Verify the truth of Rolle's Theorem for the following functions
`f(x)=|x|" in "[-1,1]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = |x| \) on the interval \([-1, 1]\), we will follow the steps outlined in the theorem:

### Step 1: Check the continuity of \( f(x) \) on the closed interval \([-1, 1]\).

The function \( f(x) = |x| \) is defined as:
\[ f(x) = \begin{cases} -x & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ x & \text{if } x > 0 \end{cases} \]

Since \( f(x) \) is composed of linear functions on the intervals \([-1, 0)\) and \((0, 1]\), and it is defined at \( x = 0 \), we can conclude that \( f(x) \) is continuous on \([-1, 1]\).

### Step 2: Check the differentiability of \( f(x) \) on the open interval \((-1, 1)\).

To check for differentiability, we need to find the derivative of \( f(x) \):
- For \( x < 0 \), \( f(x) = -x \) and \( f'(x) = -1 \).
- For \( x > 0 \), \( f(x) = x \) and \( f'(x) = 1 \).

At \( x = 0 \), we need to check the left-hand and right-hand derivatives:
- Left-hand derivative at \( x = 0 \):
\[ \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h - 0}{h} = -1 \]
- Right-hand derivative at \( x = 0 \):
\[ \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h - 0}{h} = 1 \]

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

### Step 3: Check if \( f(-1) = f(1) \).

Calculating the values:
\[ f(-1) = |-1| = 1 \]
\[ f(1) = |1| = 1 \]
Thus, \( f(-1) = f(1) \).

### Conclusion

Since \( f(x) \) is continuous on \([-1, 1]\) and \( f(-1) = f(1) \), but \( f(x) \) is not differentiable on the open interval \((-1, 1)\) due to the sharp turn at \( x = 0 \), we conclude that Rolle's Theorem does not apply to this function on the given interval.