Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 13

$$\eqalign{ & \left( a \right)L\left( x \right) = - 4x + 16 \cr & \left( b \right){\text{graph}} \cr & \left( c \right)7.6 \cr & \left( d \right)0.13\% {\text{ error}} \cr} $$

$$\eqalign{ & f\left( x \right) = 12 - {x^2},{\text{ }}a = 2{\text{ at }}f\left( {2.1} \right) \cr & \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {12 - {x^2}} \right] \cr & f'\left( x \right) = - 2x \cr & \cr & \left( a \right){\text{Use the linear approximation formula }}\left( {{\text{See page 287}}} \right) \cr & f\left( x \right) = L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Evaluate }}f\left( a \right){\text{ and }}f'\left( a \right) \cr & f\left( a \right) = f\left( 2 \right) = f\left( 2 \right) = 12 - {\left( 2 \right)^2} = 8 \cr & f'\left( a \right) = f'\left( 2 \right) = - 2\left( 2 \right) = - 4 \cr & {\text{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & f\left( x \right) = L\left( x \right) = 8 + \left( { - 4} \right)\left( {x - 2} \right) \cr & L\left( x \right) = 8 - 4\left( {x - 2} \right) \cr & L\left( x \right) = 8 - 4x + 8 \cr & L\left( x \right) = - 4x + 16 \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = a{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}f\left( {2.1} \right) \cr & L\left( {2.1} \right) = - 4\left( {2.1} \right) + 16 \cr & L\left( {2.1} \right) = 7.6 \cr & \cr & \left( d \right){\text{ The percent error is:}} \cr & \frac{{\left| {{\text{approximation}} - {\text{exact}}} \right|}}{{{\text{exact}}}} \times 100\% \cr & {\text{The exact value given by a calculator is }} \cr & f\left( {2.1} \right) = 12 - {\left( {2.1} \right)^2} \cr & f\left( {2.1} \right) = 7.59,{\text{ then}} \cr & \frac{{\left| {7.6 - 7.59} \right|}}{{7.59}} \times 100\% = 0.13\% {\text{ error}} \cr} $$