Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 22

$$\left. {\bf{a}} \right)y = - 2x + 3\,\,\,\,\,\,\,\left. {\bf{b}} \right)y = 2x + 7$$

$$\eqalign{ & f\left( x \right) = 6 - {x^2};\,\,\,\,\,\,\,\,\,{x_1} = - 1,\,\,\,\,\,\,{x_2} = 3 \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( { - 1} \right) = 6 - {\left( { - 1} \right)^2} \cr & {y_1} = 5 \cr & \left( {{x_1},{y_1}} \right) = \left( { - 1,5} \right) \cr & and \cr & {y_2} = f\left( {{x_2}} \right) = f\left( 3 \right) = 6 - {\left( 3 \right)^2} \cr & {y_2} = - 3 \cr & \left( {{x_2},{y_2}} \right) = \left( {3, - 3} \right) \cr & \cr & {\text{The slope of the secant line is }} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{ - 3 - 5}}{{3 - \left( { - 1} \right)}} = - 2 \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 5 = - 2\left( {x - \left( { - 1} \right)} \right) \cr & y - 5 = - 2\left( {x + 1} \right) \cr & y - 5 = - 2x - 2 \cr & y = - 2x + 3 \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{{6 - {{\left( {x + h} \right)}^2} - \left( {6 - {x^2}} \right)}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{6 - {x^2} - 2xh - {h^2} - 6 + {x^2}}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2xh - {h^2}}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \left( { - 2x - h} \right) \cr & f'\left( x \right) = - 2x \cr & \cr & {\text{The slope of the tangent line at the point }}\left( {{x_1},{y_1}} \right) = \left( { - 1,5} \right){\text{ is}} \cr & f'\left( { - 1} \right) = - 2\left( { - 1} \right) \cr & f'\left( { - 1} \right) = 2 \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 5 = 2\left( {x + 1} \right) \cr & y - 5 = 2x + 2 \cr & y = 2x + 7 \cr} $$