Summarize the pertinent information obtained by applying the $$ f(x)=(x-4)\left(x^{2}-8 x-32\right) $$ Select the correct choice below and, if necessary, fill in the a A. The function $$ f $$ has a local maximurn at $$ x-\square $$. (Use a comma to separate answers as needed.) B. The function $$ f $$ has no local maximum. Select the cortect choice below and, if necessary, fill in the A. The lunction $$ f $$ has a local mirimum at $$ x=\square $$. (Use a comma to separale answers as needed.) B. The function $$ f $$ has no local minimum

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We begin with the function
$$
f(x)=(x-4)\left(x^{2}-8x-32\right).
$$
It is easier to work with the fully expanded form. Expanding, we have
$$
\begin{aligned}
f(x) &= x\left(x^{2}-8x-32\right) - 4\left(x^{2}-8x-32\right) \
&= x^3 - 8x^2 -32x - 4x^2 +32x +128 \
&= x^3 - 12x^2 + 128.
\end{aligned}
$$

Next, we compute the first derivative to find potential local extrema:
$$
f'(x) = \frac{d}{dx}\left(x^3 - 12x^2 + 128\right) = 3x^2 - 24x.
$$
Factor the derivative:
$$
f'(x)= 3x(x-8).
$$
Setting $$ f'(x)=0 $$ gives the critical points:
$$
3x(x-8)=0 \quad \Longrightarrow \quad x=0 \quad \text{or} \quad x=8.
$$

To determine the nature of these critical points, we compute the second derivative:
$$
f''(x)= \frac{d}{dx}\left(3x^2-24x\right)= 6x-24 = 6(x-4).
$$

Evaluate $$ f''(x) $$ at each critical point:
$$
\begin{aligned}
f''(0) &= 6(0-4)= -24 \quad (\text{negative: local maximum}), \
f''(8) &= 6(8-4)= 24 \quad (\text{positive: local minimum}).
\end{aligned}
$$

Thus, we have:
- A local maximum at $$ x=0 $$.
- A local minimum at $$ x=8 $$.

The choices provided indicate:

For the maximum:
- Option A: The function $$ f $$ has a local maximum at $$ x=\boxed{0} $$.
- Option B: The function $$ f $$ has no local maximum.

For the minimum:
- Option A: The function $$ f $$ has a local minimum at $$ x=\boxed{8} $$.
- Option B: The function $$ f $$ has no local minimum.

The correct selections are Option A for both questions:

- The function $$ f $$ has a local maximum at $$ x=0 $$.
- The function $$ f $$ has a local minimum at $$ x=8 $$.
The function $$ f(x) $$ has a local maximum at $$ x=0 $$ and a local minimum at $$ x=8 $$.

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