Consider the function $$ f(x)=8(x-5)^{2 / 3} $$. For this function there are two important intervals: $$ (-\infty, A) $$ and $$ (A, \infty) $$ where $$ A $$ is a critical number. $$ A $$ is

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1. The function is given by
   $$
   f(x)=8(x-5)^{\frac{2}{3}}.
   $$

2. To find the critical number, we first compute the derivative $$ f'(x) $$. Using the power and chain rules,
   $$
   f'(x)=8\cdot \frac{2}{3}(x-5)^{\frac{2}{3}-1} = \frac{16}{3}(x-5)^{-\frac{1}{3}}.
   $$

3. The derivative $$ f'(x) $$ is undefined when the expression $$ (x-5)^{-\frac{1}{3}} $$ is undefined, which occurs at
   $$
   x-5=0 \quad \Rightarrow \quad x=5.
   $$

4. Since the derivative does not exist at $$ x=5 $$ and the function is defined at $$ x=5 $$, this point is a critical number.

5. Therefore, the critical number is
   $$
   A=5.
   $$
$$ A = 5 $$

Consider the function . For this
function there are two important intervals:
and where is a critical number.
is

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Consider the function . For this function there are two important intervals: and where is a critical number. is