Let `f(x)` be a polynomial of degree `3` such that `f(3)=1, f'(3)=-1, f''(3)=0, and f'''(3)=12.` Then the value of `f'(1)` is

Options

A. 12
B. 23
C. -13
D. none of these

Explanation

To solve the problem, we need to find the value of \( f'(1) \) for the polynomial \( f(x) \) of degree 3, given the conditions \( f(3) = 1 \), \( f'(3) = -1 \), \( f''(3) = 0 \), and \( f'''(3) = 12 \).

### Step 1: Define the polynomial
Let \( f(x) = Ax^3 + Bx^2 + Cx + D \).

### Step 2: Calculate the derivatives
- The first derivative is:
\[ f'(x) = 3Ax^2 + 2Bx + C \]
- The second derivative is:
\[ f''(x) = 6Ax + 2B \]
- The third derivative is:
\[ f'''(x) = 6A \]

### Step 3: Use the given conditions
1. From \( f'''(3) = 12 \):
\[ 6A = 12 \implies A = 2 \]

2. From \( f''(3) = 0 \):
\[ f''(3) = 6A(3) + 2B = 0 \implies 18 + 2B = 0 \implies 2B = -18 \implies B = -9 \]

3. From \( f'(3) = -1 \):
\[ f'(3) = 3A(3^2) + 2B(3) + C = -1 \]
Substituting \( A = 2 \) and \( B = -9 \):
\[ f'(3) = 3(2)(9) + 2(-9)(3) + C = -1 \]
\[ 54 - 54 + C = -1 \implies C = -1 \]

4. From \( f(3) = 1 \):
\[ f(3) = A(3^3) + B(3^2) + C(3) + D = 1 \]
Substituting \( A = 2 \), \( B = -9 \), and \( C = -1 \):
\[ f(3) = 2(27) - 9(9) - 3 + D = 1 \]
\[ 54 - 81 - 3 + D = 1 \implies D = 1 + 30 = 31 \]

### Step 4: Write the polynomial
Now we have:
\[ f(x) = 2x^3 - 9x^2 - x + 31 \]

### Step 5: Find \( f'(1) \)
Now we need to find \( f'(1) \):
\[ f'(1) = 3A(1^2) + 2B(1) + C \]
Substituting \( A = 2 \), \( B = -9 \), and \( C = -1 \):
\[ f'(1) = 3(2)(1) + 2(-9)(1) - 1 \]
\[ = 6 - 18 - 1 = -13 \]

### Final Answer
Thus, the value of \( f'(1) \) is \( \boxed{-13} \).