Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 21

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{local minimum at }}\left( {0,0} \right) \cr & {\text{No vertical asymptotes}} \cr & {\text{Horizontal asymptote }}y = \frac{\pi}{2} \cr} $$

$$\eqalign{ & f\left( x \right) = {\tan ^{ - 1}}{x^2} \cr & {\text{The domain of the function is }}\left( { - \infty ,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = {\tan ^{ - 1}}{\left( 0 \right)^2} \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & 0 = {\tan ^{ - 1}}{x^2} \cr & x = 0 \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}{x^2}} \right] \cr & \arctan x \cr & f'\left( x \right) = \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr & f'\left( x \right) = \frac{{2x}}{{1 + {x^4}}} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & \frac{{2x}}{{1 + {x^4}}} = 0 \cr & 2x = 0 \cr & x = 0 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{2x}}{{1 + {x^4}}}} \right] \cr & f''\left( x \right) = \frac{{\left( {1 + {x^4}} \right)\left( 2 \right) - 2x\left( {4{x^3}} \right)}}{{{{\left( {1 + {x^4}} \right)}^2}}} \cr & {\text{Simplifying}} \cr & f''\left( x \right) = \frac{{2 + 2{x^4} - 8{x^4}}}{{{{\left( {1 + {x^4}} \right)}^2}}} \cr & f''\left( x \right) = \frac{{2 - 6{x^4}}}{{{{\left( {1 + {x^4}} \right)}^2}}} \cr & {\text{Evaluate the second derivative at the critical point }}x = 0 \cr & f''\left( 0 \right) = \frac{{2 - 6{{\left( 0 \right)}^4}}}{{{{\left( {1 + {{\left( 0 \right)}^4}} \right)}^2}}} = 2 > 0 \to {\text{Local minimum at }}\left( {0,f\left( 0 \right)} \right) \cr & f\left( 0 \right) = \arctan \left( {{0^2}} \right) = 0 \cr & {\text{There is a local minimum at }}\left( {0,0} \right) \cr & \cr & *{\text{Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{{2 - 6{x^4}}}{{{{\left( {1 + {x^4}} \right)}^2}}} = 0 \cr & 2 - 6{x^4} = 0 \cr & {x^4} = \frac{1}{3} \cr & x = \pm \root 4 \of {\frac{1}{3}} \cr & {\text{Inflection points at }}x = \pm \frac{1}{{\root 4 \of 3 }} \cr & \cr & {\text{The denominator is never zero}}{\text{, then there are no}} \cr & {\text{vertical asymptotes}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {{{\tan }^{ - 1}}{x^2}} \right) = \frac{\pi }{2} \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {{{\tan }^{ - 1}}{x^2}} \right) = \frac{\pi }{2} \cr & {\text{Horizontal asymptote }}y = \frac{\pi}{2}. \cr & \cr & {\text{Graph}} \cr} $$