Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 15

$$\eqalign{ & \left( a \right)L\left( x \right) = x \cr & \left( b \right){\text{graph}} \cr & \left( c \right)0.9 \cr & \left( d \right)40\% {\text{ error}} \cr} $$

$$\eqalign{ & f\left( x \right) = \ln \left( {1 + x} \right),{\text{ }}a = 0{\text{ at }}f\left( {0.9} \right) \cr & \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {1 + x} \right)} \right] \cr & f'\left( x \right) = \frac{1}{{1 + x}} \cr & \cr & \left( a \right){\text{Use the linear approximation formula }}\left( {{\text{See page 287}}} \right) \cr & f\left( x \right) = L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Evaluate }}f\left( a \right){\text{ and }}f'\left( a \right) \cr & f\left( a \right) = f\left( 0 \right) = \ln \left( {1 + 0} \right) = 0 \cr & f'\left( a \right) = f'\left( 0 \right) = \frac{1}{{1 + 0}} = 1 \cr & {\text{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & f\left( x \right) = L\left( x \right) = 0 + 1\left( {x - 0} \right) \cr & L\left( x \right) = x \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = 0{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}f\left( {0.9} \right) \cr & L\left( {0.9} \right) = 0.9 \cr & \cr & \left( d \right){\text{ The percent error is:}} \cr & \frac{{\left| {{\text{approximation}} - {\text{exact}}} \right|}}{{{\text{exact}}}} \times 100\% \cr & {\text{The exact value given by a calculator is }} \cr & f\left( {0.9} \right) = \ln \left( {1 + 0.9} \right) \cr & f\left( {0.9} \right) = 0.6418538862,{\text{ then}} \cr & \frac{{\left| {0.9 - 0.6418538862} \right|}}{{0.6418538862}} \times 100\% = 40\% {\text{ error}} \cr} $$