If f(x) = x ln(1+x) estimate the value of ln (2.1) given that ln2 = 0.6931.
f'(x) = ln(1+x) + x/(1+x)
f'(1) = ln2 + 1/2 = 1.1931
That is the slope of the tangent line at x=1
f(1) = ln2 = 0.6931
Now we can extrapolate to y=f(1.1) = 1.1 ln2.1
(y-f(1))/(1.1-1) = 1.1931
y - .6931 = .11931
y = .8124 = 1.1 * ln2.1
ln 2.1 = 0.7385
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check: ln 2.1 = .7419