If `f(x) = log x` satisfies Lagrange's theorem on `[1,e]` then value of `c in (1,e)` such that the tangent at `c` is parallel to line joining `(1,f(1))` and `(e,f(e ))` is

Options

A. `e-(3)/(2)`
B. `(1 + e)/(2)`
C. `e-1`
D. `e-(1)/(2)`

Answer

To solve the problem, we will follow the steps outlined in the video transcript and apply Lagrange's Mean Value Theorem.

### Step 1: Identify the function and the interval
We have the function:
\[ f(x) = \log x \]
We need to consider the interval \([1, e]\).

### Step 2: Verify continuity and differentiability
The function \( f(x) = \log x \) is continuous on the closed interval \([1, e]\) and differentiable on the open interval \((1, e)\). Therefore, it satisfies the conditions of Lagrange's Mean Value Theorem.

### Step 3: Apply Lagrange's Mean Value Theorem
According to Lagrange's Mean Value Theorem, there exists at least one point \( c \) in the open interval \((1, e)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
where \( a = 1 \) and \( b = e \).

### Step 4: Calculate \( f(a) \) and \( f(b) \)
Calculate \( f(1) \) and \( f(e) \):
- \( f(1) = \log 1 = 0 \)
- \( f(e) = \log e = 1 \)

### Step 5: Substitute into the Mean Value Theorem formula
Now, substitute these values into the formula:
\[ f'(c) = \frac{f(e) - f(1)}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} \]

### Step 6: Find the derivative \( f'(x) \)
The derivative of the function \( f(x) = \log x \) is:
\[ f'(x) = \frac{1}{x} \]
Thus, at point \( c \):
\[ f'(c) = \frac{1}{c} \]

### Step 7: Set the derivatives equal
Now, we set the expressions for \( f'(c) \) equal to each other:
\[ \frac{1}{c} = \frac{1}{e - 1} \]

### Step 8: Solve for \( c \)
Cross-multiplying gives:
\[ e - 1 = c \]
Thus, we find:
\[ c = e - 1 \]

### Conclusion
The value of \( c \) in the interval \( (1, e) \) such that the tangent at \( c \) is parallel to the line joining the points \( (1, f(1)) \) and \( (e, f(e)) \) is:
\[ c = e - 1 \]