Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 16

$$\eqalign{ & \left( a \right)L\left( x \right) = \frac{1}{4}x + \frac{1}{4} \cr & \left( b \right){\text{graph}} \cr & \left( c \right)0.525 \cr & \left( d \right)0.22\% {\text{ error}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{x}{{x + 1}},{\text{ }}a = 1{\text{ at }}f\left( {1.1} \right) \cr & \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{x + 1}}} \right] \cr & f'\left( x \right) = \frac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}} \cr & f'\left( x \right) = \frac{1}{{{{\left( {x + 1} \right)}^2}}} \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( 1 \right) = \frac{1}{2} \cr & f'\left( a \right) = f'\left( 1 \right) = \frac{1}{{{{\left( {1 + 1} \right)}^2}}} = \frac{1}{4} \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) = \frac{1}{2} + \frac{1}{4}\left( {x - 1} \right) \cr & L\left( x \right) = \frac{1}{2} + \frac{1}{4}x - \frac{1}{4} \cr & L\left( x \right) = \frac{1}{4}x + \frac{1}{4} \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = 1{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}f\left( {1.1} \right) \cr & L\left( {1.1} \right) = \frac{1}{4}\left( {1.1} \right) + \frac{1}{4} \cr & L\left( {1.1} \right) = 0.525 \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( {1.1} \right) = \frac{{1.1}}{{1 + 1.1}} \cr & f\left( {1.1} \right) = 0.52338095238,{\text{ then}} \cr & \frac{{\left| {0.525 - 0.52338095238} \right|}}{{0.52338095238}} \times 100\% = 0.22\% {\text{ error}} \cr} $$