Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 78

$$\eqalign{ & {\text{Critical points: }}x = 0{\text{ and }}x = 4 \cr & {\text{local minimum at }}x = 0 \cr & {\text{local maximum at }}x = 4 \cr} $$

$$\eqalign{ & p\left( x \right) = {x^4}{e^{ - x}} \cr & {\text{Calculate the first derivative }} \cr & p'\left( x \right) = \frac{d}{{dx}}\left[ {{x^4}{e^{ - x}}} \right] \cr & p'\left( x \right) = - {x^4}{e^{ - x}} + 4{x^3}{e^{ - x}} \cr & {\text{Find the critical point}}{\text{, set }}p'\left( x \right) = 0 \cr & - {x^4}{e^{ - x}} + 4{x^3}{e^{ - x}} = 0 \cr & {x^3}{e^{ - x}}\left( {4 - x} \right) = 0 \cr & {x^3}\left( {4 - x} \right) = 0 \cr & {\text{The critical points are }}x = 0{\text{ and }}x = 4 \cr & \cr & {\text{Calculate the second derivative }} \cr & p''\left( x \right) = \frac{d}{{dx}}\left[ { - {x^4}{e^{ - x}} + 4{x^3}{e^{ - x}}} \right] \cr & p''\left( x \right) = {x^4}{e^{ - x}} - 4{x^3}{e^{ - x}} - 4{x^3}{e^{ - x}} + 12{x^2}{e^{ - x}} \cr & p''\left( x \right) = {x^4}{e^{ - x}} - 8{x^3}{e^{ - x}} + 12{x^2}{e^{ - x}} \cr & p''\left( x \right) = \left( {{x^2} - 8x + 12} \right){x^2}{e^{ - x}} \cr & \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}p''\left( 0 \right) \cr & p''\left( 0 \right) = \left( {{0^2} - 8\left( 0 \right) + 12} \right){\left( 0 \right)^2}{e^{ - 0}} \cr & p''\left( 0 \right) = 0,{\text{ then the test is inconclusive; may have a local maximum}}{\text{,}} \cr & {\text{local minimum}}{\text{, or neither at 0}}{\text{.}} \cr & {\text{Evaluating }}p'\left( { - 1} \right){\text{ and }}p'\left( 1 \right) \cr & p'\left( { - 1} \right) = - {\left( { - 1} \right)^4}{e^{ - \left( { - 1} \right)}} + 4{\left( { - 1} \right)^3}{e^{ - \left( { - 1} \right)}} = - 3e < 0 \cr & p'\left( 1 \right) = - {\left( 1 \right)^4}{e^{ - \left( 1 \right)}} + 4{\left( 1 \right)^3}{e^{ - \left( 1 \right)}} = 3e > 0 \cr & {\text{The first derivative changes negative to positive}}{\text{, then }}p\left( x \right){\text{ has a}} \cr & {\text{local minimun at }}x = 0. \cr & \cr & {\text{Evaluate }}p''\left( 4 \right) \cr & p''\left( 4 \right) = \left( {{4^2} - 8\left( 4 \right) + 12} \right){\left( 4 \right)^2}{e^{ - 4}} \cr & p''\left( 4 \right) = - 64{e^{ - 4}} \cr & p''\left( 4 \right) < 0,{\text{ then }}p\left( x \right){\text{ has a local maximum at }}x = 4 \cr & \cr & {\text{Critical points: }}x = 0{\text{ and }}x = 4 \cr & {\text{local minimum at }}x = 0 \cr & {\text{local maximum at }}x = 4 \cr} $$