If Rolle's theoram holds for the function f(x)`2x^(3)+bx^(2)+cx`,
`xepsilon[-1,1],"at the point x" = 1/2`,then 2b + c equals :

Options

A. -3
B. -1
C. 2
D. 1

Answer

To solve the problem step by step, we will apply Rolle's Theorem to the function \( f(x) = 2x^3 + bx^2 + cx \) over the interval \([-1, 1]\) with the condition that it holds at the point \( x = \frac{1}{2} \).

### Step 1: Apply Rolle's Theorem
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), and if \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

Here, we have:
- \( a = -1 \)
- \( b = 1 \)
- \( d = \frac{1}{2} \)

### Step 2: Calculate \( f(-1) \) and \( f(1) \)
First, we calculate \( f(-1) \) and \( f(1) \):
\[ f(-1) = 2(-1)^3 + b(-1)^2 + c(-1) = -2 + b - c \]
\[ f(1) = 2(1)^3 + b(1)^2 + c(1) = 2 + b + c \]

### Step 3: Set \( f(-1) = f(1) \)
According to Rolle's Theorem, since \( f(-1) = f(1) \):
\[ -2 + b - c = 2 + b + c \]
Simplifying this equation:
\[ -2 - 2 = c + c \implies -4 = 2c \implies c = -2 \]

### Step 4: Differentiate \( f(x) \)
Next, we find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 + bx^2 + cx) = 6x^2 + 2bx + c \]

### Step 5: Calculate \( f'\left(\frac{1}{2}\right) \)
Now we evaluate the derivative at \( x = \frac{1}{2} \):
\[ f'\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + 2b\left(\frac{1}{2}\right) + c \]
\[ = 6 \cdot \frac{1}{4} + b + c = \frac{3}{2} + b + c \]
Since we know \( c = -2 \):
\[ f'\left(\frac{1}{2}\right) = \frac{3}{2} + b - 2 = \frac{3}{2} + b - \frac{4}{2} = b - \frac{1}{2} \]

### Step 6: Set \( f'\left(\frac{1}{2}\right) = 0 \)
Setting the derivative equal to zero:
\[ b - \frac{1}{2} = 0 \implies b = \frac{1}{2} \]

### Step 7: Calculate \( 2b + c \)
Now we can find \( 2b + c \):
\[ 2b + c = 2\left(\frac{1}{2}\right) + (-2) = 1 - 2 = -1 \]

### Final Answer
Thus, the value of \( 2b + c \) is \(-1\).