Question

求f(x)=sin²x+cosx，x∈[-π/3,5π/6]的值域

Answer 1

求f(x)=sin²x+cosx，x∈[-π/3,5π/6]的值域

解：f(x)=1-cos²x+cosx=-cos²x+cosx+1=-(cos²x-cosx)+1
=-[(cosx-1/2)²-1/4]+1=-(cosx-1/2)²+5/4≦5/4
当-π/3≦x≦5π/6时，-(√3)/2≦cosx≦1，cosx=1/2∈[-(√3)/2，1]，故f(x)能获得最大值5/4；
当x=-5π/6时cos(5π/6)=cos(π-π/6)=-cos(π/6)=-(√3)/2，此时f(x)获得最小值;
minf(x)=f(5π/6)=-[-(√3)/2-1/2]²+5/4=-(2+√3)/2+5/4=(1-2√3)/4.
即值域为[(1-2√3)/4，5/4]

Answer 2

f(x)=sin²x+cosx
=1-cos²x+cosx
=-cos²x+cosx-1/4+5/4
=-(cosx-1/2)²+5/4
cos(-π/3)=1/2
cos0=1
cos(π/2)=0
cos(5π/6)=-√3/2
-(√3/2+1/2)²≤-(cosx-1/2)²≤0
-1-√3)/2≤-(cosx-1/2)²≤0
1/4-√3/2≤f(x)≤5/4

Answer 3

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