Chapter 3 - Derivatives - 3.2 Working with Derivatives - 3.2 Exercises - Page 143: 28

$$\left( { - 1.3065,3.6932} \right)$$

$$\eqalign{ & f\left( x \right) = {e^{ - x}};{\text{ }}Q\left( {1, - 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[ {{e^{ - a + h}}} \right]}^{f\left( {a + h} \right)} - \overbrace {{e^{ - a}}}^{f\left( a \right)}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - a + h}} - {e^{ - a}}}}{h} \cr & {\text{Recall that }}{a^{m + n}} = {e^m}{e^n} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - a}}{e^h} - {e^{ - a}}}}{h} \cr & {m_{\tan }} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - a}}\left( {{e^h} - 1} \right)}}{h} \cr & {m_{\tan }} = {e^{ - a}}\mathop {\lim }\limits_{h \to 0} \frac{{\left( {{e^h} - 1} \right)}}{h} \cr & {\text{Using the special limit }}\mathop {\lim }\limits_{h \to 0} \frac{{\left( {{e^h} - 1} \right)}}{h} = - 1,{\text{ then}} \cr & {m_{\tan }} = {e^{ - a}}\left( { - 1} \right) \cr & {m_{\tan }} = - {e^{ - a}} \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( {1, - 4} \right) \cr & {\text{with slope }}m = - {e^{ - a}} \cr & {\text{Therefore}} \cr & 1 - f\left( a \right) = - {e^{ - a}}\left( { - 4 - a} \right) \cr & {\text{Where }}f\left( a \right) = {e^{ - a}} \cr & 1 - {e^{ - a}} = - {e^{ - a}}\left( { - 4 - a} \right) \cr & 1 - {e^{ - a}} = 4{e^{ - a}} + a{e^{ - a}} \cr & {\text{Solving for }}a,{\text{ multiply both sides by }}{e^a} \cr & {e^a} - 1 = 4 + a \cr & {e^a} - a - 5 = 0 \cr & {\text{Solving by a graphing calculator}}{\text{, we obtain}} \cr & a \approx - 1.3065 \cr & \cr & {\text{We obtain the points }}P\left( {a,f\left( a \right)} \right){\text{ }} \cr & f\left( a \right) = f\left( { - 1.3065} \right) = {e^{ - \left( { - 1.3065} \right)}} \approx 3.6932 \cr & {\text{Points }}\left( { - 1.3065,3.6932} \right) \cr & \cr & {\text{Graph}} \cr} $$