F(x)=xlnx, then g(x)=lnx+ 1.
According to Lagrange mean value theorem, f' (x) = (f (x1)-f (x2))/(x1-x2). (x takes the value between x 1 and x2).
The question is g (x) = (f (x1)-f (x2))/(x1-x2).
Simplify the formula needed below.
f(x 1)-f(x2)= g(x)*(x 1-x2)x ∈( x2，x 1)
Then there is
a & lt(g(x)*(x 1-x2))/(g(x 1)-g(x2))
Divide the numerator by x 1-x2 to get the denominator.
a & ltg(x)/(g(x 1)-g(x2))/(x 1-x2)
Pay attention to the denominator, and then use Lagrange's mean value theorem, which is g'(x).
g'(x)= 1/x. x∈ (x2，x 1)
a & ltg(x)/g'(x)
G (x) = lnx+1.g' (x) =1/x, then
A & lt(xlnx+x)min makes f (x) = xlnx+x.
Deduce F(x), f' (x) =1+(1+lnx) = 2+lnx.
When f' (x) = 0, the minimum value (which is also the minimum value of the function) is obtained.
At this time X = e (-2).
Calculate f (x) min =1/(E2)-2/(E2) =-1/(E2).
So a <-1/(e 2)
Are you studying Lagrange theorem in high school now? I didn't even have this in the college entrance examination a few years ago. . .