Let ƒ (x)=x1/3.
First we need a number a as close to 30 as possible but whose cube root we know. A good candidate would be 27, because 33=27 and so
.
We know that the general formula for a linear approximation is given by
![{\displaystyle \displaystyle L(x)=f(a)+f'(a)(x-a)}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/5638886bee9d22396557ecdc8341f611897116b6)
We can find the derivative:
![{\displaystyle \displaystyle f'(x)={\frac {x^{-2/3}}{3}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b5a0525e626c1ee98287ca8453556921511b8b6b)
Knowing these values, we can plug them in to the linear approximation equation and find the approximation for x=30, using a=27:
![{\displaystyle \displaystyle {\begin{aligned}L(x)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(x-27)\\L(30)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(30-27)\\&=3+{\frac {1/9}{3}}(3)\\&=3+{\frac {1}{9}}\\&={\frac {28}{9}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/807fdf0e9f26c42526904d8b2f58e437c19d5c84)
So using linear approximation,