Chapter 4 - Applications of the Derivative - Review Exercises - Page 330: 15

$$\eqalign{ & y{\text{ - intercept }}\left( {0, - 4} \right) \cr & x{\text{ - intercepts }}\left( {\frac{1}{2},0} \right){\text{ and }}\left( {\frac{3}{2},0} \right) \cr & {\text{Local maximum at }}\left( {1,4} \right) \cr & {\text{inflection points: }}\left( {\frac{1}{2},0} \right){\text{ and }}\left( {\frac{3}{2},0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = 4\cos \left[ {\pi \left( {x - 1} \right)} \right]{\text{ on }}\left[ {0,2} \right] \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 4\cos \left[ {\pi \left( {0 - 1} \right)} \right] \cr & f\left( 0 \right) = 4\cos \left( { - \pi } \right) \cr & f\left( 0 \right) = - 4 \cr & y{\text{ - intercept }}\left( {0, - 4} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 4\cos \left[ {\pi \left( {x - 1} \right)} \right] = 0 \cr & \cos \pi \left( {x - 1} \right) = 0 \cr & {\text{Solving for the interval }}\left[ {0,2} \right]{\text{ we obtain}} \cr & x = \frac{1}{2}\,{\text{and }}x = \frac{3}{2} \cr & x{\text{ - intercepts }}\left( {\frac{1}{2},0} \right){\text{ and }}\left( {\frac{3}{2},0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {4\cos \left[ {\pi \left( {x - 1} \right)} \right]} \right] \cr & f'\left( x \right) = - 4\pi \sin \left[ {\pi \left( {x - 1} \right)} \right] \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & - 4\pi \sin \left[ {\pi \left( {x - 1} \right)} \right] = 0 \cr & \sin \left[ {\pi \left( {x - 1} \right)} \right] = 0 \cr & {\text{Solving for the interval }}\left[ {0,2} \right]{\text{ we obtain}} \cr & x = 0{\text{ }}\left( {{\text{Endpoint}}} \right),{\text{ }}x = 1,{\text{ }}x = 2{\text{ }}\left( {{\text{Endpoint}}} \right) \cr & {\text{Let }}x = 1 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ { - 4\pi \sin \left[ {\pi \left( {x - 1} \right)} \right]} \right] \cr & f''\left( x \right) = - 4{\pi ^2}\cos \left[ {\pi \left( {x - 1} \right)} \right] \cr & \cr & {\text{Evaluate the second derivative at the critical point }}x = 1 \cr & f''\left( 1 \right) = - 4{\pi ^2}\cos \left[ {\pi \left( {\left( 1 \right) - 1} \right)} \right] = - 4{\pi ^2} < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}x = 1 \cr & f\left( 1 \right) = 4\cos \left[ {\pi \left( {1 - 1} \right)} \right] = 4 \cr & {\text{Local maximum at }}\left( {1,f\left( 1 \right)} \right) \to \left( {1,4} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & - 4{\pi ^2}\cos \left[ {\pi \left( {x - 1} \right)} \right] = 0 \cr & x = \frac{1}{2}{\text{ and }}x = \frac{3}{2} \cr & {\text{The inflection points are: }}\left( {\frac{1}{2},0} \right){\text{ and }}\left( {\frac{3}{2},0} \right) \cr & {\text{Graph}} \cr} $$