Is Rolle's Theorem applicable to the function:
f(x) = |x| in the interval [-1, 1]?

Answer

To determine if Rolle's Theorem is applicable to the function \( f(x) = |x| \) on the interval \([-1, 1]\), we need to follow these steps:

### Step 1: Check the conditions of Rolle's Theorem
Rolle's Theorem states that if a function \( f \) 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 \( f'(c) = 0 \).

### Step 2: Evaluate \( f(-1) \) and \( f(1) \)
Calculate the function values at the endpoints of the interval:
\[ f(-1) = |-1| = 1 \]
\[ f(1) = |1| = 1 \]
Since \( f(-1) = f(1) \), the condition \( f(a) = f(b) \) is satisfied.

### Step 3: Check continuity of \( f(x) \)
The function \( f(x) = |x| \) is continuous everywhere, including the interval \([-1, 1]\). Therefore, \( f(x) \) is continuous on \([-1, 1]\).

### Step 4: Check differentiability of \( f(x) \)
Next, we need to check if \( f(x) \) is differentiable on the open interval \((-1, 1)\). The derivative of \( f(x) \) can be computed as follows:
- For \( x > 0 \), \( f(x) = x \) so \( f'(x) = 1 \).
- For \( x < 0 \), \( f(x) = -x \) so \( f'(x) = -1 \).
- At \( x = 0 \), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\).

Since the left-hand and right-hand derivatives at \( x = 0 \) do not match, \( f(x) \) is not differentiable at \( x = 0 \).

### Step 5: Conclusion
Since \( f(x) \) is not differentiable at \( x = 0 \), it does not satisfy all the conditions of Rolle's Theorem. Therefore, we conclude that Rolle's Theorem is **not applicable** to the function \( f(x) = |x| \) on the interval \([-1, 1]\).

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