Find the minimum value of the function $$\frac{{40}}{{3{x^4} + 8{x^3} - 18{x^2} + 60}}$$

Solution :
$$\eqalign{ & {\text{Let }}y = \frac{1}{{40}}\left( {3{x^4} + 8{x^3} - 18{x^2} + 60} \right) \cr & \Rightarrow \frac{{dy}}{{dx}} = \frac{1}{{40}}\left( {12{x^3} + 24{x^2} - 36x} \right) \cr & {\text{and }}\frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{{40}}\left( {36{x^2} + 48x - 36} \right) \cr & {\text{Now, }}\frac{{dy}}{{dx}} = 0 \Rightarrow {x^3} + 2{x^2} - 3x = 0 \cr & {\text{or }}x\left( {x - 1} \right)\left( {x + 3} \right) = 0 \cr & {\text{or}}\,\,x = 0,\,1,\, - 3 \cr & {\text{At }}x = 0,\,\frac{{{d^2}y}}{{d{x^2}}} = - 36 < 0 \cr & \therefore y{\text{ is maximum at }}x = 0 \cr & \Rightarrow {\text{The given function i}}{\text{.e}}{\text{., }}\frac{1}{y}{\text{ is minimum at }}x = 0 \cr & \therefore {\text{minimum value of the function}} = \frac{{40}}{{60}} = \frac{2}{3} \cr} $$