Question

If $f(x) = x^x$, find the critical point:

• A. $x = e$
• B. $x = e^{-1}$
• C. $x = 0$
• D. $x = 1$

Hint:

Use logarithmic differentiation for functions like $x^x$ and find critical points by setting derivative to zero.

Answer 1

The Correct Option is B

Let $f(x) = x^x = e^{x \log x}$ Then: \[ f'(x) = \frac{d}{dx}[e^{x \log x}] = e^{x \log x} \cdot \frac{d}{dx}(x \log x) = x^x \cdot (\log x + 1) \] Setting $f'(x) = 0$: \[ x^x (\log x + 1) = 0 \Rightarrow \log x = -1 \Rightarrow x = e^{-1} \]