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

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 $f \left(x\right) = \sqrt{4 - {x}^{2}}$:

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

1) Differentiate the function:

By Chain Rule**:

$f ' \left(x\right) = \frac{1}{2} {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 x\right)$

Simplifying:

$f ' \left(x\right) = - x {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}$

2) Set the derivative equal to 0:

$0 = - x {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}$

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

3) Solve for the unknown variable:

$0 = - x$ and $0 = {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}$

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

${0}^{-} 2 = {\left({\left(4 - {x}^{2}\right)}^{- \frac{1}{2}}\right)}^{- 2}$

$0 = 4 - {x}^{2}$
$0 = \left(2 - x\right) \left(2 + x\right)$
$x = - 2 , 2$

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:

$f \left(0\right) = 2$
$f \left(- 2\right) = 0$
$f \left(2\right) = 0$

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! :)