Key Concepts and Formulas

• Rolle's Theorem: If a function $f(x)$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$.
• Derivative of Logarithmic Functions: $\frac{d}{dx} \log_e(u) = \frac{1}{u} \frac{du}{dx}$, where $u$ is a function of $x$.
• Quotient Rule: $\frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$.

Step-by-Step Solution

Step 1: Apply Rolle's Theorem Conditions

Since Rolle's theorem holds for $f(x) = \log_e\left(\frac{x^2 + \alpha}{7x}\right)$ in the interval $[3, 4]$, we must have $f(3) = f(4)$. Therefore, {\log _e}\left( {{{3^2} + \alpha } \over {7(3)}}} \right) = {\log _e}\left( {{{4^2} + \alpha } \over {7(4)}}} \right) ${\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)$ Since the logarithms are equal, their arguments must be equal as well: ${{9 + \alpha } \over {21}} = {{16 + \alpha } \over {28}}$ Cross-multiplying, we get: $28(9 + \alpha) = 21(16 + \alpha)$ $252 + 28\alpha = 336 + 21\alpha$ $7\alpha = 84$ $\alpha = 12$

Step 2: Find the first derivative, f'(x)

Now that we have the value of $\alpha$, we can write the function as: f(x) = {\log _e}\left( {{{x^2} + 12} \over {7x}}} \right) Using the chain rule and quotient rule, we find the first derivative: $f'(x) = \frac{7x}{x^2 + 12} \cdot \frac{7x(2x) - 7(x^2 + 12)}{(7x)^2}$ $f'(x) = \frac{7x}{x^2 + 12} \cdot \frac{14x^2 - 7x^2 - 84}{49x^2}$ $f'(x) = \frac{7x^2 - 84}{7x(x^2 + 12)} = \frac{x^2 - 12}{x(x^2 + 12)}$

Step 3: Find the second derivative, f''(x)

We differentiate $f'(x)$ again using the quotient rule: $f''(x) = \frac{x(x^2+12)(2x) - (x^2-12)(3x^2+12)}{[x(x^2+12)]^2}$ $f''(x) = \frac{(x^3+12x)(2x) - (x^2-12)(3x^2+12)}{x^2(x^2+12)^2}$ $f''(x) = \frac{2x^4+24x^2 - (3x^4 + 12x^2 - 36x^2 - 144)}{x^2(x^2+12)^2}$ $f''(x) = \frac{2x^4+24x^2 - 3x^4 - 12x^2 + 36x^2 + 144}{x^2(x^2+12)^2}$ $f''(x) = \frac{-x^4 + 48x^2 + 144}{x^2(x^2+12)^2}$

Step 4: Find the value of c

Since Rolle's theorem holds, there exists a $c \in (3, 4)$ such that $f'(c) = 0$. $f'(c) = \frac{c^2 - 12}{c(c^2 + 12)} = 0$ This implies $c^2 - 12 = 0$, so $c^2 = 12$, and $c = \pm \sqrt{12} = \pm 2\sqrt{3}$. Since $c \in (3, 4)$, we have $c = 2\sqrt{3}$.

Step 5: Evaluate f''(c)

Now we substitute $c = 2\sqrt{3}$ into the expression for $f''(x)$: $f''(2\sqrt{3}) = \frac{-(2\sqrt{3})^4 + 48(2\sqrt{3})^2 + 144}{(2\sqrt{3})^2((2\sqrt{3})^2+12)^2}$ $f''(2\sqrt{3}) = \frac{-144 + 48(12) + 144}{12(12+12)^2}$ $f''(2\sqrt{3}) = \frac{-144 + 576 + 144}{12(24)^2}$ $f''(2\sqrt{3}) = \frac{576}{12(576)}$ $f''(2\sqrt{3}) = \frac{1}{12}$

Common Mistakes & Tips

• Remember to check that $c$ lies within the open interval $(a, b)$ after finding it.
• Be careful when applying the quotient rule, especially with complex expressions.
• Double-check your derivatives to avoid errors in subsequent calculations.

Summary

We applied Rolle's theorem to find the value of $\alpha$, then calculated the first and second derivatives of the function. By setting the first derivative to zero, we found the value of $c$ and substituted it into the second derivative to find $f''(c)$.

The final answer is $\boxed{\frac{1}{12}}$, which corresponds to option (A). Final Answer The final answer is \boxed{\frac{1}{12}}, which corresponds to option (A).