论文部分内容阅读
求函数 F(t)=dt+e/t~2+bt+c的最值,已有一些同志谈及,本文将用几何法讨论这个问题,它具有便捷、简单、直观的特点。在F(t)=dt+e/t~2+bt+c中,令y=dt+e…①x=t~2+bt+c…②于是求F(t)的最值,就是求y/x的最值。由①得t=y-e/d代入②则x=(y-e/2)~2+b·y-e/d+c 整理可得(y-2e-bd/2)~2=d~2(x-4c-b~2/4)…③它所表示的是以(1/4(4c-b~2),1/2(2c-bd)为顶点,y=1/2(2c-bd)为对称轴,开口向右的一条抛物线。由于y/x=y-0/x-0,y-0/x-0是坐标为(x,y)的点与坐标原点连线的斜率,于是求y/x的最值,就是求曲线③上的点与坐标原点连线斜率的最值。由切线的意义,显然我们有
Finding the function F(t)=dt+e/t~2+bt+c is the best value, and some comrades have already talked about it. This paper will discuss this problem using geometric methods. It is convenient, simple, and intuitive. In F(t)=dt+e/t~2+bt+c, let y=dt+e...1x=t~2+bt+c...2 then find the maximum value of F(t), that is, find y The maximum value of /x. From 1 to t=ye/d, then 2 x=(ye/2)~2+b·ye/d+c is collated (y-2e-bd/2)~2=d~2(x-4c) -b~2/4)...3 It is represented as (1/4(4c-b~2), 1/2(2c-bd) as a vertex, and y=1/2(2c-bd) as a symmetry. Axis, a parabola opening to the right, since y/x=y-0/x-0, y-0/x-0 is the slope of the point where the coordinate is (x,y) and the origin of the coordinate, so seek y The maximum value of /x is the value of the slope of the line connecting the point on the curve 3 and the origin of the coordinate. Obviously we have the meaning of the tangent line.