Chapter 3 - Derivatives - 3.2 Working with Derivatives - 3.2 Exercises - Page 143: 25

$$\left( {1,2} \right){\text{ and }}\left( {5,26} \right)$$

$$\eqalign{ & f\left( x \right) = {x^2} + 1;{\text{ }}Q\left( {3,6} \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 }}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[ {{{\left( {a + h} \right)}^2} + 1} \right]}^{f\left( {a + h} \right)} - \overbrace {\left( {{a^2} + 1} \right)}^{f\left( a \right)}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{a^2} + 2ah + {h^2} + 1 - {a^2} - 1}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{2ah + {h^2}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {2a + h} \right)}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \left( {2a + h} \right) \cr & {\text{Evaluate the limit when }}h \to 0 \cr & {m_{\tan }} = 2a \cr & \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( {3,6} \right){\text{with slope }}m = 2a \cr & {\text{Therefore}} \cr & 6 - f\left( a \right) = 2a\left( {3 - a} \right) \cr & {\text{Where }}f\left( a \right) = {a^2} + 1 \cr & 6 - \left( {{a^2} + 1} \right) = 2a\left( {3 - a} \right) \cr & 6 - {a^2} - 1 = 6a - 2{a^2} \cr & {\text{Solving for }}a \cr & 5 - {a^2} - 6a + 2{a^2} = 0 \cr & {a^2} - 6a + 5 = 0 \cr & {\text{factoring}} \cr & \left( {a - 5} \right)\left( {a - 1} \right) = 0 \cr & {a_1} = 1,{\text{ }}{a_2} = 5 \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) = {\left( 1 \right)^2} + 1 = 2 \to {P_1}\left( {1,2} \right) \cr & f\left( {{a_2}} \right) = f\left( 5 \right) = {\left( 5 \right)^2} + 1 = 26 \to {P_2}\left( {5,26} \right) \cr & {\text{Points }}\left( {1,2} \right){\text{ and }}\left( {5,26} \right) \cr & \cr & {\text{Graph}} \cr} $$