Rolle's theorem is not applicable to the function `f(x)=|x|"for"-2 le x le2` becase

Options

A. f is continuus on [ -2,2]
B. f is not derivable at x=0
C. `f(-2)=f(x)`
D. f is not a constant function

Explanation

To determine why Rolle's theorem is not applicable to the function \( f(x) = |x| \) for the interval \([-2, 2]\), we need to check the conditions required for Rolle's theorem to hold.

### Step-by-Step Solution:

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

2. **Check Continuity**:
- The function \( f(x) = |x| \) is continuous everywhere, including the interval \([-2, 2]\).
- Specifically, at \( x = 0 \), we can check the limits:
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0 \)
- Since both limits are equal and \( f(0) = 0 \), \( f(x) \) is continuous at \( x = 0 \).

3. **Check Differentiability**:
- To apply Rolle's theorem, we need to check if \( f(x) \) is differentiable on the open interval \((-2, 2)\).
- The derivative from the left at \( x = 0 \):
\[ f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{-x - 0}{x} = \lim_{x \to 0^-} -1 = -1 \]
- The derivative from the right at \( x = 0 \):
\[ f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{x - 0}{x} = \lim_{x \to 0^+} 1 = 1 \]
- Since \( f'(0^-) \neq f'(0^+) \), the function \( f(x) \) is not differentiable at \( x = 0 \).

4. **Conclusion**:
- Since \( f(x) = |x| \) is not differentiable at \( x = 0 \), it fails the differentiability condition of Rolle's theorem. Therefore, Rolle's theorem is not applicable to the function \( f(x) = |x| \) on the interval \([-2, 2]\).

### Final Answer:
Rolle's theorem is not applicable to the function \( f(x) = |x| \) for the interval \([-2, 2]\) because the function is not differentiable at \( x = 0 \).