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

Options

A. f is continuous for `-2 le x le 2`
B. f is continous for `-2 le x le 2`
C. `f (-2) =f (2)`
D. f is not a constnat function

Answer

To determine why Rolle's theorem is not applicable to the function \( f(x) = |x| \) on the interval \([-2, 2]\), we need to analyze the conditions of Rolle's theorem and the properties of the function.

### Step-by-Step Solution:

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

2. **Check Continuity**:
The function \( f(x) = |x| \) is continuous everywhere, including the interval \([-2, 2]\). Therefore, it satisfies the first condition of Rolle's theorem.

3. **Check Differentiability**:
To apply Rolle's theorem, we need to check if \( f(x) \) is differentiable on the open interval \((-2, 2)\). The function \( f(x) = |x| \) can be expressed as:
\[ f(x) = \begin{cases} -x & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases} \]
At \( x = 0 \), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\). Since these two derivatives do not match, \( f(x) \) is not differentiable at \( x = 0 \).

4. **Check Endpoint Values**:
Now, we check the values at the endpoints:
\[ f(-2) = |-2| = 2 \quad \text{and} \quad f(2) = |2| = 2 \]
Since \( f(-2) = f(2) \), the condition \( f(a) = f(b) \) is satisfied.

5. **Conclusion**:
Although \( f(x) \) is continuous on \([-2, 2]\) and \( f(-2) = f(2) \), it is not differentiable at \( x = 0 \). 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 \([-2, 2]\) because the function is not differentiable at \( x = 0 \).