Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 23

$$\left. {\bf{a}} \right)y = - \frac{1}{2}x + \frac{7}{2},\,\,\,\,\,\,\,\left. {\bf{b}} \right)y = - \frac{5}{4}x + 5$$

$$\eqalign{ & f\left( x \right) = \frac{5}{x};\,\,\,\,\,\,\,\,\,{x_1} = 2,\,\,\,\,\,\,{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( 2 \right) \cr & {y_1} = \frac{5}{2} \cr & \left( {{x_1},{y_1}} \right) = \left( {2,\frac{5}{2}} \right) \cr & and \cr & {y_2} = f\left( {{x_2}} \right) = f\left( 5 \right) \cr & {y_2} = 1 \cr & \left( {{x_2},{y_2}} \right) = \left( {5,1} \right) \cr & \cr & {\text{The slope of the secant line is }} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{1 - 5/2}}{{5 - 2}} = - \frac{1}{2} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{5}{2} = - \frac{1}{2}\left( {x - 2} \right) \cr & y - \frac{5}{2} = - \frac{1}{2}x + 1 \cr & y = - \frac{1}{2}x + \frac{7}{2} \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{{\frac{5}{{x + h}} - \frac{5}{x}}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5x - 5x - 5h}}{{hx\left( {x + h} \right)}} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 5}}{{x\left( {x + h} \right)}} \cr & f'\left( x \right) = \frac{{ - 5}}{{x\left( {x + 0} \right)}} \cr & f'\left( x \right) = - \frac{5}{{{x^2}}} \cr & \cr & {\text{The slope of the tangent line at the point }}\left( {{x_1},{y_1}} \right) = \left( {2,\frac{5}{2}} \right){\text{ is}} \cr & f'\left( 2 \right) = - \frac{5}{{{{\left( 2 \right)}^2}}} \cr & f'\left( 2 \right) = - \frac{5}{4} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - \frac{5}{2} = - \frac{5}{4}\left( {x - 2} \right) \cr & y - \frac{5}{2} = - \frac{5}{4}x + \frac{5}{2} \cr & y = - \frac{5}{4}x + 5 \cr} $$