Let f (x) = x loge x.
Clearly,
f(x) is only defined for x > 0.
For minimum value;
f’(x) = 0 and f’’(x) > 0.
f’(x) = 1 + ln x
⇒ x = 1/e
f’’(x) = 1/x,
Clearly,
f’’(x) > 0 for all x.
So, only minima is defined.
So, minima of f (x) is at x = 1/e
f(\(\frac{1}{e}\)) = \(-\frac{1}{e}\)