Step 1: Understanding the Concept:
To determine the maximum value of a function, we first identify its critical points by setting the first derivative equal to zero. The second derivative test is then employed to verify if a critical point corresponds to a maximum. Finally, the required expression is evaluated.
Step 2: Key Formula or Approach:
Employ the quotient rule for differentiation: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. \] For a critical point \( x = a \) to be a maximum, it must satisfy \( f'(a) = 0 \) and \( f''(a)<0 \).
Step 3: Detailed Explanation:
Consider the function: \( f(x) = \frac{\ln x}{x} \).
First Derivative:
Applying the quotient rule with \( u = \ln x \) and \( v = x \):
\[ f'(x) = \frac{\left(\frac{1}{x}\right) \cdot x - (\ln x) \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}. \] To find the point where the maximum occurs, set \( f'(x) = 0 \): \[ \frac{1 - \ln x}{x^2} = 0 \implies 1 - \ln x = 0 \implies \ln x = 1 \implies x = e. \] Thus, the maximum is located at \( a = e \).
Second Derivative:
Differentiate \( f'(x) \) using the quotient rule with \( u = 1 - \ln x \) and \( v = x^2 \): \[ f''(x) = \frac{\left(-\frac{1}{x}\right) \cdot x^2 - (1 - \ln x) \cdot (2x)}{(x^2)^2} \] \[ f''(x) = \frac{-x - 2x + 2x \ln x}{x^4} = \frac{-3x + 2x \ln x}{x^4} = \frac{2 \ln x - 3}{x^3}. \] Evaluate \( f''(a) = f''(e) \): \[ f''(e) = \frac{2 \ln e - 3}{e^3} = \frac{2(1) - 3}{e^3} = \frac{-1}{e^3}. \]
Calculate the final expression:
We need to compute \( a^2 f''(a) \). Substituting \( a = e \): \[ a^2 f''(a) = e^2 \cdot \left(\frac{-1}{e^3}\right) = \frac{-e^2}{e^3} = -\frac{1}{e}. \]
Step 4: Final Answer:
The computed value of \( a^2 f''(a) \) is \(-\frac{1}{e}\).