Science:Math Exam Resources/Courses/MATH104/December 2015/Question 05 (a)/Solution 1

The linear approximation to a function $f(x)$ at ${\color {OliveGreen}a}$ is $f(x)\approx f({\color {OliveGreen}a})+f'({\color {OliveGreen}a})(x-{\color {OliveGreen}a}).$

In this problem, we are asked to use a linear approximation to $f(x)={\sqrt[{3}]{x}}.$ We choose to approximate $f$ at $a={\color {OliveGreen}27}$ because $27\approx 30$ and $f({\color {OliveGreen}a})={\sqrt[{3}]{\color {OliveGreen}27}}=3$ is exactly known and easy to work with. Thus our approximation is

$f(x)\approx f({\color {OliveGreen}27})+f'({\color {OliveGreen}27})(x-{\color {OliveGreen}27}).$

Now $f'(x)={\tfrac {1}{3}}x^{-2/3},$ so $f'(27)={\tfrac {1}{3}}\cdot 27^{-2/3}={\tfrac {1}{3}}\cdot (27^{1/3})^{-2}={\tfrac {1}{3}}\cdot 3^{-2}={\color {OrangeRed}{\tfrac {1}{27}}}.$ Hence

$f(x)\approx f(27)+{\color {OrangeRed}{\frac {1}{27}}}(x-27).$

Plugging in $x={\color {Orange}30}$ gives

${\begin{aligned}f({\color {Orange}30})={\sqrt[{3}]{\color {Orange}30}}&\approx f(27)+{\frac {1}{27}}({\color {Orange}30}-27)\\&={\sqrt[{3}]{27}}+{\frac {1}{27}}\cdot 3\\&=3+{\frac {3}{27}}\\&=3+{\frac {1}{9}}\approx {\color {blue}3.11}.\end{aligned}}$