The function f(x) = xx decreases on the interval

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(0, 1/e) ext { Given }: \hspace{0.167em} f\left( x ight) = x^x ext { Applying log with base e on both sides, we get } \log \left( f\left( x ight) ight) = x \log_e x \frac{f'\left( x ight)}{f\left( x ight)} = 1 + \log_e x f'\left( x ight) = f\left( x ight)\left( 1 + \log_e x ight) = x^x \left( 1 + \log_e x ight) ext { For f(x) to be decreasing, we must have } f'\left( x ight) < 0 \Rightarrow x^x \left( 1 + \log_e x ight) < 0 ext { Here, logaritmic function is defined for positive values of x } . \Rightarrow x^x > 0 \Rightarrow 1 + \log_e x < 0 \left[ ext { Since } x^x > 0, x^x \left( 1 + \log_e x ight) < 0 \Rightarrow 1 + \log_e x < 0 ight] \Rightarrow \log_e x < - 1 \Rightarrow x < e^{- 1} \left[ \because l {og}_a x < N \Rightarrow x < a^N ext { for }a > 1 ight] ext { Here }, e > 1 \Rightarrow \log_e x < - 1 \Rightarrow x < e^{- 1} \Rightarrow x \in \left( 0, e^{- 1} ight) ext { So,f(x) is decreasing on }\left( 0, 1/e ight).

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