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 $f'(x)={\frac {d}{dx}}x^{1/3}={\frac {1}{3}}x^{-2/3}$, $f''(x)={\frac {d}{dx}}{\frac {1}{3}}x^{-2/3}=-{\frac {2}{9}}x^{-5/3}$, and finally $f^{(3)}(x)={\frac {d}{dx}}\left(-{\frac {2}{9}}x^{-5/3}\right)={\frac {10}{27}}x^{-8/3}$.

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