设函数y=sin(π/2x+π/3)若对任意x∈R,存在x1、x2使f(x1)≤f(x)≤f(x2)恒成立,则绝对值x1-x2的最_数学解答

设函数y=sin(π/2x+π/3)若对任意x∈R,存在x1、x2使f(x1)≤f(x)≤f(x2)恒成立,则绝对值x1-x2的最小值

人气：282 ℃　时间：2019-08-20 14:39:11

解答

由题意可知f(x1)=f(x)min=-1=>sin(π/2x1+π/3)=-1=>π/2x1+π/3=2k1π-π/2=>x1=1/(4k1-5/3)同理f(x2)=f(x)max=1=>sin(π/2x2+π/3)=1=>π/2x2+π/3=2k2π+π/2=>x2=1/(4k2+1/3)|x1-x2|=|1/(4k1-5/3)-1/(4k2+1/3)|...