3. 10 points Find the point on the graph of f(x) = ln x whose tangent line goes through the point (0, 1).

Question

3. 10 points Find the point on the graph of f(x) = ln x whose tangent line goes through the point (0, 1).

Answer 1

f(x) = lnx

f'(x) = 1/x = x^-1

y= lnx= lne^2 = 2lne = 2

the point on the f(x) curve is (e^2, 2) = about (7.389, 2)

It's the point of tangency with the line through (0,1) and (e^2, 2)

the line is y = x/e^2 +1

slope = 1/e^2 = 1/x where x= e^2 = about 7.389

the line is

y= about x/7.389 +1

= about y= 0.1353x + 1

or about y= 0.14x + 1