Question

In the interval $ \left( \frac{1}{e}, e \right) $, a decreasing function among the following functions is:

• A. \( f(x) = \frac{\log x}{x} \)
• B. \( f(x) = x^2 \log x \)
• C. \( f(x) = x \log x \)
• D. \( f(x) = x^{-x} \)

Hint:

To determine if a function is decreasing, compute \( f'(x) \). If \( f'(x)<0 \) over the entire interval, the function is decreasing.

Answer 1

The Correct Option is D

Step 1: Understand what makes a function decreasing.
A function is decreasing if \( f'(x)<0 \). Interval: \( \left( \frac{1}{e}, e \right) \), where \( \frac{1}{e} \approx 0.367 \), \( e \approx 2.718 \).
Step 2: Analyze each function.
Option 1: \( f(x) = \frac{\log x}{x} \), \( f'(x) = \frac{1 - \log x}{x^2} \). In \( \left( \frac{1}{e}, e \right) \), \( 1 - \log x>0 \), so \( f'(x)>0 \). Increasing.
Option 2: \( f(x) = x^2 \log x \), \( f'(x) = x (2 \log x + 1) \). Changes sign at \( x = e^{-1/2} \), not decreasing throughout.
Option 3: \( f(x) = x \log x \), \( f'(x) = \log x + 1>0 \). Increasing.
Option 4: \( f(x) = x^{-x} = e^{-x \log x} \). Let \( g(x) = -x \log x \), \( g'(x) = -(\log x + 1)<0 \), so \( f'(x)<0 \). Decreasing.
Step 3: Conclusion.
Only \( f(x) = x^{-x} \) is decreasing over the entire interval.