Given the function: 𝑓𝑓(𝑥𝑥) = 1
5
𝑥𝑥5 − 8
3
𝑥𝑥3 + 16𝑥𝑥 (10 Points)
a. What are the critical points of f(x)?

To find the critical points of the function \( f(x) = \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x \), we need to follow these steps:

### Step 1: Find the derivative \( f'(x) \)
The critical points occur where the derivative is equal to zero or undefined. Let's compute the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx} \left( \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x \right)
\]

Using the power rule for differentiation:

\[
f'(x) = \frac{1}{5} \cdot 5x^4 - \frac{8}{3} \cdot 3x^2 + 16
\]

Simplifying:

\[
f'(x) = x^4 - 8x^2 + 16
\]

### Step 2: Set the derivative equal to zero
Now, we set \( f'(x) = 0 \) to find the critical points:

\[
x^4 - 8x^2 + 16 = 0
\]

This is a quadratic equation in terms of \( x^2 \). Let \( u = x^2 \), so the equation becomes:

\[
u^2 - 8u + 16 = 0
\]

### Step 3: Solve the quadratic equation
Solve the quadratic equation \( u^2 - 8u + 16 = 0 \) using the quadratic formula:

\[
u = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(16)}}{2(1)}
\]

\[
u = \frac{8 \pm \sqrt{64 - 64}}{2}
\]

\[
u = \frac{8 \pm \sqrt{0}}{2}
\]

\[
u = \frac{8}{2} = 4
\]

Thus, \( u = x^2 = 4 \), so \( x = \pm 2 \).

### Step 4: Find the critical points
The critical points are \( x = 2 \) and \( x = -2 \).

### Conclusion:
The critical points of the function \( f(x) \) are \( x = 2 \) and \( x = -2 \).

Would you like more details on analyzing the nature of these critical points (e.g., whether they are maxima, minima, or points of inflection)?

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Here are 5 questions that follow from this topic:

1. How do you determine if a critical point is a maximum or minimum?
2. What is the second derivative test, and how can it be applied to this problem?
3. How do inflection points differ from critical points in a function?
4. Can there be critical points where the derivative is undefined? If so, what types of functions might show this behavior?
5. How can graphing a function help in visualizing its critical points?

**Tip:** Always check the second derivative or use the first derivative test to confirm the nature of critical points (maxima, minima, or inflection points).

Given the function: f(x) = (1/5)x^5 - (8/3)x^3 + 16x, what are the critical points of f(x)?

Explore how to find the critical points of the function f(x) = (1/5)x^5 - (8/3)x^3 + 16x through differentiation. This detailed step-by-step solution covers concepts in calculus, including finding critical points, solving quadratic equations, and applying the power rule. Suitable for Grades 11-12 and college-level calculus students.

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