用换元法求值域f(x)=cos2x+3sinx+3
问题描述:
用换元法求值域
f(x)=cos2x+3sinx+3
答
f(x)=cos2x+3sinx+3=-2(sinx)^2+3sinx+4=-2(sinx-3/4)+41/8.
-1所以当sinx-3/4=-1是最大,f(x)=41/8+2.=57/8
当sinx-3/4=1最小,f(x)=41/8-2=25/8
答
f(x)=cos2x+3sinx+3=1-2Sin^2x+3sinx+3
令t=sinx
f(t)=-2t^2+3t+4,其中-1