The value of c in Rolle's theorem when `f(x)=2x^(3)-5x^(2)-4x+3, x in [1//2,3]` is

Options

A. 2
B. `-(1)/(3)`
C. `-2`
D. `(2)/(3)`

Explanation

To find the value of \( c \) in Rolle's theorem for the function \( f(x) = 2x^3 - 5x^2 - 4x + 3 \) on the interval \( \left[\frac{1}{2}, 3\right] \), we follow these steps:

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

- **Function Continuity and Differentiability**: The function \( f(x) \) is a polynomial, which is continuous and differentiable everywhere.

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = \frac{1}{2} \) and \( b = 3 \).

- **Calculate \( f\left(\frac{1}{2}\right) \)**:
\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 5\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 3 \]
\[ = 2 \cdot \frac{1}{8} - 5 \cdot \frac{1}{4} - 2 + 3 \]
\[ = \frac{1}{4} - \frac{5}{4} - 2 + 3 \]
\[ = \frac{1 - 5 - 8 + 12}{4} = \frac{0}{4} = 0 \]

- **Calculate \( f(3) \)**:
\[ f(3) = 2(3)^3 - 5(3)^2 - 4(3) + 3 \]
\[ = 2 \cdot 27 - 5 \cdot 9 - 12 + 3 \]
\[ = 54 - 45 - 12 + 3 = 0 \]

Since \( f\left(\frac{1}{2}\right) = 0 \) and \( f(3) = 0 \), we have \( f(a) = f(b) \).

### Step 3: Apply Rolle's Theorem
Since the conditions of Rolle's theorem are satisfied, we can find \( c \) such that \( f'(c) = 0 \).

### Step 4: Find the derivative \( f'(x) \)
\[ f'(x) = \frac{d}{dx}(2x^3 - 5x^2 - 4x + 3) = 6x^2 - 10x - 4 \]

### Step 5: Set the derivative equal to zero and solve for \( c \)
\[ 6c^2 - 10c - 4 = 0 \]
Dividing the entire equation by 2:
\[ 3c^2 - 5c - 2 = 0 \]

### Step 6: Factor the quadratic equation
We can factor this equation:
\[ 3c^2 - 6c + c - 2 = 0 \]
\[ (3c + 1)(c - 2) = 0 \]

### Step 7: Solve for \( c \)
Setting each factor to zero gives:
1. \( 3c + 1 = 0 \) → \( c = -\frac{1}{3} \)
2. \( c - 2 = 0 \) → \( c = 2 \)

### Step 8: Determine valid \( c \) in the interval
The valid \( c \) must be in the interval \( \left(\frac{1}{2}, 3\right) \).
- \( c = -\frac{1}{3} \) is not in the interval.
- \( c = 2 \) is in the interval.

### Conclusion
Thus, the value of \( c \) in Rolle's theorem is:
\[ \boxed{2} \]