Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 28

$$a = 100,{\text{ }}\frac{1}{{\sqrt {119} }} \approx 0.0905$$

$$\eqalign{ & \frac{1}{{\sqrt {119} }} \cr & {\text{Rewrite }}\frac{1}{{\sqrt {119} }}{\text{ as }}\frac{1}{{\sqrt {100 + 19} }} \cr & {\text{Using the function }}f\left( x \right) = \frac{1}{{\sqrt x }},{\text{ }} \cr & f\left( x \right) = \frac{1}{{\sqrt x }} \cr & {\text{we have that }}119 = 100 + 19,{\text{ let }}a = 100 \cr & {\text{* Evaluate }}f\left( {100} \right) \cr & f\left( {100} \right) = \frac{1}{{\sqrt {100} }} \cr & f\left( {100} \right) = \frac{1}{{10}} \cr & \cr & {\text{Differentiating we obtain}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{\sqrt x }}} \right] = \frac{d}{{dx}}\left[ {{x^{ - 1/2}}} \right] \cr & f'\left( x \right) = - \frac{1}{2}{x^{ - 3/2}} \cr & {\text{* Evaluate }}f'\left( {100} \right) \cr & f'\left( {100} \right) = - \frac{1}{2}{\left( {100} \right)^{ - 3/2}} \cr & f'\left( {100} \right) = - \frac{1}{{2000}} \cr & \cr & {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }} \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr & \cr & {\text{The linear approximation }}L\left( x \right){\text{ at }}a{\text{ is:}} \cr & L\left( x \right) = \frac{1}{{10}} - \frac{1}{{2000}}\left( {x - 100} \right) \cr & \cr & {\text{Estimating the given value at }}f\left( {119} \right) \cr & L\left( {119} \right) = \frac{1}{{10}} - \frac{1}{{2000}}\left( {119 - 100} \right) \cr & L\left( {119} \right) = 0.0905 \cr & {\text{Therefore}}{\text{,}} \cr & \frac{1}{{\sqrt {119} }} \approx L\left( {119} \right) \approx 0.0905 \cr} $$