Question

The maximum value of the function \( f(x) = \frac{\log x}{x} \) is

• A. \( e \)
• B. \( \frac{1}{e} \)
• C. \( e^2 \)
• D. \( \frac{1}{e^2} \)

Hint:

Maximize functions using first and second derivative tests; simplify logarithmic expressions.

Answer 1

The Correct Option is B

Find critical points: 
\[ f'(x) = \frac{\frac{1}{x} \cdot x - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}. \] Set \( f'(x) = 0 \): \( 1 - \log x = 0 \Rightarrow \log x = 1 \Rightarrow x = e \). 
Second derivative: 
\[ f''(x) = \frac{-\frac{1}{x} \cdot x^2 - (1 - \log x) \cdot 2x}{x^4} = \frac{-x - 2x(1 - \log x)}{x^4} = \frac{-1 - 2(1 - \log x)}{x^3}. \] At \( x = e \): \( f''(e) = \frac{-1 - 2(1 - 1)}{e^3} = \frac{-1}{e^3} < 0 \), maximum. 
\[ f(e) = \frac{\log e}{e} = \frac{1}{e}. \] Answer: \( \frac{1}{e} \).