Posted by Jake on .
for f(x)=lnx we know that f(e)=1. Use the tangent line at(e,1) to compute the tangent line approximation of f(3). what does the difference between the actual value and the tangent line approximation have to do with the concavity (2nd derivative) of the function at (e,1)?

ap calculus ab 
Steve,
f'(x) = 1/x
so, at (e,1), the slope is 1/e
the line is thus
(y1)/(xe) = 1/e
y = 1/e (xe) + 1
y = x/e
at x=3, the line has yvalue
y = 3/e
y = 1.104
Actual value of ln3 = 1.099
Since the curve is concave down, it will lie below the tangent line, so any linear approximation using just the tangent line will be too high.