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: $f'(x)={\frac {d}{dx}}x^{1/3}={\frac {1}{3}}x^{-2/3}$, and then $f''(x)={\frac {d}{dx}}{\frac {1}{3}}x^{-2/3}=-{\frac {2}{9}}x^{-5/3}$

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 $(7)^{1/3}\approx Q(7)=(8)^{1/3}+{\frac {1}{3}}(8)^{-2/3}(7-8)-{\frac {2}{9}}\cdot {\frac {8^{-5/3}}{2}}(7-8)^{2}$.