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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
asked in Calculus by Veteran (47.9k points)
edited by | 227 views

Just observe loge(x) curve carefully..

2 Answers

+6 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 ans is c
answered by Veteran (34.3k 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.
answered by Boss (5.9k points)

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