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

The correct option is B

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

Step 1: Differentiate the given function

Given,$x^x$

Let $y=x^x$

Applying $\log$ on both sides

$\log y=x\log x$

By differentiating w.r.t $x$,we get

$$\begin{gathered} \frac{1}{y}\frac{dy}{dx}=\log x+\frac{1}{x}\times x \\ \Rightarrow \frac{dy}{dx}=y(1+\log x) \\ \Rightarrow \frac{dy}{dx}=x^x(1+\log x) \end{gathered}$$

Step 2: Find the required condition

For increasing function, $\frac{dy}{dx}>0$

$$\begin{gathered} \Rightarrow x^x(1+\log x)>0 \\ \Rightarrow 1+\log x>0 \\ \Rightarrow 1+\log x>0 \\ \Rightarrow l{\mathrm{og}}_{e}x>{\log }_e\frac{1}{e} \\ \Rightarrow x>\frac{1}{e} \end{gathered}$$

Hence, function$x^x$ is increasing when $x>\frac{1}{e}$, so, option (B) is the correct answer.