求函数f(x)=x-cosx(sinx+cosx),x∈[-π/4,3π/4]的最大值和最小值
热心网友
求f(x)=x-cosx(sinx+cosx),x∈[-π/4,3π/4]的最大值和最小值 令导数f'(x)=1-cosx(cosx-sinx)+sinx(sinx+cosx)=1-cos^x+sin^x+2sinxcosx=2sin^x+2sinxcosx=2sinx(sinx+cosx)=2√2sinxsin(x+π/4)=√2cos(π/4)-√2cos(2x+π/4)=1-√2cos(2x+π/4)=0cos(2x+π/4)=√2/2∵x∈[-π/4,3π/4],∴2x+π/4∈[-π/4,7π/4],∴2x+π/4=-π/4、π/4、7π/4，∴函数f(x)的极值点为x=-π/4、0、3π/4二阶导数f''(x)=-2√2sin(2x+π/4)∵f''(-π/4)=f''(3π/4)=20，f''(0)=-2<0∴f(x)在x=-π/4或3π/4时有最小值min[f(-π/4),f(3π/4)]=f(-π/4)=-π/4f(x)在x=-0时有最大值f(0)=-1。
热心网友
f(x)=x-cosx(sinx+cosx) = x - (根号2)*[cos(2x-π/4)]/2 - 1/2做函数 y1 = x -1/2 及 y2 = (根号2)*[cos(2x-π/4)]/2 的图象。在区间 x∈[-π/4,3π/4] 内，可得 y1-y2 的极值为：函数最大值 = 5π/8 -1/2 +(根号2)/2，（x = 5π/8 时）函数最小值 = π/8 -1/2 -(根号2)/2，（x = π/8 时）