The value of c in Lagrange 's mean value theorem for the function `f(x)=x^(2)+x+1, x in[0,4]`

Options

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

Answer

To find the value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = x^2 + x + 1 \) on the interval \([0, 4]\), we can follow these steps:

### Step 1: Verify the conditions of the theorem
Lagrange's Mean Value Theorem states that if a function is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \) in \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
Since \( f(x) = x^2 + x + 1 \) is a polynomial, it is continuous and differentiable everywhere, including the interval \([0, 4]\).

### Step 2: Calculate \( f(a) \) and \( f(b) \)
Let \( a = 0 \) and \( b = 4 \).

Calculate \( f(0) \):
\[ f(0) = 0^2 + 0 + 1 = 1 \]

Calculate \( f(4) \):
\[ f(4) = 4^2 + 4 + 1 = 16 + 4 + 1 = 21 \]

### Step 3: Apply the Mean Value Theorem
Now we can apply the Mean Value Theorem:
\[ f'(c) = \frac{f(4) - f(0)}{4 - 0} = \frac{21 - 1}{4 - 0} = \frac{20}{4} = 5 \]

### Step 4: Find \( f'(x) \)
Next, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \]

### Step 5: Set \( f'(c) \) equal to the average rate of change
Set \( f'(c) = 5 \):
\[ 2c + 1 = 5 \]

### Step 6: Solve for \( c \)
Now, solve for \( c \):
\[ 2c = 5 - 1 = 4 \\ c = \frac{4}{2} = 2 \]

### Conclusion
Thus, the value of \( c \) is \( 2 \).