https://file.aiquickdraw.com/mathbot/file/1727802843894x1pc0u69.jpg

To find the relative maximum and minimum of the given functions, we need to perform the following steps for each function:

1. **Find the derivative** of the function.
2. **Set the derivative equal to zero** and solve for \( x \). These values of \( x \) are the critical points.
3. **Determine the nature** of each critical point (whether it's a maximum, minimum, or saddle point) using the second derivative test or by analyzing the sign changes in the first derivative.

Let's begin by solving for the first few functions from the image you provided. I'll start by finding the derivative of each function and then we will solve for the critical points.

### Function 22: \( f(x) = 3x^3 - 6x^2 + 8 \)
1. **First derivative:**
   \[
   f'(x) = 9x^2 - 12x
   \]
2. **Set the derivative to zero and solve for \( x \):**
   \[
   9x^2 - 12x = 0
   \]
   Factor out \( 3x \):
   \[
   3x(3x - 4) = 0
   \]
   This gives us \( x = 0 \) and \( x = \frac{4}{3} \).

3. **Second derivative:**
   \[
   f''(x) = 18x - 12
   \]
   - At \( x = 0 \), \( f''(0) = -12 \), so this is a **relative maximum**.
   - At \( x = \frac{4}{3} \), \( f''\left(\frac{4}{3}\right) = 12 \), so this is a **relative minimum**.

### Function 23: \( g(x) = -2x^3 + 7x - 5 \)
1. **First derivative:**
   \[
   g'(x) = -6x^2 + 7
   \]
2. **Set the derivative to zero and solve for \( x \):**
   \[
   -6x^2 + 7 = 0 \quad \Rightarrow \quad x^2 = \frac{7}{6}
   \quad \Rightarrow \quad x = \pm \sqrt{\frac{7}{6}} \approx \pm 1.08
   \]

3. **Second derivative:**
   \[
   g''(x) = -12x
   \]
   - At \( x = \sqrt{\frac{7}{6}} \), \( g''(x) < 0 \), indicating a **relative maximum**.
   - At \( x = -\sqrt{\frac{7}{6}} \), \( g''(x) > 0 \), indicating a **relative minimum**.

### Function 24: \( f(x) = -x^4 + 3x^3 - 2 \)
1. **First derivative:**
   \[
   f'(x) = -4x^3 + 9x^2
   \]
2. **Set the derivative to zero and solve for \( x \):**
   \[
   -4x^3 + 9x^2 = 0
   \quad \Rightarrow \quad x^2(-4x + 9) = 0
   \quad \Rightarrow \quad x = 0, \, x = \frac{9}{4} = 2.25
   \]

3. **Second derivative:**
   \[
   f''(x) = -12x^2 + 18x
   \]
   - At \( x = 0 \), \( f''(0) = 0 \) (use other methods like sign analysis).
   - At \( x = 2.25 \), \( f''(2.25) = -54 \), so this is a **relative maximum**.

Let me know if you would like me to continue with more functions or clarify any step!

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Here are 5 related questions:
1. What is the importance of the second derivative test in finding extrema?
2. Can a function have both absolute and relative extrema at the same point?
3. How do we determine if a critical point is a saddle point?
4. What happens if the second derivative is zero at a critical point?
5. What are the roles of concavity in determining the shape of the graph near critical points?

**Tip:** Always double-check your critical points by plugging them back into the original function to understand their nature and ensure accuracy.

Find the relative maximum and minimum for the given functions.

This page provides a step-by-step guide to finding the relative maxima and minima of polynomial functions using derivatives and the second derivative test. Concepts covered include finding the first and second derivatives, solving for critical points, and analyzing concavity. Suitable for high school and college-level calculus students.

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