If $f(x)=x^{\mathrm{x}}$, then the interval in which $f(x)$ decreases is

If $f(x)=x^{\mathrm{x}}$, then the interval in which $f(x)$ decreases is
  1. $\left[0, \frac{1}{e}\right]$
  2. $[0, e]$
  3. $\left[\frac{1}{e}, \infty\right]$
  4. $\left[0, e^e\right]$

Solution

$\begin{aligned} & \text { } f(x)=x^x \Rightarrow y=x^x \\ & \log y=x \log x \\ & \frac{1}{y} \frac{d y}{d x}=\log x+1 \Rightarrow \frac{d y}{d x}=x^x(1+\log x) \end{aligned}$
Since, $x^x$ is always +ve $\therefore f(x)$ decreases when $1+\log \leq 0$ $\Rightarrow \log x \leq 1 \Rightarrow x \leq \frac{1}{e}$ $\therefore f(x)$ decreases when $x \in\left[0, \frac{1}{e}\right]$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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