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

Let $y=x^x$

$\Rightarrow \quad \frac{d y}{d x}=x^x(1+\log x)$

For increasing function $\frac{d y}{d x}>0$

$\therefore x^x(1+\log x)>0 \Rightarrow 1+\log x>0$

$\Rightarrow \quad \log _e x>\log _e \frac{1}{e} \Rightarrow x>\frac{1}{e}$

Function is increasing when $x>\frac{1}{e}$

So, The correct option will be (A).