Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 21

$$\left. {\bf{a}} \right)y = 10x - 15,\,\,\,\,\,\,\,\left. {\bf{b}} \right)y = 8x - 9$$

$$\eqalign{ & f\left( x \right) = {x^2} + 2x;\,\,\,\,\,\,\,\,\,{x_1} = 3,\,\,\,\,\,\,{x_2} = 5 \cr & \cr & \left. {\bf{a}} \right) \cr & {\text{Evaluate }}f\left( {{x_1}} \right){\text{ and }}f\left( {{x_2}} \right) \cr & {y_1} = f\left( {{x_1}} \right) = f\left( 3 \right) = {\left( 3 \right)^2} + 2\left( 3 \right) \cr & {y_1} = 15 \cr & \left( {{x_1},{y_1}} \right) = \left( {3,15} \right) \cr & and \cr & {y_2} = f\left( {{x_2}} \right) = f\left( 5 \right) = {\left( 5 \right)^2} + 2\left( 5 \right) \cr & {y_2} = 35 \cr & \left( {{x_2},{y_2}} \right) = \left( {5,35} \right) \cr & \cr & {\text{The slope of the secant line is }} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{35 - 15}}{{5 - 3}} = 10 \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 15 = 10\left( {x - 3} \right) \cr & y - 15 = 10x - 30 \cr & y = 10x - 15 \cr & \cr & \left. {\bf{b}} \right) \cr & {\text{Using the definition of the derivative}}: \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} + 2\left( {x + h} \right) - \left( {{x^2} + 2x} \right)}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{x^2} + 2xh + {h^2} + 2x + 2h - {x^2} - 2x}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2xh + {h^2} + 2h}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( {2x + h + 2} \right) \cr & f'\left( x \right) = 2x + 2 \cr & \cr & {\text{The slope of the tangent line at the point }}\left( {{x_1},{y_1}} \right) = \left( {3,15} \right){\text{ is}} \cr & f'\left( 3 \right) = 2\left( 3 \right) + 2 \cr & f'\left( 3 \right) = 8 \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 15 = 8\left( {x - 3} \right) \cr & y - 15 = 8x - 24 \cr & y = 8x - 9 \cr & \cr} $$