The curve C has the equation y=3x/(9+x^2 ) (a) Find the turning points of the curve C (b) Using the fact that (d^2 y)/(dx^2 )=(6x(x^2-27))/(x^2+9)^3 or otherwise, classify the nature of each turning point of C

(a)To find the turning points of a curve, need to solve dy/dx=0. Using the quotient rule one can differentiate y:y=f(x)/g(x) dy/dx=(f'(x)g(x)-f(x)g'(x))/(g(x))²f(x)=3x, f'(x)=3, g(x)=(9+x² ), g'(x)=2x→ dy/dx=(3(9+x² )-(3x)(2x))/(9+x²)² =(3(9-x²))/(9+x² )²When dy/dx=0→ 9-x²=0→x²=9→x=±3x₁=3 y₁=(3)(3)/(9+3² )=9/18=1/2 x₂=-3 y₂=(3)(-3)/(9+(-3)² )=-9/18=-1/2 Two turning points are P₁=(3,1/2) and P₂=(-3,-1/2)(b) (d² y)/(dx² ) (x₁ )=(6)(3)(3²-27)/(3²+9)³=(18)(9-27)/(9+9)³=-18²/18³=-1/18<0→P₁ is a maximum turning point (d² y)/(dx² ) (x₂ )=(6)(-3)(-3²-27)/(-3²+9)³=(-18)(9-27)/(9+9)³=18²/18³=1/18>0→P₂ is a minimum turning point