f(x) = x^x has a stationary point at: - Vidyadip

Question

$f(x) = x^x$ has a stationary point at:

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Answer

We have, $f(x) = x^x$
Let us suppose $y = x^x$
Taking logarithm on both sides, we get
$\log\text{y}=\text{x}\log\text{x}$
$\therefore\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\cdot1$ $\big[\because(\text{fg})'=\text{fg}'+\text{gf}'\big]$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(1+\log\text{x})\cdot\text{x}^\text{x}$
Find the critical points by equating $\frac{\text{dy}}{\text{dx}}$ to $0.$
$\therefore\ \frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\ (1+\log\text{x})\text{x}^\text{x}=0$
$\Rightarrow\ \log\text{x}=-1$ as $\text{x}^\text{x}\neq0$
$\Rightarrow\ \log\text{x}=\log\text{e}^{-1}$
$\Rightarrow\ \text{x}=\text{e}^{-1}$
$\Rightarrow\ \text{x}=\frac{1}{\text{e}}$
Hence, f(x) has a stationary point at $\text{x}=\frac{1}{\text{e}}.$

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