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

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

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

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 $| f (2) - T_{2} (2) | \leq \frac{4 \cdot 6}{3 \cdot 8} = 1$ , as required.