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
LC
