First consider the domain, x>0.

A=- 1 replacement

f(x)= lnx+x+2/x- 1x & gt； 0

When x= 1, f(x)=0+ 1+2- 1=2.

The derivative f'(x)= 1/x+ 1-(2/x squared) =(x+x squared -2)/x squared x>0.

F'(x)=0 when x= 1.

So the tangent is a straight line parallel to the X axis and passing through the (1, 2) point.

So the tangent equation is y=2.

2f' (x) =1/x+a-[(1-a)/x square ]=(x+ax square-1+a)/x square x>0.

X > 0

Let f' (x) = 0, then ax +x-( 1-a)=0.

Find x 1, x2 with the root formula, and consider the symbol of f'x in combination with parabolic images.