y(x) = (9lnx)/x
y'(x) = 9(1-lnx)/x^2
y'(1) = 9
y'(e) = 0
Now just use the point-slope form of the lines.
Find the equation of the tangent line to the curve y = (9 ln x)/x at the points (1, 0) and (e, 9/e).
3 answers
y=0
s