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+6 votes

The minimum of the function $f(x) = x \log_{e}(x)$ over the interval $[\frac{1}{2}, \infty )$ is

  1. $0$
  2. $-e$
  3. $\frac{-\log_{e}(2)}{2}$
  4. $\frac{-1}{e}$
  5. None of the above
in Calculus by Boss (29.7k points)
edited by | 346 views

Just observe loge(x) curve carefully..


we know $log\ x$ is a monotonically increasing function. So we need to minimize only $x$ and $x$ will be minimum at $\frac{1}{2}$

So find $f(\frac{1}{2})$


           $=\frac{1}{2}(log_e\ 1-log_e\ 2)$

           $=\frac{-log_e\ 2}{2}$

2 Answers

+8 votes
Minimum value of function occurs at end points or critical points

$f'(x)=1+\log x$

Equate it to $40$

$x=\dfrac {1}{e}$


Put $x=\dfrac{1}{e}$   $f''(x)=e$ so minima at $\dfrac {1}{e}.$

But $\dfrac{1}{e}=0.36$

But $x\in \left[\frac{1}{2},\infty \right]$

So min occurs at $\frac{1}{2}$

So min value=$\frac{1}{2} \log \frac{1}{2}=\frac{1}{2}* {\log_{e} -2}$

So, answer is $C$
by Boss (31k points)
edited by
we need the function to be strictly increasing also :)
In given interval f'(x)>0 so function is increasing....
yes, here it is correct :)
0 votes
$f'(x) = 1 + ln x$

Now, $f''(x) = \frac{1}{x}$ , if we put $x = \frac{1}{2}$ , it will be greater than 0 , so we have minima here. Minimum value will be $\frac{1}{2}*[ln1-ln2] = \frac{1}{2}*[0-ln2] = \frac{1}{2}*[-ln2]$


option c .

Please correct me if I am wrong.
by Active (3.7k points)

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