Let `p(x)` be a real polynomial of least degree which has a local maximum at `x=1` and a local minimum at `x=3.` If `p(1)=6a n dp(3)=2,` then `p^(prime)(0)` is_____

Explanation

To find \( p'(0) \) for the polynomial \( p(x) \) that has a local maximum at \( x = 1 \) and a local minimum at \( x = 3 \), we can follow these steps:

### Step 1: Determine the form of \( p'(x) \)

Since \( p(x) \) has a local maximum at \( x = 1 \) and a local minimum at \( x = 3 \), the derivative \( p'(x) \) must have roots at these points. Therefore, we can express \( p'(x) \) as:

\[ p'(x) = k(x - 1)(x - 3) \]

where \( k \) is a constant.

### Step 2: Expand \( p'(x) \)

Expanding \( p'(x) \):

\[ p'(x) = k(x^2 - 4x + 3) = kx^2 - 4kx + 3k \]

### Step 3: Integrate to find \( p(x) \)

To find \( p(x) \), we integrate \( p'(x) \):

\[ p(x) = \int p'(x) \, dx = \int (kx^2 - 4kx + 3k) \, dx \]

This gives:

\[ p(x) = \frac{k}{3}x^3 - 2kx^2 + 3kx + C \]

where \( C \) is the constant of integration.

### Step 4: Use the given conditions

We know \( p(1) = 6 \) and \( p(3) = 2 \).

1. **Using \( p(1) = 6 \)**:

\[ p(1) = \frac{k}{3}(1)^3 - 2k(1)^2 + 3k(1) + C = 6 \]
\[ \frac{k}{3} - 2k + 3k + C = 6 \]
\[ \frac{k}{3} + k + C = 6 \]
\[ \frac{4k}{3} + C = 6 \quad \text{(Equation 1)} \]

2. **Using \( p(3) = 2 \)**:

\[ p(3) = \frac{k}{3}(3)^3 - 2k(3)^2 + 3k(3) + C = 2 \]
\[ \frac{k}{3}(27) - 2k(9) + 9k + C = 2 \]
\[ 9k - 18k + 9k + C = 2 \]
\[ 0 + C = 2 \quad \text{(Equation 2)} \]

From Equation 2, we find:

\[ C = 2 \]

### Step 5: Substitute \( C \) back into Equation 1

Substituting \( C = 2 \) into Equation 1:

\[ \frac{4k}{3} + 2 = 6 \]
\[ \frac{4k}{3} = 4 \]
\[ 4k = 12 \quad \Rightarrow \quad k = 3 \]

### Step 6: Write the polynomial \( p(x) \)

Now substituting \( k \) and \( C \) back into \( p(x) \):

\[ p(x) = \frac{3}{3}x^3 - 2(3)x^2 + 3(3)x + 2 \]
\[ p(x) = x^3 - 6x^2 + 9x + 2 \]

### Step 7: Find \( p'(0) \)

Now we can find \( p'(0) \):

\[ p'(x) = 3x^2 - 12x + 9 \]

Substituting \( x = 0 \):

\[ p'(0) = 3(0)^2 - 12(0) + 9 = 9 \]

### Final Answer

Thus, \( p'(0) = 9 \).

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