Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 21

$$a = 200,{\text{ }}\frac{1}{{203}} \approx 0.004925$$

$$\eqalign{ & \frac{1}{{203}} \cr & \cr & {\text{Using the function }}f\left( x \right) = \frac{1}{x},{\text{ }} \cr & f\left( x \right) = \frac{1}{x} \cr & {\text{we have that }}\frac{1}{{203}} = \frac{1}{{200 + 3}},{\text{ let }}a = 200 \cr & {\text{* Evaluate }}f\left( {200} \right) \cr & {\text{ }}f\left( {200} \right) = \frac{1}{{200}} \cr & \cr & {\text{Differentiating we obtain}} \cr & f'\left( x \right) = - {x^{ - 2}} \cr & f'\left( x \right) = - \frac{1}{{{x^2}}} \cr & {\text{* Evaluate }}f'\left( {200} \right) \cr & f'\left( {200} \right) = - \frac{1}{{{{\left( {200} \right)}^2}}} \cr & \cr & {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }}\left( {page{\text{ 282}}} \right) \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr & \cr & {\text{The linear approximation }}L\left( x \right){\text{ at }}a{\text{ is:}} \cr & L\left( x \right) = \frac{1}{{200}} - \frac{1}{{{{\left( {200} \right)}^2}}}\left( {x - 200} \right) \cr & \cr & {\text{Estimating the given value }}\frac{1}{{203}} = \frac{1}{{200 + 3}}{\text{ at }}f\left( {203} \right) \cr & L\left( {203} \right) = \frac{1}{{200}} - \frac{1}{{{{\left( {200} \right)}^2}}}\left( {203 - 200} \right) \cr & L\left( {203} \right) = \frac{1}{{200}} - \frac{3}{{40000}} \cr & L\left( {203} \right) = 4.925 \times {10^{ - 3}} \cr & L\left( {203} \right) = 0.004925 \cr & {\text{Therefore}}{\text{,}} \cr & \frac{1}{{203}} \approx 0.004925 \cr} $$