Compute the first 3 derivatives of f(x):
[tex]f(x)=\ln x\implies f'(x)=\dfrac1x\implies f''(x)=-\dfrac1{x^2}\implies f'''(x)=\dfrac2{x^3}[/tex]
The 3rd degree Taylor polynomial about [tex]x=1[/tex] is then
[tex]T_3(x)=f(1)+\dfrac{f'(1)}{1!}(x-1)+\dfrac{f''(1)}{2!}(x-1)^2+\dfrac{f'''(1)}{3!}(x-1)^3[/tex]
[tex]T_3(x)=(x-1)-(x-1)^2+2(x-1)^3[/tex]
[tex]T_3(x)=-4+9x-7x^2+2x^3[/tex]
Now use [tex]T_3[/tex] to approximate ln(1.2):
[tex]f(x)\approx T_3(x)\implies \ln1.2\approx T_3(1.2)=\boxed{0.17600}[/tex]
(Compare to the actual value, closer to 0.18232)
