In which of the following interval the function $f(x) = x^x,x>0$ is strictly increasing?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $(\frac{1}{e},∞)$
Given function
$f(x)=x^x,\ x>0$
Differentiating
$f'(x)=x^x(\log x+1)$
For strictly increasing function
$f'(x)>0$
$x^x(\log x+1)>0$
Since $x^x>0$ for $x>0$, condition becomes
$\log x+1>0$
$\log x>-1$
$x>\frac{1}{e}$
Hence $f(x)$ is strictly increasing in $\left(\frac{1}{e},\infty\right)$