Chapter 3 - Applications of Differentiation - 3.9 Exercises - Page 236: 1

\[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{3.61}&{3.9601}&4&{4.0401}&{4.41} \\ {T\left( x \right)}&{3.6}&{3.96}&4&{4.04}&{4.4} \end{array}}\]

$$\eqalign{ & f\left( x \right) = {x^2},{\text{ }}\left( {2,4} \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = 2x \cr & f'\left( 2 \right) = 2\left( 2 \right) = 4 \cr & {\text{The equation for the tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is:}} \cr & y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr & {\text{We have the point }}\left( {2,4} \right) \to c = 2,{\text{ }}f\left( c \right) = 4,{\text{ and }}f'\left( c \right) = 4 \cr & y = 4 + 4\left( {x - 2} \right) \cr & y = 4 + 4x - 8 \cr & y = 4x - 4 \cr & T\left( x \right) = 4x - 4 \cr & {\text{Completing the table for }}f\left( x \right){\text{:}} \cr & x = 1.9 \to f\left( {1.9} \right) = {\left( {1.9} \right)^2} = 3.61 \cr & x = 1.99 \to f\left( {1.99} \right) = {\left( {1.99} \right)^2} = 3.9601 \cr & x = 2 \to f\left( 2 \right) = {\left( 2 \right)^2} = 4 \cr & x = 2.01 \to f\left( {2.01} \right) = {\left( {2.01} \right)^2} = 4.0401 \cr & x = 2.1 \to f\left( {2.1} \right) = {\left( {2.1} \right)^2} = 4.41 \cr & {\text{Completing the table for }}T\left( x \right){\text{:}} \cr & x = 1.9 \to T\left( {1.9} \right) = 4\left( {1.9} \right) - 4 = 3.6 \cr & x = 1.99 \to T\left( {1.99} \right) = 4\left( {1.99} \right) - 4 = 3.96 \cr & x = 2 \to T\left( 2 \right) = 4\left( 2 \right) - 4 = 4 \cr & x = 2.01 \to T\left( {2.01} \right) = 4\left( {2.01} \right) - 4 = 4.04 \cr & x = 2.1 \to T\left( {2.1} \right) = 4\left( {2.01} \right) - 4 = 4.4 \cr & \cr & {\text{Therefore}} \cr} $$ \[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{3.61}&{3.9601}&4&{4.0401}&{4.41} \\ {T\left( x \right)}&{3.6}&{3.96}&4&{4.04}&{4.4} \end{array}}\]