Science:Math Exam Resources/Courses/MATH110/April 2017/Question 08 (b)/Solution 1

To estimate $f(0.1)$, we compute $T_2(0.1)$ since $T_2(0.1)$ is an approximation of $f(0.1)$.

$T_2(0.1)=3+0.1-8\times (0.1{)}^2=3+0.1-8\times 0.01=3+0.1-0.08=3.02$

It follows that the approximation for $f(0.1)$ is

$3.02$