Browse Questions

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

$(A)\;x=e\quad(B)\;x=\frac{1}{e}\quad(C)\;x=1\quad(D)\;x=\sqrt e$

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
$f(x)=x^x$
Let $y=x^x$
taking log on both sides,
$\log y= x \log x$
differentiating w.r.t $x$ on both sides we get,
$\large\frac{1}{y}.\large\frac{dy}{dx}=x.\large\frac{1}{x}+ \log x.1$
$\Rightarrow \large\frac{dy}{dx}=y [ 1+ \log x]$
$x^x [1+ \log x]$
Step 2
when $\large\frac{dy}{dx}=0$
$\Rightarrow x^x[1+ \log x]=0$
i.e., $1+ \log x=0$
$\Rightarrow \log_ex=-1$
$e^{-1}=x$
$\Rightarrow x= \large\frac{1}{e}$
Hence $x^x$ has stationary point at $x=\large\frac{1}{e}$
Hence the correct option is B.