Question 61512
Use calculus to determine the relative extrema of 
 f(x)=x-lnx 
DF/DX =1-1/X = 0 AT EXTREMUM
X =1
D^F/DX^2 = 1/X^2 = 1/1=+VE 
HENCE X=1 IS A MINIMUM
SLOPE OF TANGENT AT X=1 DY/DX AT X=1........THAT IS M = 0

AT X=1 ..Y = 1-LN1 =1
HENCE EQN.OF TANGENT AT X=1,Y=1 IS 
Y-1=M(X-1)=0
Y = 1 IS THE EQN. OF THE TANGENT AT X=1

Find the equation of the tangent line at x=1 
please help, i'm so confused!