Identify the open intervals) where the function f of F(x)= x√10-x^2 is increasing.

I assume you meant

f(x) = x √(10 - x^2) and not the way you typed it

Of course the function would only be defined for

-√10 ≤ x ≤ √10

f' (x) = x (1/2)(10-x^2)^(-1/2) (-2x ) + (10-x^2)^(1/2)
= (10 - x^2)^(-1/2) [ -x^2 + 10-x^2]
= (10-2x^2)/√(10-x^2)

now the bottom is positive for our domain of x
so let's look only at the numerator

to be increasing the F' (x) has to be positive
so when is 10 - 2x^2 > 0

- 2x^2 > -10
2x^2 < 10
± x < √5
x < √5 and x ≥ -√5

the function increases for -√5 < x ≤ √5

confirmation:

http://www.wolframalpha.com/input/?i=plot+y+%3D+x√%2810-x%5E2%29

To identify the open intervals where the function f(x) = x√(10 - x^2) is increasing, we need to find the values of x for which the derivative of f(x) is positive.

Step 1: Find the derivative of f(x)
To find the derivative, we can use the product rule for differentiation. Let's call the square root part g(x) = √(10 - x^2). The derivative of f(x) can be calculated as:

f'(x) = x * g'(x) + g(x) * 1

The derivative of g(x) can be found using the chain rule:

g'(x) = -x / √(10 - x^2)

Substituting g(x) and g'(x) in the equation for f'(x), we get:

f'(x) = x * (-x / √(10 - x^2)) + √(10 - x^2)

Simplifying this equation, we have:

f'(x) = (-x^2) / √(10 - x^2) + √(10 - x^2)

Now we have the derivative of f(x), denoted as f'(x).

Step 2: Determine the values of x where f'(x) > 0
To find the intervals where f(x) is increasing, we need to determine the values of x for which f'(x) is greater than zero: f'(x) > 0.

Let's solve for the numerator first:

(-x^2) / √(10 - x^2) > 0

Since the denominator is always positive, the inequality will hold true if and only if the numerator is less than zero:

-x^2 < 0

Multiplying by -1 and reversing the direction of the inequality, we have:

x^2 > 0

This inequality is always true, except when x = 0.

So, we need to determine the sign of the denominator, which is given by √(10 - x^2).

For the term √(10 - x^2) to be defined, we must have:

10 - x^2 > 0

Rearranging the inequality, we get:

x^2 < 10

Taking the square root of both sides, we obtain:

-√10 < x < √10

Therefore, the open interval where f(x) is increasing is (-√10, √10), excluding x = 0.

Hence, the answer is (-√10, √10).