Identify the intervals where f\left(x\right)=\frac{1}{4}\cdot x^4+\frac{1}{3}\cdot x^3-8\cdot x^2-16\cdot x is increasing and decreasing on the interval \ci{-4,4}. Then determine the local extrema of f(x). Lastly, determine the absolute extrema of f(x). fill in this chart                                             
f(x) is increasing:  
f(x) is decreasing:  

Local Minimum:  

Local Maximum:
Absolute Minimum: 

Absolute Maximum: 


We are given the function:

\[
f(x) = \frac{1}{4}x^4 + \frac{1}{3}x^3 - 8x^2 - 16x
\]

### Step 1: Find the critical points

To find the intervals where \( f(x) \) is increasing or decreasing, we need to compute the first derivative \( f'(x) \) and solve for critical points.

\[
f'(x) = \frac{d}{dx} \left( \frac{1}{4}x^4 + \frac{1}{3}x^3 - 8x^2 - 16x \right)
\]

Using power rule for differentiation:

\[
f'(x) = x^3 + x^2 - 16x - 16
\]

Now, set \( f'(x) = 0 \) to find the critical points:

\[
x^3 + x^2 - 16x - 16 = 0
\]

This is a cubic equation. To solve it, we first check for possible rational roots using the Rational Root Theorem. Testing \( x = -2 \):

\[
(-2)^3 + (-2)^2 - 16(-2) - 16 = -8 + 4 + 32 - 16 = 12 \neq 0
\]

We check \( x = 2 \):

\[
2^3 + 2^2 - 16(2) - 16 = 8 + 4 - 32 - 16 = -36 \neq 0
\]

Since testing the Rational Root Theorem is not fruitful, we need to solve the cubic numerically. I'll proceed to approximate solutions for critical points.

### Step 2: Test for increasing/decreasing behavior

Once critical points are found, we need to check the sign of \( f'(x) \) on intervals around these critical points to determine where \( f(x) \) is increasing or decreasing.

### Step 3: Determine local extrema

At each critical point where \( f'(x) = 0 \), we will apply the second derivative test or use sign changes in \( f'(x) \) to determine if the point is a local maximum or minimum.

### Step 4: Determine absolute extrema

Lastly, we will evaluate \( f(x) \) at the endpoints \( x = -4 \) and \( x = 4 \), as well as at any critical points, to find the absolute maximum and minimum on the interval \([-4, 4]\).

Let me compute this.### Step 1: Critical Points

We found the critical points by solving \( f'(x) = 0 \), which gives:

\[
x = -4, -1, 4
\]

### Step 2: Increasing/Decreasing Intervals

The first derivative of the function is:

\[
f'(x) = x^3 + x^2 - 16x - 16
\]

We will use these critical points to test the sign of \( f'(x) \) in the intervals:

- \( (-\infty, -4) \)
- \( (-4, -1) \)
- \( (-1, 4) \)
- \( (4, \infty) \)

By analyzing the sign of \( f'(x) \) in these intervals, we determine where the function is increasing or decreasing.

### Step 3: Local Extrema

Using the second derivative test \( f''(x) = 3x^2 + 2x - 16 \), we can determine the nature of the critical points:

- \( x = -4 \): Local minimum, \( f(-4) = -21.33 \)
- \( x = -1 \): Local maximum, \( f(-1) = 7.92 \)
- \( x = 4 \): Local minimum, \( f(4) = -106.67 \)

### Step 4: Absolute Extrema

We evaluate \( f(x) \) at the critical points and endpoints:

- Absolute maximum: \( f(-1) = 7.92 \)
- Absolute minimum: \( f(4) = -106.67 \)

### Final Chart:

- **f(x) is increasing**: \( (-4, -1) \)
- **f(x) is decreasing**: \( (-\infty, -4) \cup (-1, 4) \cup (4, \infty) \)
- **Local Minimum**: \( x = -4, x = 4 \)
- **Local Maximum**: \( x = -1 \)
- **Absolute Minimum**: \( x = 4, f(4) = -106.67 \)
- **Absolute Maximum**: \( x = -1, f(-1) = 7.92 \)

Would you like more details on any of the steps?

### Relative Questions:
1. How do you interpret the second derivative test for extrema?
2. Why does the function change from increasing to decreasing at critical points?
3. What is the significance of absolute extrema compared to local extrema?
4. Can the cubic function \( f'(x) \) have more than three real roots?
5. How can you graphically confirm the increasing and decreasing behavior?

### Tip:
When solving for critical points, using numerical methods can simplify the process of solving complex polynomials.

Identify the intervals where f(x)=\frac{1}{4}x^4+\frac{1}{3}x^3-8x^2-16x is increasing and decreasing on the interval [-4,4]. Then determine the local extrema of f(x). Lastly, determine the absolute extrema of f(x).

Explore how to find intervals where f(x) = \frac{1}{4}x^4 + \frac{1}{3}x^3 - 8x^2 - 16x is increasing or decreasing and determine the local and absolute extrema. This detailed solution uses first and second derivative tests, covering core calculus concepts for Grades 11-12 or early college students.

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