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Question
The function f(x) = xx decreases on the interval
Options
(0, e)
(0, 1)
(0, 1/e)
none of these
Solution
(0, 1/e)
\[\text { Given }: \hspace{0.167em} f\left( x \right) = x^x \]
\[\text { Applying log with base e on both sides, we get }\]
\[\log \left( f\left( x \right) \right) = x \log_e x\]
\[\frac{f'\left( x \right)}{f\left( x \right)} = 1 + \log_e x\]
\[f'\left( x \right) = f\left( x \right)\left( 1 + \log_e x \right) = x^x \left( 1 + \log_e x \right)\]
\[\text { For f(x) to be decreasing, we must have }\]
\[f'\left( x \right) < 0\]
\[ \Rightarrow x^x \left( 1 + \log_e x \right) < 0\]
\[\text { Here, logaritmic function is defined for positive values of x } . \]
\[ \Rightarrow x^x > 0\]
\[ \Rightarrow 1 + \log_e x < 0 \left[ \text { Since } x^x > 0, x^x \left( 1 + \log_e x \right) < 0 \Rightarrow 1 + \log_e x < 0 \right] \]
\[ \Rightarrow \log_e x < - 1\]
\[ \Rightarrow x < e^{- 1} \left[ \because l {og}_a x < N \Rightarrow x < a^N \text { for }a > 1 \right]\]
\[\text { Here }, \]
\[e > 1\]
\[ \Rightarrow \log_e x < - 1 \Rightarrow x < e^{- 1} \]
\[ \Rightarrow x \in \left( 0, e^{- 1} \right)\]
\[\text { So,f(x) is decreasing on }\left( 0, \frac{1}{e} \right).\]
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