已知f(x)=sin2x-x,则在区间【-2分之派,2分之派】上的最大值为?最小值为?
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解:
f'(x)=2cos2x-1
令f'(x)>0,得cos2x>1/2,即2x∈[-π/3+2kπ,π/3+2kπ],即x∈[-π/6+kπ,π/6+kπ]
又∵x∈[-π/2,π/2],∴x∈[-π/6,π/6]
令f'(x)=0,得x=±π/6
令f'(x)<0,x∈[-π/2,-π/6]∪[π/6,π/2]
∴当x=-π/6时,f(x)取极小值点,极小值为π/6-√3/2
当x=π/6时,f(x)取极大值点,极大值为√3/2-π/6
端点值f(-π/2)=π/2,f(π/2)=-π/2
∵极小值>-π/2,所以最小值为端点值f(π/2)=-π/2
∵极大值<π/2,所以最大值为端点值f(-π/2)=π/2
综上所述:
f(x)=sin2x-x在[-π/2,π/2]上的最大值为π/2,最小值为-π/2
f'(x)=2cos2x-1
令f'(x)>0,得cos2x>1/2,即2x∈[-π/3+2kπ,π/3+2kπ],即x∈[-π/6+kπ,π/6+kπ]
又∵x∈[-π/2,π/2],∴x∈[-π/6,π/6]
令f'(x)=0,得x=±π/6
令f'(x)<0,x∈[-π/2,-π/6]∪[π/6,π/2]
∴当x=-π/6时,f(x)取极小值点,极小值为π/6-√3/2
当x=π/6时,f(x)取极大值点,极大值为√3/2-π/6
端点值f(-π/2)=π/2,f(π/2)=-π/2
∵极小值>-π/2,所以最小值为端点值f(π/2)=-π/2
∵极大值<π/2,所以最大值为端点值f(-π/2)=π/2
综上所述:
f(x)=sin2x-x在[-π/2,π/2]上的最大值为π/2,最小值为-π/2
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