Science:Math Exam Resources/Courses/MATH110/April 2019/Question 07 (c)/Solution 1

We are asked to find the Taylor polynomial of degree 3 for $f (x) = x^{1 / 3}$ centred at $a = 8$ . The formula for this quadratic approximation, which we denote by $C (x)$ , is given by $C (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2} (x - a)^{2} + \frac{f^{(3)} (a)}{6} (x - a)^{3}$ . Remember that $f^{(3)}$ is the third derivative of $f$ . We already know that $a = 8, f (x) = x^{1 / 3}$ , so we need only find $f^{'} (a), f^{″} (a), f^{(3)} (a)$ . To do so, we apply the power rule three times, and find that $f^{'} (x) = \frac{d}{d x} x^{1 / 3} = \frac{1}{3} x^{- 2 / 3}$ , $f^{″} (x) = \frac{d}{d x} \frac{1}{3} x^{- 2 / 3} = - \frac{2}{9} x^{- 5 / 3}$ , and finally $f^{(3)} (x) = \frac{d}{d x} (- \frac{2}{9} x^{- 5 / 3}) = \frac{10}{27} x^{- 8 / 3}$ .

Substituting this information into our formula for $C (x)$ , we have $C (x) = (8)^{1 / 3} + \frac{1}{3} (8)^{- 2 / 3} (x - 8) - \frac{2}{9} \cdot \frac{8^{- 5 / 3}}{2} (x - 8)^{2} + \frac{10}{27} \cdot \frac{8^{- 8 / 3}}{6} (x - 8)^{3}$ .