Question

已知函数f(x)=sinx/2cosx/2+cos^2x/2-2,求f(x)在[-π/2,7π/12]的最大值和最小值

Answer 1

f(x)=1/2sinx+(1+cosx)/2-2=√2/2sin(x+π/4)-3/2
x属于[-π/2,7π/12] 则x+π/4属于[-π/4,5π/6] 则sin(x+π/4)属于[-√2/2,1]
函数最小值-2 最大值(√2-3)/2

Answer 2

解：f(x)=sin(x/2)cos(x/2)+cos^2(x/2)-2
=(1/2)×sinx+(1/2)×(cosx+1)-2
=(1/2)×(sinx+cosx)-3/2
=(√2/2)×sin(x+π/4)-3/2

T=2π/1=2π

当a∈[π，17π/12]时，
x+π/4∈[5π/4,5π/3]，
则当x+(π/4)=3π/2,
即x=5π/4时，f(x)min=-√2/2-3/2

当x+(π/4)=(5π)/4,即x=π时，
f(x)max=-1/2-3/2=-2。