Answer
$${\text{a}}){f_L}\left( x \right) = 8 - 0.3x{\text{ b}}){\text{ }}{f_L}\left( {10.4} \right) = 4.88$$
Work Step by Step
$$\eqalign{
& {\text{We have}}: \cr
& f\left( {10} \right) = 5,{\text{ }}f'\left( {10} \right) = - 0.3;{\text{ }}x = 10.4 \cr
& \cr
& \left( {\bf{a}} \right){\text{Using the formula }}\left( {{\text{see page 253}}} \right){\text{ }} \cr
& {f_L}\left( x \right) = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr
& {\text{Let }}c = 10 \cr
& {f_L}\left( x \right) = f\left( {10} \right) + f'\left( {10} \right)\left( {x - 10} \right) \cr
& {\text{replace }}f\left( {10} \right){\text{ and }}f'\left( {10} \right) \cr
& {f_L}\left( x \right) = 5 - 0.3\left( {x - 10} \right) \cr
& {\text{multiply}} \cr
& {f_L}\left( x \right) = 5 - 0.3x + 3 \cr
& {f_L}\left( x \right) = 8 - 0.3x \cr
& \cr
& \left( {\bf{b}} \right){\text{use the linearization to estimate }}f\left( x \right){\text{ at }}x = 10.4 \cr
& {f_L}\left( {10.4} \right) = 8 - 0.3\left( {10.4} \right) \cr
& {f_L}\left( {10.4} \right) = 4.88 \cr} $$
