Posted by **Jake** on Sunday, January 8, 2012 at 4:43pm.

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**, Sunday, January 8, 2012 at 5:54pm
f'(x) = 1/x

so, at (e,1), the slope is 1/e

the line is thus

(y-1)/(x-e) = 1/e

y = 1/e (x-e) + 1

y = x/e

at x=3, the line has y-value

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.

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