Find all the critical numbers for the following function.

F(x) = (16 - x^2)^(3/5)

f' = (3/5)(16-x^2)^(-2/5)(-2x)

That is undefined when x = 4 or x = -4 and is zero when x = 0

f'(x) = (3/5)(16 - x^2)^(-2/5) (-2x)

= 0 for max/min of f(x)
-2x = 0
x = 0 , then y = 16^(3/5)

find f''(x) using the product rule
set that equal to zero to find any points of inflection

x-intercepts, let y = 0
(16-x^2)^(3/5) = 0
x^2 = 16
x = Β± 4

y-intercept, let x = 0
y = 16^(3/5)

To find the critical numbers of a function, we need to first find the derivative of the function and then solve for the values of x where the derivative is equal to zero or does not exist. In this case, we need to find the derivative of the function F(x) = (16 - x^2)^(3/5).

Step 1: Find the derivative of F(x):
To find the derivative of F(x), we can use the chain rule. Let's start by rewriting the function as a power:
F(x) = (16 - x^2)^(3/5) = 16^(3/5) - x^(6/5)
Now let's find the derivative of the second term using the power rule:
d/dx (x^(6/5)) = (6/5)x^(1/5)

Step 2: Set the derivative equal to zero and solve for x:
(6/5)x^(1/5) = 0
Since a fraction is equal to zero only when the numerator is equal to zero, we have:
6/5 = 0 (This equation has no solution)

Step 3: Determine if there are values of x where the derivative does not exist:
The derivative of F(x) exists for all values of x because it is a polynomial function.

Therefore, the function F(x) = (16 - x^2)^(3/5) does not have any critical numbers because the derivative is never equal to zero or undefined.