2. y=x/(x^2+9) yx^2-x+9y=0 D=1^2-4*9y^2>=0 y^2<=1/36 -1/6<=y<=1/6 From solution of a), -1/6<=x/(x^2+9)<=1/6 <+(*) Let M=|x|/(2x^2+18) 2(x^2+9)M=|x| 2(x^2+9)M=x OR 2(x^2+9)M=-x (provided yhat M>=0) 2M=x/(x^2+9) OR -2M=x/(x^2+9) From (*), ±2M<=1/6 -1/12<M<=1/12 -1/12<=|x|/(2x^2+18)<=1/12 But M>=0 0>=|x|/(2x^2+18)>=1/12 <+(**) (|x|+x)/(2x^2+18) |x|/(2x^2+18)+x/(2x^2+18) |x|/(2x^2+18)+1/2 * x/(x^2+9) 1/2(*) + (**), -1/12-1/12<=x/(2x^2+18)+(|x|+x)/(2x^2+18)<=1/12+1/12 -1/12<=(|x|+x)/(2x^2+18)<=1/6 [ Last edited by 落雷 on 2004-10-19 at 07:48 PM ] |