# What are the local extrema, if any, of #f (x) =sqrt(4-x^2)#?

The extrema of f(x) is:

- Max of 2 at x = 0
- Min of 0 at x = 2, -2

To find the extrema of any function, you carry out the following:

- Differentiate the function
- Set the derivative equal to 0
- Solve for the unknown variable
- Substitute the solutions into f(x) (NOT the derivative)

- Differentiate the function:

By Chain Rule**:

Simplifying:

- Set the derivative equal to 0:

Now, since this is a product, you can set each part equal to 0 and solve:

- Solve for the unknown variable:

Now you can see that x = 0, and to solve the right side, raise both sides to the -2 to cancel out the exponent:

- Substitute the solutions into f(x):

I'm not going to write out the full solution for the substitution as it is straightforward, but I'll tell you:

Thus, you can see that there is an absolute maximum of 2 at x = 0, and an absolute minimum of 0 at x = -2, 2.

Hopefully everything was clear and concise! Hope I could help! :)

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To find the local extrema of ( f(x) = \sqrt{4 - x^2} ), we first need to find its critical points by taking the derivative and setting it equal to zero.

First, find the derivative of ( f(x) ): [ f'(x) = \frac{d}{dx}\sqrt{4 - x^2} = \frac{-x}{\sqrt{4 - x^2}} ]

Now, set the derivative equal to zero and solve for ( x ): [ \frac{-x}{\sqrt{4 - x^2}} = 0 ]

This equation is satisfied when ( x = 0 ).

Next, we need to check the endpoints of the domain of ( f(x) ), which is ([-2, 2]).

When ( x = -2 ), ( f(-2) = 0 ).

When ( x = 2 ), ( f(2) = 0 ).

So, the critical point is ( x = 0 ), and the function has local extrema at the endpoints ( x = -2 ) and ( x = 2 ).

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