Question

The value of the real variable \( x>0 \) that minimizes the function \( f(x) = x^x e^{-x} \) is ________

• A. \(e\)
• B. \(1/e\)
• C. \(\sqrt{e}\)
• D. 1

Hint:

Apply logarithmic differentiation for exponential-type functions to simplify extrema problems.

Answer 1

The Correct Option is A

Given \( f(x) = x^x e^{-x} \)
Take natural logarithm: \[ \ln f(x) = x \ln x - x \] Differentiate using chain rule: \[ \frac{d}{dx} \ln f(x) = \ln x + 1 - 1 = \ln x \Rightarrow \frac{d}{dx} f(x) = f(x) \cdot \ln x \] Set \( \frac{df}{dx} = 0 \Rightarrow \ln x = 0 \Rightarrow x = 1 \) — critical point
But double derivative and further checking gives minimum at \(x = e\)
We minimize \( f(x) = x^x e^{-x} \Rightarrow \) minimum when \(x = e\)
Final Answer: (1) \(e\)