Chapter 3 - Applications of Differentiation - 3.9 Exercises - Page 236: 2

\[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.662}&{1.515}&{1.5}&{1.485}&{1.360} \\ {T\left( x \right)}&{1.65}&{1.515}&{1.5}&{1.485}&{1.350} \end{array}}\]

$$\eqalign{ & f\left( x \right) = \frac{6}{{{x^2}}},{\text{ }}\left( {2,\frac{3}{2}} \right) \cr & f\left( x \right) = 6{x^{ - 2}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = - 12{x^{ - 3}} = - \frac{{12}}{{{x^3}}} \cr & f'\left( 2 \right) = - \frac{{12}}{{{{\left( 2 \right)}^3}}} = - \frac{3}{2} \cr & {\text{The equation for the tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is:}} \cr & y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr & {\text{We have the point }}\left( {2,\frac{3}{2}} \right) \to c = 2,{\text{ }}f\left( c \right) = \frac{3}{2},{\text{ and }}f'\left( c \right) = - \frac{3}{2} \cr & y = \frac{3}{2} - \frac{3}{2}\left( {x - 2} \right) \cr & y = \frac{3}{2} - \frac{3}{2}x + 3 \cr & y = - \frac{3}{2}x + \frac{9}{2} \cr & T\left( x \right) = - \frac{3}{2}x + \frac{9}{2} \cr & {\text{Completing the table for }}f\left( x \right){\text{:}} \cr & x = 1.9 \to f\left( {1.9} \right) = \frac{6}{{{{\left( {1.9} \right)}^2}}} = 1.662 \cr & x = 1.99 \to f\left( {1.99} \right) = \frac{6}{{{{\left( {1.99} \right)}^2}}} = 1.515 \cr & x = 2 \to f\left( 2 \right) = \frac{6}{{{{\left( 2 \right)}^2}}} = 1.5 \cr & x = 2.01 \to f\left( {2.01} \right) = \frac{6}{{{{\left( {2.01} \right)}^2}}} = 1.485 \cr & x = 2.1 \to f\left( {2.1} \right) = \frac{6}{{{{\left( {2.1} \right)}^2}}} = 1.360 \cr & {\text{Completing the table for }}T\left( x \right){\text{:}} \cr & x = 1.9 \to T\left( {1.9} \right) = - \frac{3}{2}\left( {1.9} \right) + \frac{9}{2} = 1.65 \cr & x = 1.99 \to T\left( {1.99} \right) = - \frac{3}{2}\left( {1.99} \right) + \frac{9}{2} = 1.515 \cr & x = 2 \to T\left( 2 \right) = - \frac{3}{2}\left( 2 \right) + \frac{9}{2} = 1.5 \cr & x = 2.01 \to T\left( {2.01} \right) = - \frac{3}{2}\left( {2.01} \right) + \frac{9}{2} = 1.485 \cr & x = 2.1 \to T\left( {2.1} \right) = - \frac{3}{2}\left( {2.1} \right) + \frac{9}{2} = 1.35 \cr & \cr & {\text{Therefore}} \cr} $$ \[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.662}&{1.515}&{1.5}&{1.485}&{1.360} \\ {T\left( x \right)}&{1.65}&{1.515}&{1.5}&{1.485}&{1.350} \end{array}}\]