Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 74

$$\eqalign{ & x{\text{ - intercept }}\left( {0,0} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Relative minimum at }}\left( { - 1, - 1} \right) \cr & {\text{Relative maximum at }}\left( {1,1} \right) \cr & \cr & {\text{Horizontal asymptote }}y = 0 \cr & {\text{Inflection points at: }}\left( {0,0} \right),{\text{ }}\left( { - \sqrt 3 , - \frac{{\sqrt 3 }}{2}} \right),{\text{ }}\left( {\sqrt 3 ,\frac{{\sqrt 3 }}{2}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{2x}}{{1 + {x^2}}} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = \frac{{2x}}{{1 + {x^2}}} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = \frac{{2x}}{{1 + {x^2}}} \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\frac{{2x}}{{1 + {x^2}}}} \right] \cr & y' = \frac{{\left( {1 + {x^2}} \right)\left( 2 \right) - 2x\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & y' = \frac{{2 + 2{x^2} - 4{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & y' = \frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}} \cr & {\text{Let }}y' = 0 \cr & 2 - 2{x^2} = 0 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & \cr & {\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {\frac{{2 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^2}}}} \right] \cr & y'' = \frac{{{{\left( {1 + {x^2}} \right)}^2}\left( { - 4x} \right) - 2\left( {2 - 2{x^2}} \right)\left( {1 + {x^2}} \right)\left( {2x} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}} \cr & y'' = \frac{{ - 4x{{\left( {1 + {x^2}} \right)}^2} - 4x\left( {2 - 2{x^2}} \right)\left( {1 + {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}} \cr & y'' = \frac{{4x\left( {1 + {x^2}} \right)\left( { - {x^2} - 1 - 2 + 2{x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^4}}} \cr & y'' = \frac{{4x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & {\text{Evaluate }}y''\left( { - 1} \right) \cr & y''\left( { - 1} \right) = 1 > 0 \cr & {\text{Relative minimum at }}\left( { - 1,f\left( { - 1} \right)} \right) \to \left( { - 1, - 1} \right) \cr & y''\left( 1 \right) = - 1 < 0 \cr & {\text{Relative maximum at }}\left( {1,f\left( 1 \right)} \right) \to \left( {1,1} \right) \cr & \cr & *{\text{Find the Inflection points}} \cr & y'' = \frac{{4x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} \cr & \frac{{4x\left( {{x^2} - 3} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}} = 0 \cr & x = 0,{\text{ }}x = - \sqrt 3 ,{\text{ }}x = \sqrt 3 \cr & f\left( 0 \right) = 0,{\text{ }}f\left( { - \sqrt 3 } \right) = - \frac{{\sqrt 3 }}{2},{\text{ }}f\left( {\sqrt 3 } \right) = \frac{{\sqrt 3 }}{2} \cr & {\text{Inflection points at: }}\left( {0,0} \right),{\text{ }}\left( { - \sqrt 3 , - \frac{{\sqrt 3 }}{2}} \right),{\text{ }}\left( {\sqrt 3 ,\frac{{\sqrt 3 }}{2}} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & y = \frac{{2x}}{{1 + {x^2}}} \cr & {x^2} + 1 = 0,{\text{ no real solutions, then}} \cr & {\text{No vertical asymptotes }} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2x}}{{1 + {x^2}}}} \right) = 0 \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {\frac{x}{{{x^2} + 1}}} \right) = 0 \cr & {\text{Horizontal asymptote }}y = 0 \cr & \cr } $$