Verify Lagrange's Mean Value Theorem for the functions :
f(x) = |x| in the interval [-1, 1].

Answer

To verify Lagrange's Mean Value Theorem (LMVT) for the function \( f(x) = |x| \) on the interval \([-1, 1]\), we will follow these steps:

### Step 1: Check if the function is continuous on the 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 a piecewise function and both pieces (\(-x\) and \(x\)) are continuous, and since the limit from the left and right at \( x = 0 \) is equal to \( f(0) \), we conclude that \( f(x) \) is continuous on \([-1, 1]\).

### Step 2: Check if the function is differentiable on the interval \((-1, 1)\).

The function \( f(x) = |x| \) is differentiable everywhere in \((-1, 1)\) except at \( x = 0 \). To see this, we calculate the derivative:
\[ f'(x) = \begin{cases} -1 & \text{if } x < 0 \\ 1 & \text{if } x > 0 \end{cases} \]

At \( x = 0 \), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\). Since these two values are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

### Step 3: Apply Lagrange's Mean Value Theorem.

Lagrange's Mean Value Theorem states that if a function 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, \( a = -1 \) and \( b = 1 \):
\[ f(-1) = |-1| = 1, \quad f(1) = |1| = 1 \]
Thus,
\[ f(b) - f(a) = 1 - 1 = 0 \]
And,
\[ b - a = 1 - (-1) = 2 \]
So,
\[ \frac{f(b) - f(a)}{b - a} = \frac{0}{2} = 0 \]

### Step 4: Find \( c \) such that \( f'(c) = 0 \).

From our earlier derivative calculation:
- \( f'(x) = -1 \) when \( x < 0 \)
- \( f'(x) = 1 \) when \( x > 0 \)

Since \( f'(x) \) does not equal \( 0 \) for any \( x \) in the interval \((-1, 1)\), there is no \( c \) such that \( f'(c) = 0 \).

### Conclusion

Since \( f(x) = |x| \) is continuous on \([-1, 1]\) but not differentiable at \( x = 0 \), and we cannot find a \( c \) such that \( f'(c) = 0 \), we conclude that Lagrange's Mean Value Theorem does not hold for this function on the given interval.

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