证明:曲线y=(x-1)/(x^2+1)有三个拐点,且这三个拐点位于同一直线上,
问题描述:
证明:曲线y=(x-1)/(x^2+1)有三个拐点,且这三个拐点位于同一直线上,
答
y=(x-1)/(x^2+1),y'=(x^2+1-2x(x-1))/(x^2+1)^2=(-x^2+2x+1)/(x^2+1)^2y''=[(-2x+2)(x^2+1)^2-(-x^2+2x+1)(2(x^2+1)*2x]/(x^2+1)^4=[(2-2x)(x^2+1)-4x(-x^2+2x+1)]/(x^2+1)^3=(2x^2+2-2x^3-2x+4x^3-8x^2-4x}/(x^2+1)...