Question

The function $x^x$, $x > 0$ is strictly increasing at:

• A. $\forall x \in \mathbb{R}$
• B. $x < \frac{1}{e}$
• C. $x > \frac{1}{e}$
• D. $x < 0$

Answer 1

The Correct Option is C

Rewrite $y = x^x$ as $y = e^{x \ln x}$. Differentiate: \[ \frac{dy}{dx} = y (\ln x + 1). \] The sign of $\ln x + 1$ determines monotonicity. Solve $\ln x + 1 > 0$: \[ \ln x > -1 \implies x > \frac{1}{e}. \] Thus, $x^x$ is strictly increasing for $x > \frac{1}{e}$.