Consider a polynomial function 
𝑝
(
𝑥
)
=
−
(
𝑥
2
−
16
)
(
𝑥
−
3
)
2
(
2
−
𝑥
)
2
(
𝑥
+
9
)
p(x)=−(x 
2
 −16)(x−3) 
2
 (2−x) 
2
 (x+9). Choose the set of correct options.

𝑝
(
𝑥
)
p(x) first increases then decreases in the interval 
(
2
,
3
)
(2,3)

There are exactly 7 points on 
𝑝
(
𝑥
)
p(x) where the slope of the tangent is 0.

There are exactly 6 points on 
𝑝
(
𝑥
)
p(x) where the slope of the tangent is 0.

𝑝
(
𝑥
)
p(x) is one-one function when 
𝑥
∈
(
2
,
4
)
x∈(2,4)

To analyze the polynomial function \( p(x) = -(x^2 - 16)(x - 3)^2(2 - x)^2(x + 9) \) and choose the correct set of options, we need to study the critical points, behavior in intervals, and the nature of the function. Here is a step-by-step approach:

### 1. Determine the Critical Points
Critical points occur where the derivative \( p'(x) = 0 \) or where \( p(x) \) is not differentiable.

\[ p(x) = -(x^2 - 16)(x - 3)^2(2 - x)^2(x + 9) \]

First, factorize \( x^2 - 16 \):

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

So,

\[ p(x) = - (x - 4)(x + 4)(x - 3)^2(2 - x)^2(x + 9) \]

### 2. Finding Points where \( p'(x) = 0 \)
The derivative \( p'(x) \) is zero where any of the factors of \( p(x) \) is zero, except where the factor is raised to an even power (since this would not change sign).

Set each factor to zero:
\[ x - 4 = 0 \implies x = 4 \]
\[ x + 4 = 0 \implies x = -4 \]
\[ x - 3 = 0 \implies x = 3 \]
\[ 2 - x = 0 \implies x = 2 \]
\[ x + 9 = 0 \implies x = -9 \]

Also consider the multiplicity of roots:
- \( (x - 3)^2 \) and \( (2 - x)^2 \) have a multiplicity of 2, so these points are likely not where the derivative changes sign.
- The points \( x = 4 \), \( x = -4 \), and \( x = -9 \) have a multiplicity of 1 and should be checked for changing sign.

Thus, the critical points are \( x = -9, -4, 2, 3, 4 \).

### 3. Check the Behavior of the Function in Intervals
#### In the interval \( (2, 3) \):
Since \( p(x) = -(x - 4)(x + 4)(x - 3)^2(2 - x)^2(x + 9) \):
- At \( x = 2 \), the factor \( (2 - x)^2 \) goes to zero and changes concavity because of the squared term.
- As \( x \) increases from 2 to 3, the sign of \( p(x) \) may first increase and then decrease due to the multiplicity of the factors.

Thus, \( p(x) \) does indeed first increase and then decrease in the interval \( (2, 3) \).

### 4. Points where the Tangent Slope is Zero
From the factors, we observe the zero slopes at:
\[ x = 4, -4, -9, 3, 2 \]

Since multiplicity-2 points do not change the sign:
\[ x = 3, 2 \]

There are exactly 5 points where \( p'(x) = 0 \):
\[ x = 4, -4, -9, 3, 2 \]

So, there are exactly 6 points on \( p(x) \) where the slope of the tangent is 0 (considering multiplicity):
\[ x = 4, -4, -9, 3 (twice), 2 (twice) \]

### 5. One-to-One Function in \( (2, 4) \)
To check if \( p(x) \) is one-to-one in the interval \( (2, 4) \):
- The function must be strictly increasing or decreasing.
- Given the critical points, \( p(x) \) does not remain strictly increasing or decreasing in \( (2, 4) \).

Thus, \( p(x) \) is not one-to-one in \( (2, 4) \).

### Correct Options:
1. \( p(x) \) first increases then decreases in the interval \( (2, 3) \).
2. There are exactly 6 points on \( p(x) \) where the slope of the tangent is 0.

Therefore, the correct options are:
\[ \boxed{1 \text{ and 3}} \]

Would you like more details on this analysis or have any further questions?

Here are some questions you can ask:
1. How to calculate the derivative of a polynomial function?
2. What is the significance of multiplicity of roots?
3. How to determine if a function is one-to-one?
4. Can you explain the first and second derivative tests in detail?
5. How to analyze the concavity of a function graphically?

**Tip:** Always analyze the multiplicity of roots to understand how they influence the derivative and the behavior of polynomial functions.

Explore the analysis of the polynomial function p(x) = -(x^2 - 16)(x - 3)^2(2 - x)^2(x + 9), focusing on critical points, slopes of tangents, and function behavior in specific intervals. Suitable for advanced high school students studying calculus and polynomial functions.

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