Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 25

$$a = 1,{\text{ }}\ln \left( {1.05} \right) \approx 0.05$$

$$\eqalign{ & \ln \left( {1.05} \right) \cr & {\text{Using the function }}f\left( x \right) = \ln x,{\text{ }} \cr & f\left( x \right) = \ln x \cr & {\text{we have that }}1.05 = 1 + 0.05,{\text{ let }}a = 1 \cr & {\text{* Evaluate }}f\left( 1 \right) \cr & f\left( 1 \right) = \ln \left( 1 \right) \cr & f\left( 1 \right) = 0 \cr & \cr & {\text{Differentiating we obtain}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x} \right] \cr & f'\left( x \right) = \frac{1}{x} \cr & {\text{* Evaluate }}f'\left( 1 \right) \cr & f'\left( 1 \right) = \frac{1}{1} \cr & f'\left( 1 \right) = 1 \cr & \cr & {\text{Using the definition of a linear approximation to }}f{\text{ at }}a{\text{ }}\left( {page{\text{ 282}}} \right) \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right),{\text{ for an interval }}I{\text{ containing }}a \cr & \cr & {\text{The linear approximation }}L\left( x \right){\text{ at }}a{\text{ is:}} \cr & L\left( x \right) = 0 + 1\left( {x - 1} \right) \cr & L\left( x \right) = x - 1 \cr & \cr & {\text{Estimating the given value at }}f\left( {1.05} \right) \cr & L\left( x \right) = 1 - x \cr & L\left( {1.05} \right) = 1.05 - 1 \cr & L\left( {1.05} \right) = 0.05 \cr & {\text{Therefore}}{\text{,}} \cr & \ln \left( {1.05} \right) \approx L\left( {1.05} \right) \approx 0.05 \cr} $$