If from Largrange's mean value theorem, we have `f'(x(1))=(f'(b)-f(a))/(b-a)` then,

Options

A. `altx_(1) leb`
B. `alex_(1) ltb`
C. `altx_(1)ltb`
D. `a le x_(1) leb`

Answer

To solve the problem using Lagrange's Mean Value Theorem, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Lagrange's Mean Value Theorem (LMVT)**:
- The theorem states that if a function \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one point \( c \) in the interval \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

2. **Identify the given information**:
- From the problem, we have:
\[ f'(x_1) = \frac{f(b) - f(a)}{b - a} \]
- This is similar to the conclusion of LMVT, where \( c \) is replaced by \( x_1 \).

3. **Determine the range of \( x_1 \)**:
- According to LMVT, the point \( c \) (or \( x_1 \) in this case) must lie within the open interval \((a, b)\).
- Therefore, we conclude that:
\[ x_1 \in (a, b) \]

4. **Select the correct option**:
- Based on our conclusion, the correct answer is that \( x_1 \) belongs to the open interval \((a, b)\).

### Final Answer:
The range of \( x_1 \) is \( (a, b) \).

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