find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer…

Explanation:

Step1: Find the derivative using the product rule

The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u = x^{4/5}) and (v=(x - 3)^2).

First, find (u^\prime): Using the power rule ((x^n)^\prime=nx^{n - 1}), (u^\prime=\frac{4}{5}x^{-1/5}).

Second, find (v^\prime): Using the chain rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)), where (f(u)=u^2) and (u = x - 3). So (v^\prime = 2(x - 3)\cdot1=2(x - 3)).

Then (F^\prime(x)=\frac{4}{5}x^{-1/5}(x - 3)^2+x^{4/5}\cdot2(x - 3)).

Step2: Simplify the derivative

Factor out (\frac{2}{5}x^{-1/5}(x - 3)):

[ \begin{align*} F^\prime(x)&=\frac{2}{5}x^{-1/5}(x - 3)[2(x - 3)+5x]\ &=\frac{2}{5}x^{-1/5}(x - 3)(2x-6 + 5x)\ &=\frac{2}{5}x^{-1/5}(x - 3)(7x-6) \end{align*} ]

Step3: Set the derivative equal to zero

(F^\prime(x)=0) when:

• (x^{-1/5}=0) (no solution since (x^{-1/5}=\frac{1}{x^{1/5}}), and (\frac{1}{x^{1/5}} = 0) has no solution for (x\in R))

• (x - 3=0), then (x = 3)

• (7x-6=0), then (x=\frac{6}{7})

Also, (F^\prime(x)) is undefined when (x = 0) (because of the (x^{-1/5}) term, (x^{1/5}=0) when (x = 0))

Answer:

(0,\frac{6}{7},3)