The value of c in Lagrange's mean value theorem for the function `f(x) = |x|` in the interval `[-1, 1]` is

Options

A. 0
B. `(1)/(2)`
C. `-(1)/(2)`
D. Non-existent in the interval

Answer

To find the value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = |x| \) in the interval \([-1, 1]\), we follow these steps:

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

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

In our case, \( f(x) = |x| \), \( a = -1 \), and \( b = 1 \).

### Step 2: Check continuity
The function \( f(x) = |x| \) is continuous everywhere, including the interval \([-1, 1]\).

### Step 3: Check differentiability
Next, we need to check if \( f(x) \) is differentiable on the open interval \((-1, 1)\). The function \( f(x) = |x| \) is not differentiable at \( x = 0 \) because it has a sharp corner there. Therefore, \( f(x) \) is not differentiable on the entire open interval \((-1, 1)\).

### Step 4: Conclusion
Since the function is not differentiable on the open interval \((-1, 1)\), we cannot apply Lagrange's Mean Value Theorem. Thus, there is no value of \( c \) that satisfies the theorem for the function \( f(x) = |x| \) on the interval \([-1, 1]\).

### Final Answer
The value of \( c \) in Lagrange's Mean Value Theorem for the function \( f(x) = |x| \) in the interval \([-1, 1]\) does not exist.

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