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

$L(x)=f(a)+f^{\prime }(a)(x-a)$

We can find the derivative:

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

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

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

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