Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 67

$$\eqalign{ & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = 4x - {x^2} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = 4\left( 0 \right) - {\left( 0 \right)^2} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = 4x - {x^2} \cr & {x^2} - 4x = 0 \cr & x\left( {x - 4} \right) = 0 \cr & x = 0,{\text{ }}x = 4 \cr & x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {4,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {4x - {x^2}} \right] \cr & y' = 4 - 2x \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & 4 - 2x = 0 \cr & x = 2 \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {4 - 2x} \right] \cr & y'' = - 2 \cr & {\text{Evaluate }}y''{\text{ at the critical point }}x = 2 \cr & y''\left( 2 \right) = - 2 < 0,{\text{ there are a relative maximun at }}f\left( 2 \right) \cr & f\left( 2 \right) = 4\left( 2 \right) - {\left( 2 \right)^2} = 4 \cr & {\text{Relative maximum at }}\left( {2,4} \right) \cr & {\text{Let }}y''\left( x \right) = 0 \cr & - 2 \ne 0 \cr & {\text{There are no inflection points}} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {4x - {x^2}} \right) = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {4x - {x^2}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr } $$