Does the function `f(x) = {:{(x" if "x lt 1),(1//x " if "x ge1):}`
satisfy the conditions of the Lagrange theorem on the interval [0,2] ?

Answer

To determine whether the function \( f(x) \) satisfies the conditions of the Lagrange Mean Value Theorem (LMVT) on the interval \([0, 2]\), we need to check two main conditions: continuity and differentiability of the function on the given interval.

### Step-by-Step Solution:

1. **Define the Function**:
The function is defined as:
\[ f(x) = \begin{cases} x & \text{if } x < 1 \\ \frac{1}{x} & \text{if } x \geq 1 \end{cases} \]

2. **Check Continuity on the Interval [0, 2]**:
- **At \( x = 0 \)**: \( f(0) = 0 \) (which is continuous).
- **For \( 0 < x < 1 \)**: \( f(x) = x \) is continuous as it is a polynomial function.
- **At \( x = 1 \)**: Check the limit from both sides:
- \( \lim_{x \to 1^-} f(x) = 1 \) (from the left).
- \( \lim_{x \to 1^+} f(x) = \frac{1}{1} = 1 \) (from the right).
- Since both limits equal \( f(1) = 1 \), the function is continuous at \( x = 1 \).
- **For \( 1 < x < 2 \)**: \( f(x) = \frac{1}{x} \) is continuous as it is a rational function and defined for \( x > 1 \).
- Therefore, \( f(x) \) is continuous on the interval \([0, 2]\).

3. **Check Differentiability on the Interval [0, 2]**:
- **For \( 0 < x < 1 \)**: The derivative \( f'(x) = 1 \) (which exists).
- **For \( 1 < x < 2 \)**: The derivative \( f'(x) = -\frac{1}{x^2} \) (which exists).
- **At \( x = 1 \)**: Check differentiability:
- From the left: \( f'(1^-) = 1 \).
- From the right: \( f'(1^+) = -\frac{1}{1^2} = -1 \).
- Since the left-hand derivative and right-hand derivative at \( x = 1 \) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

4. **Conclusion**:
Since \( f(x) \) is not differentiable at \( x = 1 \), it does not satisfy the conditions of the Lagrange Mean Value Theorem on the interval \([0, 2]\).

### Final Answer:
The function \( f(x) \) does not satisfy the conditions of the Lagrange Mean Value Theorem on the interval \([0, 2]\).

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