What is the critical y=sqrt(4-x^2) ?

The critical y=sqrt(4-x^2) is x=-2,x=0,x=2

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critical y=sqrt(4-x^2)

critical points

y = 4 - x^{2}

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

Solution steps

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

4 - x^{2} : - 2 \leq x \leq 2

All critical points are in domain

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

f (x) = \frac{( x + 4 )}{( 2 x - 7 )}

\frac{x}{3 x + 6}

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

x = 3

f (x) = \frac{( x + 5 )}{( x + 7 )}