If Rolle's theorem holds for the function `f(x) =x^(3) + bx^(2) +ax -6` for `x in [1,3]`, then a+4b=

Options

A. 13
B. `-26`
C. `-13`
D. 26

Answer

To solve the problem, we need to apply Rolle's theorem to the function \( f(x) = x^3 + bx^2 + ax - 6 \) over the interval \([1, 3]\). According to Rolle's theorem, if the function is continuous on the closed interval and differentiable on the open interval, and if \( f(1) = f(3) \), then there exists at least one \( c \) in \( (1, 3) \) such that \( f'(c) = 0 \).

### Step-by-step Solution:

1. **Evaluate \( f(1) \) and \( f(3) \)**:
\[ f(1) = 1^3 + b \cdot 1^2 + a \cdot 1 - 6 = 1 + b + a - 6 = b + a - 5 \]
\[ f(3) = 3^3 + b \cdot 3^2 + a \cdot 3 - 6 = 27 + 9b + 3a - 6 = 21 + 9b + 3a \]

2. **Set \( f(1) = f(3) \)**:
\[ b + a - 5 = 21 + 9b + 3a \]

3. **Rearranging the equation**:
\[ b + a - 5 - 21 - 9b - 3a = 0 \]
\[ -8b - 2a - 26 = 0 \]

4. **Simplifying the equation**:
\[ 2a + 8b = -26 \]
\[ a + 4b = -13 \]

Thus, the value of \( a + 4b \) is \( -13 \).

### Final Answer:
\[ a + 4b = -13 \]