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

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

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

critical points

f (x) = 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{5 x + 4}{2 x ^{2} - 4 x - 16}

5 x^{2} - 2 x + 4

f (x) = x^{3} - 3 x^{2} + 2 x + 5

y = - 2 x^{2} + 4 x - 3

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

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