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

We are asked to use the quadratic approximation of $f (x) = x^{1 / 3}$ at the point $a = 8$ to approximate $(7)^{1 / 3}$ . The formula for this quadratic approximation, which we denote by $Q (x)$ , is given by $Q (x) = f (a) + f^{'} (a) (x - a) + \frac{f^{″} (a)}{2} (x - a)^{2}$ . We already know that $a = 8, f (x) = x^{1 / 3}$ , so we need only find $f^{'} (a), f^{″} (a)$ . To do so, we apply the power rule twice, and find that: $f^{'} (x) = \frac{d}{d x} x^{1 / 3} = \frac{1}{3} x^{- 2 / 3}$ , and then $f^{″} (x) = \frac{d}{d x} \frac{1}{3} x^{- 2 / 3} = - \frac{2}{9} x^{- 5 / 3}$

Substituting this information into our formula for $Q (x)$ , we have $Q (x) = (8)^{1 / 3} + \frac{1}{3} (8)^{- 2 / 3} (x - 8) - \frac{2}{9} \cdot \frac{8^{- 5 / 3}}{2} (x - 8)^{2}$ . Then, to estimate the value of $7^{1 / 3}$ , we plug-in $x = 7$ into our quadratic approximation. This gives $(7)^{1 / 3} \approx Q (7) = (8)^{1 / 3} + \frac{1}{3} (8)^{- 2 / 3} (7 - 8) - \frac{2}{9} \cdot \frac{8^{- 5 / 3}}{2} (7 - 8)^{2}$ .