Science:Math Exam Resources/Courses/MATH104/December 2014/Question 01 (m)/Solution 1

We are looking at the linear approximation of $f(x)=\sin x$ about the point $0$. Hence $a=0$ and the linear approximation is given by

${\begin{aligned}L(x)&=f(a)+f'(a)(x-a)\\&=\sin(0)+\cos(0)(x-0)\\&=0+(1)(x)\\&=x\end{aligned}}$

We now want to estimate the error of this approximation at the point $x=0.12$. From the formula for the error of a linear approximation we find that

${\begin{aligned}|\sin(0.12)-0.12|&=|\sin(0.12)-L(0.12)|\\&=|f(x)-L(x)|\\&\leq {\frac {M(x-a)^{2}}{2}}\\&={\frac {M(0.12-0)^{2}}{2}}\\&=0.0072M\end{aligned}}$

Finally it remains to calculate $M$, which is the largest absolute value of the second derivative of $f(x)$ in the interval $[a,x]=[0,0.12]$. Doing the math we obtain $|f''(x)|=|-\sin(x)|\leq \sin(0.12)$ (the number 1 could also be used as an upper bound though it is not as tight of an upper bound), therefore the final answer is

$|\sin(0.12)-0.12|\leq 0.0072M=0.0072\sin(0.12)$