 - x ^{2}} : - 2 < x < 2

\frac{1}{4 - x ^{2}}

is not defined at

x = - 2, x = 2

, therefore:

x = 0

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

t^{3} - 9 x^{2} + 27 x

5 - x + 1

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

x^{2} + 4 x y + y^{2} - 40 x - 56 y + 1

What is the critical f(x)= 1/(sqrt(4-x^2)) ?
The critical f(x)= 1/(sqrt(4-x^2)) is x=0