Chapter 3 - Applications of Differentiation - 3.9 Exercises - Page 236: 4

\[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.3784}&{1.4106}&{1.4142}&{1.4177}&{1.4491} \\ {T\left( x \right)}&{1.3788}&{1.4106}&{1.4142}&{1.4177}&{1.4495} \end{array}}\]

$$\eqalign{ & f\left( x \right) = \sqrt x ,{\text{ }}\left( {2,\sqrt 2 } \right) \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr & f'\left( 2 \right) = \frac{1}{{2\sqrt 2 }} \cr & {\text{The equation for the tangent line at the point }}\left( {c,f\left( c \right)} \right){\text{ is:}} \cr & y = f\left( c \right) + f'\left( c \right)\left( {x - c} \right) \cr & {\text{We have the point }}\left( {2,\sqrt 2 } \right) \to c = 2,{\text{ }}f\left( c \right) = \sqrt 2 ,{\text{ }}f'\left( c \right) = \frac{1}{{2\sqrt 2 }} \cr & y = \sqrt 2 + \frac{1}{{2\sqrt 2 }}\left( {x - 2} \right) \cr & y = \sqrt 2 + \frac{1}{{2\sqrt 2 }}x - \frac{1}{{\sqrt 2 }} \cr & T\left( x \right) = \frac{1}{{2\sqrt 2 }}x + \frac{{\sqrt 2 }}{2} \cr & {\text{Completing the table for }}f\left( x \right){\text{:}} \cr & x = 1.9 \to f\left( {1.9} \right) = \sqrt {1.9} = 1.3784 \cr & x = 1.99 \to f\left( {1.99} \right) = \sqrt {1.99} = 1.4106 \cr & x = 2 \to f\left( 2 \right) = \sqrt 2 = 1.4142 \cr & x = 2.01 \to f\left( {2.01} \right) = \sqrt {2.01} = 1.4177 \cr & x = 2.1 \to f\left( {2.1} \right) = \sqrt {2.1} = 1.4491 \cr & {\text{Completing the table for }}T\left( x \right){\text{:}} \cr & x = 1.9 \to T\left( {1.9} \right) = \frac{1}{{2\sqrt 2 }}\left( {1.9} \right) + \frac{{\sqrt 2 }}{2} = 1.3788 \cr & x = 1.99 \to T\left( {1.99} \right) = \frac{1}{{2\sqrt 2 }}\left( {1.99} \right) + \frac{{\sqrt 2 }}{2} = 1.4106 \cr & x = 2 \to T\left( 2 \right) = \frac{1}{{2\sqrt 2 }}\left( 2 \right) + \frac{{\sqrt 2 }}{2} = 1.4142 \cr & x = 2.01 \to T\left( {2.01} \right) = \frac{1}{{2\sqrt 2 }}\left( {2.01} \right) + \frac{{\sqrt 2 }}{2} = 1.4177 \cr & x = 2.1 \to T\left( {2.1} \right) = \frac{1}{{2\sqrt 2 }}\left( {2.1} \right) + \frac{{\sqrt 2 }}{2} = 1.4495 \cr & \cr & {\text{Therefore}} \cr} $$ \[\boxed{\begin{array}{*{20}{c}} x&{1.9}&{1.99}&2&{2.01}&{2.1} \\ {f\left( x \right)}&{1.3784}&{1.4106}&{1.4142}&{1.4177}&{1.4491} \\ {T\left( x \right)}&{1.3788}&{1.4106}&{1.4142}&{1.4177}&{1.4495} \end{array}}\]