Question

The minimum value of the function $f(x)=x\log x$ is

• A. $-\frac{1}{e}$
• B. $-e$
• C. $\frac{1}{e}$
• D. e

Answer 1

Answer: (A)

We have,
$f(x)=x\log x$
$f^{\prime }(x)=1+\log x$
For maxima or minima, put $f^{\prime }(x)=0$
$\therefore 1+\log x=0$
$\Rightarrow \log x=-1$
$\Rightarrow x=e^{-1}$
Now,
$f^{\prime \prime }(x)=\frac{1}{x}$
$\Rightarrow f^{\prime \prime }\left(e^{-1}\right)=\frac{1}{e^{-1}}=e>0$
$\therefore f(x)$ is decreasing. Hence, minimum value of
$f(x)$ at $x=e^{-1}$ is
$f\left(e^{-1}\right)=e^{-1}\log e^{-1}=\frac{1}{e}(-\log e)=-\frac{1}{e}$