Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Solution 1

We know that $f$ is at least three times differentiable, so we will use the second degree Maclaurin polynomial, which is

$A=T_{2}(x)=f(0)+f'(0)x+{\frac {f''(0)}{2}}\cdot x^{2}=1-2x+{\frac {3x^{2}}{2}}.$

That is, $f(2)$ is approximated by $\color {blue}T(2)=1-2\cdot 2+{\frac {3\cdot 2^{2}}{2}}=3$.

To estimate the error of our approximation, by the Lagrange Remainder Theorem we have

$f(2)-T_{2}(2)={\frac {f'''(c)}{3!}}\cdot (2-0)^{3}={\frac {\frac {6-c^{2}}{9-\cos(c)}}{6}}\cdot 8={\frac {4(6-c^{2})}{3(9-\cos(c))}},$

where $c$ is in the interval $[0,2]$. Note that this expression is always positive for $c$ in $[0,2]$. Furthermore, the numerator is decreasing on this interval and the denominator attains its minimum at $c=0$, which implies that this error is maximised at $c=0$. We conclude that the error $\color {blue}|f(2)-T_{2}(2)|\leq {\frac {4\cdot 6}{3\cdot 8}}=1$, as required.