Question

The function $x^{x}$ is increasing, when

• A. $x > \frac{1}{e}$
• B. $x < \frac{1}{e}$
• C. $x < 0$
• D. For all real x

Answer 1

Answer: (A)

$x > \frac{1}{e}$

Answer 2

Let $y = x^{x}$ ⇒ $\frac{dy}{dx} = x^{x}(1 + \log x)$; For $\frac{dy}{dx} > 0$

$x^{x}(1 + \log x) > 0$ ⇒$1 + \log x > 0$ ⇒ $\log_{e}x > \log_{e}\frac{1}{e}$

For this to be positive, x should be greater than $\frac{1}{e}$.