Find the absolute maximum and absolute minimum values of f(x)=x^5+5x^4-35x^3 on [-1,4] and where they occur.

To find the absolute maximum and minimum values of the function \( f(x) = x^5 + 5x^4 - 35x^3 \) on the interval \([-1, 4]\), we will follow these steps:

1. **Find the critical points** by taking the derivative and setting it to zero.
2. **Evaluate the function** at the critical points and at the endpoints of the interval.
3. **Determine the absolute maximum and minimum** values.

### Step 1: Find the derivative and critical points

First, we compute the derivative of \( f(x) \):

\[
f'(x) = 5x^4 + 20x^3 - 105x^2
\]

Next, we set the derivative equal to zero:

\[
5x^4 + 20x^3 - 105x^2 = 0
\]

Factoring out the common term:

\[
5x^2(x^2 + 4x - 21) = 0
\]

This gives us:

\[
5x^2 = 0 \quad \text{or} \quad x^2 + 4x - 21 = 0
\]

From \( 5x^2 = 0 \), we find \( x = 0 \).

For the quadratic equation \( x^2 + 4x - 21 = 0 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 84}}{2} = \frac{-4 \pm \sqrt{100}}{2} = \frac{-4 \pm 10}{2}
\]

Calculating the two roots:

1. \( x = \frac{6}{2} = 3 \)
2. \( x = \frac{-14}{2} = -7 \) (not in the interval \([-1, 4]\))

Thus, the critical points within the interval \([-1, 4]\) are \( x = 0 \) and \( x = 3 \).

### Step 2: Evaluate the function at critical points and endpoints

We now evaluate \( f(x) \) at the critical points and at the endpoints \( x = -1 \) and \( x = 4 \).

- **At \( x = -1 \)**:
\[
f(-1) = (-1)^5 + 5(-1)^4 - 35(-1)^3 = -1 + 5 + 35 = 39
\]

- **At \( x = 0 \)**:
\[
f(0) = 0^5 + 5(0)^4 - 35(0)^3 = 0
\]

- **At \( x = 3 \)**:
\[
f(3) = 3^5 + 5(3)^4 - 35(3)^3 = 243 + 405 - 945 = -297
\]

- **At \( x = 4 \)**:
\[
f(4) = 4^5 + 5(4)^4 - 35(4)^3 = 1024 + 1280 - 5600 = -4296
\]

### Step 3: Determine absolute maximum and minimum

Now we compare the values:

- \( f(-1) = 39 \)
- \( f(0) = 0 \)
- \( f(3) = -297 \)
- \( f(4) = -4296 \)

The absolute maximum value is \( 39 \) at \( x = -1 \).

The absolute minimum value is \( -4296 \) at \( x = 4 \).

### Summary

- **Absolute maximum**: \( 39 \) at \( x = -1 \)
- **Absolute minimum**: \( -4296 \) at \( x = 4 \)

Do you want more details or have any questions? Here are some follow-up questions:

1. How does the behavior of the function change outside the interval \([-1, 4]\)?
2. What is the second derivative of the function, and how can it be used to determine concavity?
3. Can you analyze the end behavior of the function as \( x \) approaches infinity?
4. What are the implications of finding a maximum or minimum on a given interval?
5. How would you approach finding maxima and minima for more complex functions? 

**Tip**: Always check the endpoints along with critical points when looking for absolute extrema on a closed interval!

Explore the solution to finding the absolute maximum and minimum values of the function f(x)=x^5+5x^4-35x^3 on the interval [-1,4]. The function's critical points and evaluations lead to the maximum value of 39 at x=-1 and the minimum value of -4296 at x=4. Ideal for advanced high school students.

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