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

##### 1 Answer

#### Answer:

The extrema of f(x) is:

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

#### Explanation:

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

1) **Differentiate the function**

2) **Set the derivative equal to 0**

3) **Solve for the unknown variable**

4) **Substitute the solutions into f(x) (NOT the derivative)**

In your example of

1) **Differentiate the function**:

** By **Chain Rule**:

Simplifying:

2) **Set the derivative equal to 0:**

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

3) **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:

4) **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! :)