Chapter 3 - Applications of Differentiation - 3.9 Exercises - Page 236: 6

\[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.0567}&{1.0947}&{1.0997}&{1.1048}&{1.5840} \\ {T\left( x \right)}&{1.0494}&{1.0947}&{1.0997}&{1.1047}&{1.1500} \end{array}}\]

$$\eqalign{ & \left( x \right) = \csc x,{\text{ }}\left( {2,\csc 2} \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = - \csc x\cot x \cr & f'\left( 2 \right) = - \csc \left( 2 \right)\cot \left( 2 \right) \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,\csc 2} \right) \cr & \to c = 2,{\text{ }}f\left( c \right) = \csc 2,{\text{ }}f'\left( c \right) = - \csc \left( 2 \right)\cot \left( 2 \right) \cr & y = \csc \left( 2 \right) - \csc \left( 2 \right)\cot \left( 2 \right)\left( {x - 2} \right) \cr & T\left( x \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {x - 2} \right)} \right] \cr & {\text{Completing the table for }}f\left( x \right){\text{:}} \cr & x = 1.9 \to f\left( {1.9} \right) = \csc \left( {1.9} \right) = 1.0567 \cr & x = 1.99 \to f\left( {1.99} \right) = \csc \left( {1.99} \right) = 1.0947 \cr & x = 2 \to f\left( 2 \right) = \csc \left( 2 \right) = 1.0997 \cr & x = 2.01 \to f\left( {2.01} \right) = \csc \left( {2.01} \right) = 1.1048 \cr & x = 2.1 \to f\left( {2.1} \right) = \csc \left( {2.1} \right) = 1.584 \cr & {\text{Completing the table for }}T\left( x \right){\text{:}} \cr & x = 1.9 \to T\left( {1.9} \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {1.9 - 2} \right)} \right] = 1.0494 \cr & x = 1.99 \to T\left( {1.99} \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {1.99 - 2} \right)} \right] = 1.0947 \cr & x = 2 \to T\left( 2 \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {2 - 2} \right)} \right] = 1.0997 \cr & x = 2.01 \to T\left( {2.01} \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {2.01 - 2} \right)} \right] = 1.1047 \cr & x = 2.1 \to T\left( {2.1} \right) = \csc \left( 2 \right)\left[ {1 - \cot \left( 2 \right)\left( {2.1 - 2} \right)} \right] = 1.1500 \cr & \cr & {\text{Therefore}} \cr} $$ \[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.0567}&{1.0947}&{1.0997}&{1.1048}&{1.5840} \\ {T\left( x \right)}&{1.0494}&{1.0947}&{1.0997}&{1.1047}&{1.1500} \end{array}}\]