The polynomial p is given by p(x) = 3x3 – 9x + 8. Which of the following is true about the graph of p?
 The graph of p is increasing on the interval (–1, 1) and has a local maximum at x = –1.
 The graph of p is increasing on the interval (–1, 1) and has a global minimum at x = –1.
 The graph of p is decreasing on the interval (–1, 1) and has a local maximum at x = –1.
 The graph of p is decreasing on the interval (–1, 1) and has a global minimum at x = –1.

To determine which of the statements is true about the graph of the polynomial \( p(x) = 3x^3 - 9x + 8 \), we need to analyze its behavior by finding the critical points and determining the intervals where the function is increasing or decreasing.

### Step 1: Find the derivative

The first derivative, \( p'(x) \), will help us identify the critical points and determine where the function is increasing or decreasing:

\[
p'(x) = \frac{d}{dx}(3x^3 - 9x + 8) = 9x^2 - 9
\]

Simplify the derivative:

\[
p'(x) = 9(x^2 - 1) = 9(x - 1)(x + 1)
\]

### Step 2: Find the critical points

The critical points occur where the first derivative is equal to zero:

\[
9(x - 1)(x + 1) = 0
\]

Solving for \( x \), we get:

\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

So, the critical points are \( x = -1 \) and \( x = 1 \).

### Step 3: Determine where the function is increasing or decreasing

To determine the intervals where the function is increasing or decreasing, we analyze the sign of \( p'(x) \) around the critical points:

1. **Interval: \( (-\infty, -1) \)**  
   Choose \( x = -2 \):  
   \( p'(-2) = 9(-2 - 1)(-2 + 1) = 9(-3)(-1) = 27 > 0 \)  
   So, the function is increasing on \( (-\infty, -1) \).

2. **Interval: \( (-1, 1) \)**  
   Choose \( x = 0 \):  
   \( p'(0) = 9(0 - 1)(0 + 1) = 9(-1)(1) = -9 < 0 \)  
   So, the function is decreasing on \( (-1, 1) \).

3. **Interval: \( (1, \infty) \)**  
   Choose \( x = 2 \):  
   \( p'(2) = 9(2 - 1)(2 + 1) = 9(1)(3) = 27 > 0 \)  
   So, the function is increasing on \( (1, \infty) \).

### Step 4: Determine the nature of critical points

- At \( x = -1 \), the function changes from increasing to decreasing, so \( x = -1 \) is a **local maximum**.
- At \( x = 1 \), the function changes from decreasing to increasing, so \( x = 1 \) is a **local minimum**.

### Conclusion

Based on the analysis:

- The function is **decreasing** on the interval \( (-1, 1) \).
- The function has a **local maximum** at \( x = -1 \).

The correct answer is:

**The graph of p is decreasing on the interval (-1, 1) and has a local maximum at x = -1.**

Would you like to know more details or have any questions?

### Related Questions:
1. How can you determine whether a critical point is a local maximum or minimum using the second derivative?
2. What is the significance of the global maximum or minimum for a cubic polynomial?
3. How would the graph change if the coefficient of \( x^3 \) was negative?
4. How do inflection points affect the shape of a cubic polynomial?
5. How can you find the y-intercept of the graph of \( p(x) = 3x^3 - 9x + 8 \)?

### Tip:
For any polynomial function, the degree of the function determines the maximum number of turning points it can have. A cubic polynomial, like this one, can have up to 2 turning points.

The polynomial p is given by p(x) = 3x^3 – 9x + 8. Which of the following is true about the graph of p? The graph of p is increasing on the interval (–1, 1) and has a local maximum at x = –1. The graph of p is increasing on the interval (–1, 1) and has a global minimum at x = –1. The graph of p is decreasing on the interval (–1, 1) and has a local maximum at x = –1. The graph of p is decreasing on the interval (–1, 1) and has a global minimum at x = –1.

Explore the detailed solution for analyzing the polynomial p(x) = 3x^3 - 9x + 8. This problem involves finding the derivative, identifying critical points, and determining where the function is increasing or decreasing. Learn how to apply calculus concepts, including the first derivative test, to determine local maximums and minimums. Suitable for high school students in Grades 11-12.

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