Summanze the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ \mathrm{y}=\mathrm{I}(\mathrm{x}) $$. $$ f(x)=(x-4)\left(x^{2}-8 x-32\right) $$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function $$ I $$ is increasing on the subinterval(s) $$ \square $$ . (Type your answer in interval notation. Use a comma lo separate answers as needed.) B. The function $$ f $$ is never increasing Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The function $$ f $$ is decreasing on the subinterval(s) $$ \square $$ ]. (Type your answer in interval notation. Use a comma to separale answers as needed.) B. The function $$ f $$ is never decreasing. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

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To analyze the function $$ f(x) = (x-4)(x^2 - 8x - 32) $$, we will follow these steps:

1. **Find the critical points** by taking the derivative of $$ f(x) $$ and setting it to zero.
2. **Determine the intervals of increase and decrease** by testing the sign of the derivative in the intervals defined by the critical points.
3. **Sketch the graph** based on the information obtained.

### Step 1: Find the Derivative

First, we need to find the derivative $$ f'(x) $$.

Using the product rule:
$$
f'(x) = (x-4) \cdot \frac{d}{dx}(x^2 - 8x - 32) + (x^2 - 8x - 32) \cdot \frac{d}{dx}(x-4)
$$

Calculating the derivative of $$ x^2 - 8x - 32 $$:
$$
\frac{d}{dx}(x^2 - 8x - 32) = 2x - 8
$$

Now substituting back into the product rule:
$$
f'(x) = (x-4)(2x - 8) + (x^2 - 8x - 32)(1)
$$

### Step 2: Set the Derivative to Zero

Now we will set $$ f'(x) = 0 $$ and solve for $$ x $$:
$$
f'(x) = (x-4)(2x - 8) + (x^2 - 8x - 32) = 0
$$

Let's simplify $$ f'(x) $$:
1. Expand $$ (x-4)(2x - 8) $$:
   $$
   = 2x^2 - 8x - 8x + 32 = 2x^2 - 16x + 32
   $$
2. Combine with $$ (x^2 - 8x - 32) $$:
   $$
   f'(x) = 2x^2 - 16x + 32 + x^2 - 8x - 32 = 3x^2 - 24x
   $$

Setting $$ f'(x) = 0 $$:
$$
3x^2 - 24x = 0
$$
Factoring out $$ 3x $$:
$$
3x(x - 8) = 0
$$
Thus, the critical points are $$ x = 0 $$ and $$ x = 8 $$.

### Step 3: Test Intervals

We will test the intervals defined by the critical points $$ (-\infty, 0) $$, $$ (0, 8) $$, and $$ (8, \infty) $$.

1. **Interval $$ (-\infty, 0) $$**: Choose $$ x = -1 $$:
   $$
   f'(-1) = 3(-1)^2 - 24(-1) = 3 + 24 = 27 > 0 \quad \text{(increasing)}
   $$

2. **Interval $$ (0, 8) $$**: Choose $$ x = 4 $$:
   $$
   f'(4) = 3(4)^2 - 24(4) = 48 - 96 = -48 < 0 \quad \text{(decreasing)}
   $$

3. **Interval $$ (8, \infty) $$**: Choose $$ x = 9 $$:
   $$
   f'(9) = 3(9)^2 - 24(9) = 243 - 216 = 27 > 0 \quad \text{(increasing)}
   $$

### Summary of Results

- The function $$ f $$ is increasing on the intervals $$ (-\infty, 0) $$ and $$ (8, \infty) $$.
- The function $$ f $$ is decreasing on the interval $$ (0, 8) $$.

### Final Answers

A. The function $$ f $$ is increasing on the subinterval(s) $$ (-\infty, 0) \cup (8, \infty) $$.

B. The function $$ f $$ is decreasing on the subinterval(s) $$ (0, 8) $$.

Now, let's summarize the choices:

1. **Increasing Intervals**: $$ (-\infty, 0) \cup (8, \infty) $$
2. **Decreasing Intervals**: $$ (0, 8) $$

If you need to sketch the graph, you can plot the critical points and the behavior of the function in the identified intervals.
The function $$ f(x) = (x-4)(x^2 - 8x - 32) $$ is increasing on the intervals $$ (-\infty, 0) $$ and $$ (8, \infty) $$, and decreasing on the interval $$ (0, 8) $$.

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