Question

The function f(x) = x^x decreases on the interval
(a) (0, e)
(b) (0, 1)
(c) (0, 1/e)
(d) none of these

Answer 1

(c) (0, 1/e)

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}f\left(x\right)=x^x \\ \text{Applying log with base}\mathit{\text{e}}\text{on both sides, we get} \\ \log \left(f\left(x\right)\right)=x{\log }_{e}x \\ \frac{f\text{'}\left(x\right)}{f\left(x\right)}=1+{\log }_{e}x \\ f\text{'}\left(x\right)=f\left(x\right)\left(1+{\log }_{e}x\right)=x^x\left(1+{\log }_{e}x\right) \\ \text{For}\mathit{\text{f}}\text{(}\mathit{\text{x}}\text{) to be decreasing, we must have} \\ f\text{'}\left(x\right)<0 \\ \Rightarrow x^x\left(1+{\log }_{e}x\right)<0 \\ \mathrm{Here},\mathrm{logaritmic}\mathrm{function}\mathrm{is}\mathrm{defined}\mathrm{for}\mathrm{positive}\mathrm{values}\mathrm{of}x. \\ \Rightarrow x^x>0 \\ \Rightarrow 1+{\log }_{e}x<0\left[\mathrm{Since}{x}^x>0,x^x\left(1+{\log }_{e}x\right)<0\Rightarrow 1+{\log }_{e}x<0\right] \\ \Rightarrow {\log }_{e}x<-1 \\ \Rightarrow x<e^{-1}\left[\because l{\mathrm{og}}_{a}x<N\Rightarrow x<a^N\mathrm{for}a>1\right] \\ \mathrm{Here}, \\ e>1 \\ \Rightarrow {\log }_{e}x<-1\Rightarrow x<e^{-1} \\ \Rightarrow x\in \left(0,e^{-1}\right) \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(0,\frac{1}{e}\right)\text{.} \end{gathered}$$