f(x)=xlogexmonotonically decreases in
(0,1/e)
(1/e,1)
(1,∞)
(1/e,∞)
f(x)=xlogex For g(x)=x logex,g′(x)=x1x+logex=1+logexThus, g (x) increases for 1e,∞and decreases for 0,1e.
From the graph,f(x)=xlogex decreases in 0, 1e.