Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 28

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts: }}\left( {0,0} \right),\left( { - 3,0} \right) \cr & {\text{Critical points }}x = - 3{\text{ and }}x = - 2 \cr & {\text{local minimum at }}\left( { - 2, - 2} \right) \cr & {\text{No inflection points}}{\text{.}} \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = x\sqrt {x + 3} \cr & {\text{The domain of the function is }}\left[ { - 3,\infty } \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 0\sqrt {0 + 3} \cr & f\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & x\sqrt {x + 3} = 0 \cr & x = 0,{\text{ and }}x = - 3 \cr & x{\text{ - intercepts: }}\left( {0,0} \right),\left( { - 3,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {x\sqrt {x + 3} } \right] \cr & f'\left( x \right) = x\left( {\frac{1}{{2\sqrt {x + 3} }}} \right) + \sqrt {x + 3} \cr & f'\left( x \right) = \frac{x}{{2\sqrt {x + 3} }} + \sqrt {x + 3} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{x}{{2\sqrt {x + 3} }} + \sqrt {x + 3} = 0 \cr & \frac{{x + 2\left( {x + 3} \right)}}{{2\sqrt {x + 3} }} = 0 \cr & \frac{{3x + 6}}{{2\sqrt {x + 3} }} = 0 \cr & x = - 2 \cr & {\text{And the derivative is undefined at }}x = - 3,{\text{ then}} \cr & {\text{the critical points are }}x = - 2{\text{ and }}x = - 3 \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{{2\sqrt {x + 3} }} + \sqrt {x + 3} } \right] \cr & f''\left( x \right) = \frac{{2\sqrt {x + 3} - 2x\left( {\frac{1}{{2\sqrt {x + 3} }}} \right)}}{{{{\left( {2\sqrt {x + 3} } \right)}^2}}} + \frac{1}{{2\sqrt {x + 3} }} \cr & f''\left( x \right) = \frac{{4\left( {x + 3} \right) - x}}{{4\left( {x + 3} \right)}} + \frac{1}{{2\sqrt {x + 3} }} \cr & f''\left( x \right) = \frac{{3x + 12}}{{4\left( {x + 3} \right)}} + \frac{1}{{2\sqrt {x + 3} }} \cr & f''\left( x \right) = \frac{{3x + 12 + 2\sqrt {x + 3} }}{{4\left( {x + 3} \right)}} \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical point }}x = - 3 \cr & f''\left( 3 \right) = {\text{undefined}}{\text{, use the first derivative test}} \cr & f'\left( { - 4} \right) = \frac{{ - 4}}{{2\sqrt { - 4 + 3} }} + \sqrt { - 4 + 3} ,{\text{ The value }}x = - 3{\text{ is an endpoint}} \cr & {\text{of the domain of }}f\left( x \right){\text{ }}{\text{, then we can conclude that there is no }} \cr & {\text{relative extrema at }}x = - 3 \cr & \cr & {\text{*Evaluate }}f''\left( x \right){\text{ at the critical point }}x = - 2 \cr & f''\left( { - 2} \right) = \frac{{3\left( { - 2} \right) + 12 + 2\sqrt { - 2 + 3} }}{{4\left( { - 2 + 3} \right)}} = 2 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}\left( { - 2,f\left( { - 2} \right)} \right) \cr & \left( { - 2,f\left( { - 2} \right)} \right) = \left( { - 2. - 2} \right) \cr & \to {\text{local minimum at }}\left( { - 2, - 2} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{3x + 12 + 2\sqrt {x + 3} }}{{4\left( {x + 3} \right)}} = 0 \cr & {\text{No real solutions}}{\text{, then there are no inflection points}}{\text{.}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$