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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 | 146 views

Just observe loge(x) curve carefully..

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
we need the function to be strictly increasing also :)
In given interval f'(x)>0 so function is increasing....
yes, here it is correct :)
$f'(x) = 1 + ln x$

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.