find the extreme values of the function and where they occur y=1/(sqrt 1-x^2).

So far I have y'=x/((1-x^2)(1-x^2)^(1/2)) but I am stuck.

Look at that fraction. The domain is (-1,1). As long as x ≠ ±1, the denominator is not zero.

So, the fraction is zero when x=0. On either side of that, y increases, so (0,1) is a minimum.

To check, see the graph at

http://www.wolframalpha.com/input/?i=1%2F%E2%88%9A%281-x^2%29

Lots of times the derivative is a very messy fraction, but to find the extremes, all you need to work with is the numerator.

Thanks i got it I think I just hiccuped there

To find the extreme values of the function y = 1 / √(1 - x^2), you need to find the critical points by finding the derivative of the function and setting it equal to zero.

Let's start by finding the derivative of y with respect to x:

y = 1 / √(1 - x^2)

To simplify the derivative, we can rewrite the function as:

y = (1 - x^2)^(-1/2)

Using the chain rule, the derivative can be computed as:

y' = (-1/2)(1 - x^2)^(-3/2)(-2x)

Simplifying further, we have:

y' = x / [(1 - x^2)^(3/2)]

Now, to find the critical points, we will set the derivative equal to zero and solve for x:

x / [(1 - x^2)^(3/2)] = 0

Since we cannot divide by zero, the numerator should equal zero:

x = 0

Now, we need to check the endpoints of the domain, which is -1 ≤ x ≤ 1.

When x = -1:

y = 1 / √(1 - (-1)^2) = 1 / √2

When x = 1:

y = 1 / √(1 - (1)^2) = 1 / √0

At x = -1, the function has a finite value, but at x = 1, the function approaches infinity. This means that the function has no maximum value, but it has a minimum value.

Therefore, the extreme value of the function occurs at the minimum point when x = -1, where y = 1 / √2.

To find the extreme values of the function y = 1/(sqrt(1-x^2)), you have correctly taken the derivative, which is y' = x/((1-x^2)(1-x^2)^(1/2)).

To proceed, we need to find the critical points of the function. Critical points occur when the derivative is equal to zero or undefined. In this case, the derivative is never undefined, but it could be zero.

Setting y' equal to zero and solving for x:

x/((1-x^2)(1-x^2)^(1/2)) = 0

Since the numerator x cannot be equal to zero to make the entire expression zero, we focus on the denominator:

(1 - x^2)(1 - x^2)^(1/2) = 0

Here, we have two factors, (1 - x^2) and (1 - x^2)^(1/2). Setting each factor equal to zero and solving for x individually:

1 - x^2 = 0
(1 - x^2)^(1/2) = 0

For the first equation, we have:

x^2 = 1
x = ± 1

For the second equation, we have:

(1 - x^2)^(1/2) = 0
1 - x^2 = 0
x^2 = 1
x = ± 1

So, we have three possible critical points: x = -1, x = 0, and x = 1.

Now, we need to determine which of these values produce extreme values. To do that, we can evaluate y at each critical point and also at the endpoints of the function's domain.

The given function is y = 1/(sqrt(1-x^2)). The domain of the function is -1 ≤ x ≤ 1, as the denominator cannot be zero.

Evaluating the function at -1, 0, 1, and the endpoints (-1 and 1):

y(-1) = 1/(sqrt(1-(-1)^2)) = 1/(sqrt(1-1)) = 1/(sqrt(0)) = undefined
y(0) = 1/(sqrt(1-0^2)) = 1/(sqrt(1)) = 1
y(1) = 1/(sqrt(1-1^2)) = 1/(sqrt(1-1)) = 1/(sqrt(0)) = undefined
y(-1) is undefined, but since it is not in the domain, it can be ignored.

Comparing y(0) and y(1), we find that they are both equal to 1, which means they are candidates for the maximum or minimum values.

Therefore, the extreme values of the function y = 1/(sqrt(1-x^2)) occur at x = 0 and x = 1, where y is equal to 1.