Chapter 4 - Applications of the Derivative - Review Exercises - Page 330: 13

$$\eqalign{ & y{\text{ - intercept }}\left( {0,1} \right) \cr & x{\text{ - intercepts: }}\left( { - 2.9168,0} \right){\text{ and }}\left( { - 0.2154,0} \right) \cr & {\text{local minimum at }}\left( { - 2, - 11} \right) \cr & {\text{inflection points: }}\left( { - 1, - \frac{{11}}{2}} \right){\text{ and}}\left( {1,\frac{5}{2}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{{x^4}}}{2} - 3{x^2} + 4x + 1 \cr & {\text{The function is a polynomial then the domain is }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \frac{{{{\left( 0 \right)}^4}}}{2} - 3{\left( 0 \right)^2} + 4\left( 0 \right) + 1 \cr & f\left( 0 \right) = 1 \cr & y{\text{ - intercept }}\left( {0,1} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & \frac{{{x^4}}}{2} - 3{x^2} + 4x + 1 = 0 \cr & {\text{Using a graphing utility we obtain:}} \cr & {x_1} = - 2.9168{\text{ and }}{x_2} = - 0.2154 \cr & x{\text{ - intercepts: }}\left( { - 2.9168,0} \right){\text{ and }}\left( { - 0.2154,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{2} - 3{x^2} + 4x + 1} \right] \cr & f'\left( x \right) = 2{x^3} - 6x + 4 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & 2{x^3} - 6x + 4 = 0 \cr & {\text{Factoring}} \cr & 2\left( {x + 2} \right){\left( {x - 1} \right)^2} = 0 \cr & x = - 2{\text{ and }}x = 1 \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} - 6x + 4} \right] \cr & f''\left( x \right) = 6{x^2} - 6 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = - 2{\text{ and }}x = 1 \cr & *f''\left( { - 2} \right) = 6{\left( { - 2} \right)^2} - 6 = 18 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}\left( { - 2,f\left( { - 2} \right)} \right) \cr & f\left( { - 2} \right) = \frac{{{{\left( { - 2} \right)}^4}}}{2} - 3{\left( { - 2} \right)^2} + 4\left( { - 2} \right) + 1 = - 11 \cr & \to {\text{local minimum at }}\left( { - 2, - 11} \right) \cr & *f''\left( 1 \right) = 6{\left( 1 \right)^2} - 6 = 0,{\text{ Inconclusive test}}{\text{, use the first}} \cr & {\text{derivative test:}} \cr & f'\left( 0 \right) = 2{\left( 0 \right)^3} - 6\left( 0 \right) + 4 = 4 > 0,{\text{ Increasing}} \cr & f'\left( 1 \right) = 2{\left( 1 \right)^3} - 6\left( 1 \right) + 4 = 0,{\text{ Stationary point}} \cr & f'\left( 2 \right) = 2{\left( 2 \right)^3} - 6\left( 2 \right) + 4 = 8 > 0,{\text{ Increasing}} \cr & {\text{No relative extrema at }}x = 1 \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & 6{x^2} - 6 = 0 \cr & x = \pm 1 \cr & f\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^4}}}{2} - 3{\left( { - 1} \right)^2} + 4\left( { - 1} \right) + 1 = - \frac{{11}}{2} \cr & f\left( 1 \right) = \frac{{{{\left( 1 \right)}^4}}}{2} - 3{\left( 1 \right)^2} + 4\left( 1 \right) + 1 = \frac{5}{2} \cr & {\text{The inflection points are: }} \cr & \left( { - 1, - \frac{{11}}{2}} \right),\left( {1,\frac{5}{2}} \right) \cr & \cr & {\text{Graph}} \cr} $$