# The maximum value of $\Large\frac{1}{x}$ is:

$(A)\;e\quad(B)\;e^e\quad(C)\;\frac{1}{e^e}\quad(D)\;{\frac{1}{e}}^{\Large\frac{1}{e}}$

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$y= \bigg( \large\frac{1}{x} \bigg)^x$
take log on both sides.
$\log y=x\: \log \bigg( \large\frac{1}{x} \bigg)$
differentiating w.r.t $x$ on both sides we get,
$\large\frac{1}{y}. \large\frac{dy}{dx}=x. \bigg( \large\frac{1}{\large\frac{1}{x}}\bigg).\bigg( -\large\frac{1}{x^2} \bigg)+ \log \bigg( \large\frac{1}{x} \bigg).1$
$\large\frac{dy}{dx}=y \bigg( \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg)$
$=\bigg( \large\frac{1}{x} \bigg)^x \bigg[ \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg]$
when $\large\frac{dy}{dx}=0$
$\Rightarrow \bigg( \large\frac{1}{x} \bigg)^x \bigg[ \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg]$
But $\bigg( \large\frac{1}{x} \bigg)^x \neq 0$
Hence $\Rightarrow \log_e \large\frac{1}{x}-1=0$
$\Rightarrow \log_e \large\frac{1}{x}=1$
$e^1=\large\frac{1}{x}$
$\therefore x=\large\frac{1}{e}$
At $f'(x)=0 \: at\: x=\large\frac{1}{e}$
$f''(x) = \bigg[ -\large\frac{1}{x}.y \bigg]_{x=\large\frac{1}{e}}= -\large\frac{1}{x} \bigg( \large\frac{1}{x} \bigg)^x$
$=-e.(e)^{\large\frac{1}{e}} <0$
$\therefore \: At \: x=\large\frac{1}{e},\: y$ has maximum value and the maximum value of $f(x)\: is \: e^{\Large\frac{1}{e}}.$