Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (i)/Solution 1

Let ƒ (x)=x^1/3. 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 $27^{1/3}=3$.

We know that the general formula for a linear approximation is given by

$\displaystyle L(x)=f(a)+f'(a)(x-a)$

We can find the derivative:

$\displaystyle f'(x)={\frac {x^{-2/3}}{3}}$

Knowing these values, we can plug them in to the linear approximation equation and find the approximation for x=30, using a=27:

$\displaystyle {\begin{aligned}L(x)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(x-27)\\L(30)&=(27)^{1/3}+{\frac {27^{-2/3}}{3}}(30-27)\\&=3+{\frac {1/9}{3}}(3)\\&=3+{\frac {1}{9}}\\&={\frac {28}{9}}\end{aligned}}$

So using linear approximation, $(30)^{1/3}\approx 28/9$