Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Solution 1

Let $f(x)=x^{\frac{1}{3}}$, then $f^{\prime }(x)=\frac{1}{3}{x}^{-\frac{2}{3}}$. Since we know that $f(27)=3$, this is where we will centre the Taylor polynomial. Then we have

$T_1(26)=f(27)+f^{\prime }(27)\cdot (26-27)=3+\frac{1}{27}\cdot (-1)=\frac{80}{27}.$

We conclude that $\sqrt[3]{26}\approx \frac{80}{27}$.