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(2011年江苏卷8)在平面直角坐标系xOy中,过坐标原点的一条直线与函数f(x)=2x的图象交于P、Q两点,则线段PQ长的最小值是.解:P、Q两点关于原点O对称,不妨设P(m,n)为第一象限中的点,则m>0,n>0,n=2m,所以|PQ|2=4|OP|2=4(m2+n2)=4(m2+4m2)≥16(当且仅当m2=44m2,即m=2时取等号)故线段PQ的长的最小值是4.本题上述解法主要考查函数、基本不等式性质等基础知识,换一个思维视角,实际上函数y=2x即xy=
(2011, Jiangsu Volume 8) In the plane rectangular coordinate system xOy, a straight line passing through the origin of the coordinate and the image of the function f(x)=2x intersect at two points P and Q. The minimum value of the PQ length of the line segment is the solution. : P, Q are symmetric about the origin O. Let P(m,n) be the point in the first quadrant. Then m>0, n>0, n=2m, so |PQ|2=4|OP| 2=4(m2+n2)=4(m2+4m2)≥16(If and only if m2=44m2, ie m=2, the equal sign is used) The long minimum value of the line segment PQ is 4. The above solution to the problem is mainly Examine the basic knowledge of functions, basic inequality properties, etc. For a thinking perspective, in fact the function y=2x is xy=