f(x)= xlnx-xe^x impliesf(x)=xlnx−xex⇒
g(x) equiv f^'(x) = 1+lnx - (x+1)e^x
For x to be be a local extremum, g(x) must be zero. We will now show that this does not occur for any real value of x.
Note that
g^'(x) = 1/x-(x+2)e^x,qquad g^{''}(x) = -1/x^2-(x+3)e^x
Thus g^'(x) will vanish if
e^x = 1/(x(x+2))
This is a transcendental equation which can be solved numerically. Since g^'(0) = +oo and g^'(1)=1-3e<0, the root lies between 0 and 1. And since g^{''}(0) <0 for all positive x, this is the only root and it corresponds to a maximum for g(x)
It is quite easy to solve the equation numerically, and this shows that g(x) has a maximum at x=0.3152 and the maximum value is g(0.3152) = -1.957. Since the maximum value of g(x) is negative, there is no value of x at which g(x) vanishes.
It may be instructive to look at this graphically:
graph{xlog(x)-xe^x [-0.105, 1, -1.175, 0.075]}
As you can see from the graph above, the function f(x) actually has a maximum at x=0 - but this is not a local maximum. The graph below shows that g(x)equiv f^'(x) never takes the value zero.
graph{1+log(x)-(x+1)*e^x [-0.105, 1, -3, 0.075]}
