Chapter 3 - Applications of Differentiation - 3.9 Exercises - Page 236: 3

\[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{24.760}&{31.207}&{32}&{32.808}&{40.841} \\ {T\left( x \right)}&{24}&{31.2}&{32}&{32.8}&{40} \end{array}}\]

$$\eqalign{ & f\left( x \right) = {x^5},{\text{ }}\left( {2,32} \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = 5{x^4} \cr & f'\left( 2 \right) = 5{\left( 2 \right)^4} = 80 \cr & {\text{The equation for the tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is:}} \cr & y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr & {\text{We have the point }}\left( {2,32} \right) \to c = 2,{\text{ }}f\left( c \right) = 32,{\text{ and }}f'\left( c \right) = 80 \cr & y = 32 + 80\left( {x - 2} \right) \cr & y = 32 + 80x - 160 \cr & y = 80x - 128 \cr & T\left( x \right) = 80x - 128 \cr & {\text{Completing the table for }}f\left( x \right){\text{:}} \cr & x = 1.9 \to f\left( {1.9} \right) = {\left( {1.9} \right)^5} = 24.760 \cr & x = 1.99 \to f\left( {1.99} \right) = {\left( {1.99} \right)^5} = 31.207 \cr & x = 2 \to f\left( 2 \right) = {\left( 2 \right)^5} = 32 \cr & x = 2.01 \to f\left( {2.01} \right) = {\left( {2.01} \right)^5} = 32.808 \cr & x = 2.1 \to f\left( {2.1} \right) = {\left( {2.1} \right)^5} = 40.841 \cr & {\text{Completing the table for }}T\left( x \right){\text{:}} \cr & x = 1.9 \to T\left( {1.9} \right) = 80\left( {1.9} \right) - 128 = 24 \cr & x = 1.99 \to T\left( {1.99} \right) = 80\left( {1.99} \right) - 128 = 31.2 \cr & x = 2 \to T\left( 2 \right) = 80\left( 2 \right) - 128 = 32 \cr & x = 2.01 \to T\left( {2.01} \right) = 80\left( {2.01} \right) - 128 = 32.8 \cr & x = 2.1 \to T\left( {2.1} \right) = 80\left( {2.1} \right) - 128 = 40 \cr & \cr & {\text{Therefore}} \cr} $$ \[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{24.760}&{31.207}&{32}&{32.808}&{40.841} \\ {T\left( x \right)}&{24}&{31.2}&{32}&{32.8}&{40} \end{array}}\]