= Solution
For the <natural logarithm>, $f(1)=0$, $f'(1)=1$, $f''(1)=-1$, and $f^{(3)}(x)=2/x^3$. The quadratic <Taylor polynomial> therefore gives
$$
\ln(1.1)\approx0.1-\frac{0.1^2}{2}=\boxed{0.095}.
$$
The <Taylor remainder> is $R_3=2(0.1)^3/(6\xi^3)$ with $1<\xi<1.1$. It is positive, so the approximation is \b[an underestimate]. Moreover,
$$
\boxed{0<\ln(1.1)-0.095<\frac{(0.1)^3}{3}=\frac1{3000}.}
$$
Thus $1/3000\approx0.0003334$ is a valid upper bound on the <absolute error>.
Back to article page
