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  • f (x) equals x^x has a stationary point at A. x =e

f (x) = x^x has a stationary point at
A. x = e
B. x=\frac{1}{e}
C. x = 1
D. x=\sqrt e

Answers (1)

Given equation is f (x) = x^x

Let y= x^x………(i)

Take logarithm on both side

\log y=\log (x^x)

⇒ log y=x log x

Apply first derivative and get

\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(\mathrm{x} \log \mathrm{x})}{\mathrm{d} \mathrm{x}}$
Apply product rule and get
\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{d} \mathrm{x}}+\log \mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{d} \mathrm{x}}$

Apply first derivative and get
\\ \frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{x}+\log x \cdot 1$ \\$\Rightarrow \frac{d y}{d x}=y(1+\log x)$
Substitute value of y from (i) and get
\Rightarrow \frac{d y}{d x}=x^{x}(1+\log x)$

Now we find the critical point by equating (i) to 0 and get

\\ {x\textsuperscript{x} (1+log x)=0}\\ { \Rightarrow 1+log x =0 \: \: as \: \: x\textsuperscript{x}$ cannot be equal to 0}\\

\\ { \Rightarrow log x=-1}\\ {But -1=log e\textsuperscript{-1}}\\ { \Rightarrow log x=log e\textsuperscript{-1}}\\

Equate the terms and get

x= e\textsuperscript{-1}

\Rightarrow X=\frac{1}{e}

Therefore f(x) has a stationary point at X=\frac{1}{e}

So, the correct answer is option B.

 

Posted by

infoexpert22

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