help!!1 plzz math -calc?

Question

identify the critical point and determine the local extreme values.
y=xroot4-x^2

THank you, thank YOu, thank You

Answer 1

y = x√4 - x² = 2x - x²

critical point:
y' = 2 - 2x = 0;
x = 1 -- critical point and local extreme because it's only one.

If you mean y = x√(4 - x²),
then:
y' = x / [2√(4 - x²)] + √(4 - x²) = 0;
x = -2(4 - x²);
2x² - x - 8 = 0;

Solve for x, check for extremes. Done.

WTF: people below are copying/pasting my answer :o

Answer 2

f(x) = x [sqrt(4 - x^2)]

To identify the critical points, we have to take the derivative, and make it 0. The critical points are defined to be where f(x) = 0 or where f(x) is undefined.

f'(x) = sqrt(4 - x^2) + x [ 1/ [2sqrt(4 - x^2)] ] [-2x]

Cancelling some terms,

f'(x) = sqrt(4 - x^2) + x [ -x / [sqrt(4 - x^2)] ]

Simplifying some more,

f'(x) = sqrt(4 - x^2) - (x^2)/[sqrt(4 - x^2)]

And putting the two terms under a common denominator,

f'(x) = [(4 - x^2) - x^2] / [sqrt(4 - x^2)]

Simplifying

f'(x) = [4 - 2x^2] / [sqrt (4 - x^2)]

Making f'(x) = 0,

0 = [4 - 2x^2] / [sqrt (4 - x^2)]

Therefore, we make the numerator equal to 0 to get some critical values.

4 - 2x^2 = 0
-2x^2 = -4
x^2 = 2, therefore x = +/- sqrt(2).

We also need to determine what makes f'(x) undefined. f'(x) is undefined if the denominator is equal to 0.

sqrt (4 - x^2) = 0
4 - x^2 = 0
x^2 = 4, x = +/- 2

Therefore, our critical numbers are
x = {-2, -sqrt(2), sqrt(2), 2}

To determine our intervals of increase/decrease, we test a single point between each critical number. If it is positive, it is increasing on that interval; if it's negative, it is decreasing.

Without showing you the details, the intervals of increase/decrease are as follows:

Decreasing on (-2, -sqrt(2)] U (sqrt(2), 2)
Increasing on [-sqrt(2), sqrt(2)]

This means we have a local minimum at x = -sqrt(2) and a local maximum at sqrt(2).

f(-sqrt(2)) = -sqrt(2) sqrt(4 - 2) = -2
f(sqrt(2)) = sqrt(2) sqrt(4 - 2) = 2

We have a local minimum at (-sqrt(2), -2).
Local maximum at (sqrt(2), 2)

Answer 3

y = x√4 - x² = 2x - x²

critical point:
y' = 2 - 2x = 0;
x = 1 -- critical point and local extreme because it's only one.




∫½√²×÷πΔφ

Answer 4

I don't know