Final answer:
The critical numbers of the function F(x) are found by setting the derivative of the function to zero and solving for x. After applying the product rule and simplifying, we find that x = 0 and x = 6 are the critical numbers, since these make the derivative zero.
Step-by-step explanation:
To find the critical numbers of the function F(x) = x4/5(x-6)2, we need to find where the derivative of the function equals zero or does not exist. First, let's find the derivative of F(x).
Let u = x4/5 and v = (x-6)2. Then we can use the product rule to differentiate F(x): F'(x) = u'v + uv'.
Calculating the derivatives, u' = (4/5)x-1/5 and v' = 2(x-6). Plugging these into the product rule formula, we get:
F'(x) = (4/5)x-1/5(x-6)2 + x4/52(x-6)
Now, we need to set F'(x) = 0 and solve for x. After simplification, we will get:
(4/5)x-1/5(x-6)2 + 2x4/5(x-6) = 0
We can see that this equation equals zero when x = 0 or x = 6 because the first term contains x raised to a power, which makes the term zero at x = 0; and the second term contains (x-6), which becomes zero when x = 6. Additionally, we need to check if the derivative does not exist; this happens when x4/5 is undefined, which is never the case for real numbers as the function is defined for all x >= 0. Thus, the critical numbers are 0 and 6.