If c is a point at which Roll's theorem holds for the function, `f(x) = log_(e) ((x^(2) + alpha)/(7x))` in the interval [3,4], where `alpha in R` then `f^('')( c)` is equal to:

Options

A. `-1/(24)`
B. `-1/(12)`
C. `sqrt(3)/7`
D. `1/12`

Answer

To solve the problem, we need to find \( f''(c) \) for the function \( f(x) = \log_e\left(\frac{x^2 + \alpha}{7x}\right) \) in the interval [3, 4] where \( c \) is a point satisfying Rolle's theorem.

### Step 1: Verify conditions for Rolle's Theorem
Rolle's theorem requires that the function is continuous and differentiable on the interval [3, 4], and that \( f(3) = f(4) \).

### Step 2: Calculate \( f(3) \) and \( f(4) \)
We compute:
\[ f(3) = \log_e\left(\frac{3^2 + \alpha}{7 \cdot 3}\right) = \log_e\left(\frac{9 + \alpha}{21}\right) \]
\[ f(4) = \log_e\left(\frac{4^2 + \alpha}{7 \cdot 4}\right) = \log_e\left(\frac{16 + \alpha}{28}\right) \]

### Step 3: Set \( f(3) = f(4) \)
Setting \( f(3) = f(4) \) gives us:
\[ \log_e\left(\frac{9 + \alpha}{21}\right) = \log_e\left(\frac{16 + \alpha}{28}\right) \]
This implies:
\[ \frac{9 + \alpha}{21} = \frac{16 + \alpha}{28} \]

### Step 4: Cross-multiply to solve for \( \alpha \)
Cross-multiplying gives:
\[ 28(9 + \alpha) = 21(16 + \alpha) \]
Expanding both sides:
\[ 252 + 28\alpha = 336 + 21\alpha \]
Rearranging terms:
\[ 28\alpha - 21\alpha = 336 - 252 \]
\[ 7\alpha = 84 \implies \alpha = 12 \]

### Step 5: Substitute \( \alpha \) back into \( f(x) \)
Now substituting \( \alpha = 12 \) into \( f(x) \):
\[ f(x) = \log_e\left(\frac{x^2 + 12}{7x}\right) \]

### Step 6: Find the first derivative \( f'(x) \)
Using the quotient rule:
\[ f'(x) = \frac{(7x)(2x) - (x^2 + 12)(7)}{(7x)^2} \]
Simplifying:
\[ f'(x) = \frac{14x^2 - 7x^2 - 84}{49x^2} = \frac{7x^2 - 84}{49x^2} = \frac{x^2 - 12}{7x^2} \]

### Step 7: Find the second derivative \( f''(x) \)
Using the quotient rule again:
\[ f''(x) = \frac{(7x^2)(2x) - (x^2 - 12)(14x)}{(7x^2)^2} \]
Simplifying:
\[ f''(x) = \frac{14x^3 - (14x^3 - 168x)}{49x^4} = \frac{168x}{49x^4} = \frac{24}{7x^3} \]

### Step 8: Evaluate \( f''(c) \) at \( c \)
Since \( c \) is in the interval [3, 4], we can choose \( c = \sqrt{12} \) (as derived from \( f'(c) = 0 \)):
\[ f''(c) = \frac{24}{7(\sqrt{12})^3} = \frac{24}{7 \cdot 12\sqrt{12}} = \frac{2}{7\sqrt{12}} \]

### Final Answer
Thus, the value of \( f''(c) \) is:
\[ f''(c) = \frac{2}{7\sqrt{12}} \]