Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 18

$$\eqalign{ & \left( a \right)L\left( x \right) = x + 1 \cr & \left( b \right){\text{graph}} \cr & \left( c \right)1.05 \cr & \left( d \right)0.12\% {\text{ error}} \cr} $$

$$\eqalign{ & f\left( x \right) = {e^x},{\text{ }}a = 0{\text{ at }}f\left( {0.05} \right) \cr & \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}} \right] \cr & f'\left( x \right) = {e^x} \cr & \cr & \left( a \right){\text{Use the linear approximation formula }}\left( {{\text{See page 287}}} \right) \cr & f\left( x \right) = L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Evaluate }}f\left( a \right){\text{ and }}f'\left( a \right) \cr & f\left( a \right) = f\left( 0 \right) = {e^0} = 1 \cr & f'\left( a \right) = f'\left( 0 \right) = {e^0} = 1 \cr & {\text{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & f\left( x \right) = L\left( x \right) = 1 + 1\left( {x - 0} \right) \cr & L\left( x \right) = x + 1 \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = 0{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}f\left( {0.05} \right) \cr & L\left( x \right) = x + 1 \cr & L\left( {0.05} \right) = 1.05 \cr & \cr & \left( d \right){\text{ The percent error is:}} \cr & \frac{{\left| {{\text{approximation}} - {\text{exact}}} \right|}}{{{\text{exact}}}} \times 100\% \cr & {\text{The exact value given by a calculator is }} \cr & f\left( {0.05} \right) = {e^{0.05}} \cr & f\left( {0.05} \right) = 1.051271096,{\text{ then}} \cr & \frac{{\left| {1.05 - 1.051271096} \right|}}{{1.051271096}} \times 100\% = 0.12\% {\text{ error}} \cr} $$