Let f ( x ) = x 1 3 {\displaystyle f(x)=x^{\frac {1}{3}}} , then f ′ ( x ) = 1 3 x − 2 3 {\displaystyle f'(x)={\frac {1}{3}}x^{-{\frac {2}{3}}}} . Since we know that f ( 27 ) = 3 {\displaystyle f(27)=3} , this is where we will centre the Taylor polynomial. Then we have
T 1 ( 26 ) = f ( 27 ) + f ′ ( 27 ) ⋅ ( 26 − 27 ) = 3 + 1 27 ⋅ ( − 1 ) = 80 27 . {\displaystyle T_{1}(26)=f(27)+f'(27)\cdot (26-27)=3+{\frac {1}{27}}\cdot (-1)={\frac {80}{27}}.}
We conclude that 26 3 ≈ 80 27 {\displaystyle \color {blue}{\sqrt[{3}]{26}}\approx {\frac {80}{27}}} .
![{\displaystyle \color {blue}{\sqrt[{3}]{26}}\approx {\frac {80}{27}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7fcbdda8a077ef1e3e5f70b7932b75d44e4314d3)