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.