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

  1. $0$
  2. $-e$
  3. $-\log_{e}(2)/2$
  4. $-1/e$
  5. None of the above
asked in Calculus by Veteran (29k points)   | 144 views

Just observe loge(x) curve carefully..

2 Answers

+3 votes
Minimum value of function occurs at end points or critical points

f'(x)=1+logx

Equate it to 0

x=1/e

f''(x)=1/x

Put x=1/e f''(x)=e so minima at 1/e

But 1/e=0.36

But x∈[1/2,infinity)

So min occurs at 1/2

So min value=1/2 log 1/2

So ans is c
answered by Veteran (30.7k points)  
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 (5.3k points)  
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