Chapter 3 - Derivatives - 3.2 Working with Derivatives - 3.2 Exercises - Page 143: 27

$$\left( {1,1} \right){\text{ and }}\left( { - \frac{1}{2}, - 2} \right)$$

$$\eqalign{ & f\left( x \right) = \frac{1}{x};{\text{ }}Q\left( { - 2,4} \right) \cr & {\text{Let the point }}P\left( {a,f\left( a \right)} \right){\text{ tangent to the graph }}f\left( x \right),{\text{ the}} \cr & {\text{slope of the tangent line at the point }}\left( {a,f\left( a \right)} \right){\text{ is given by}} \cr & {\text{the derivative of }}f\left( x \right){\text{ at }}x = a \cr & {\text{Find }}f'\left( a \right){\text{ using }}{m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}{\text{ }}\left( {{\text{See page 129}}} \right) \cr & {\text{Therefore}}{\text{,}} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\overbrace {\left[ {\frac{1}{{a + h}}} \right]}^{f\left( {a + h} \right)} - \overbrace {\frac{1}{a}}^{f\left( a \right)}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{a + h}} - \frac{1}{a}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{a - a - h}}{{a\left( {a + h} \right)}}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{a\left( {a + h} \right)}}}}{h} \cr & {m_{\tan }} = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{a\left( {a + h} \right)}} \cr & {\text{Evaluate the limit when }}h \to 0 \cr & {m_{\tan }} = - \frac{1}{{a\left( {a + 0} \right)}} \cr & {m_{\tan }} = - \frac{1}{{{a^2}}} \cr & {\text{The equation of the tangent line at the points }}\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right) \cr & {\text{is:}} \cr & {y_2} - {y_1} = m\left( {{x_2} - {x_1}} \right) \cr & {\text{We have the point }}P\left( {a,f\left( a \right)} \right){\text{ and }}Q\left( { - 2,4} \right) \cr & {\text{with slope }}m = - \frac{1}{{{a^2}}} \cr & {\text{Therefore}} \cr & 4 - f\left( a \right) = - \frac{1}{{{a^2}}}\left( { - 2 - a} \right) \cr & {\text{Where }}f\left( a \right) = \frac{1}{a} \cr & 4 - \frac{1}{a} = - \frac{1}{{{a^2}}}\left( { - 2 - a} \right) \cr & 4 - \frac{1}{a} = \frac{2}{{{a^2}}} + \frac{1}{a} \cr & {\text{Solving for }}a,{\text{ multiply both sides by }}{a^2} \cr & 4{a^2} - a = 2 + a \cr & 4{a^2} - 2a - 2 = 0 \cr & 2{a^2} - a - 1 = 0 \cr & {\text{factoring}} \cr & \left( {a - 1} \right)\left( {2a + 1} \right) = 0 \cr & {a_1} = 1,{\text{ }}{a_2} = - \frac{1}{2} \cr & \cr & {\text{We obtain the points }}{P_1}\left( {{a_1},f\left( {{a_1}} \right)} \right){\text{ and }}{P_2}\left( {{a_2},f\left( {{a_2}} \right)} \right) \cr & f\left( {{a_1}} \right) = f\left( 1 \right) = \frac{1}{1} = 1 \to {P_1}\left( {1,1} \right) \cr & f\left( {{a_2}} \right) = f\left( { - \frac{1}{2}} \right) = \frac{1}{{ - 1/2}} = - 2 \to {P_2}\left( { - \frac{1}{2}, - 2} \right) \cr & {\text{Points }}\left( {1,1} \right){\text{ and }}\left( { - \frac{1}{2}, - 2} \right) \cr & \cr & {\text{Graph}} \cr} $$