设f(x)=sin(2x-π/6)
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设f(x)=sin(2x-π/6)
求f(x)的最小正周期
求f(x)的最值及相应的x的值
求f(x)在[-π/4,π/4]上的最大值和最小值
求f(x)的最小正周期
求f(x)的最值及相应的x的值
求f(x)在[-π/4,π/4]上的最大值和最小值
(1)Tmin=2π\2=π
(2)当x=π/2+2kπ时,即x=π/3+kπ时,f(x)max=1;
当x=3π/2+2kπ时,即x=5π/3+kπ时,f(x)min=-1;
(3)x∈[-π/4,π/4],则2x-π/6∈[-2π/3,π/3],所以
f(x)min=f(-π/6)=sin(-π/2)=-1;
f(x)max=f(π/4])=sin(π/3)=二分之根号3
(2)当x=π/2+2kπ时,即x=π/3+kπ时,f(x)max=1;
当x=3π/2+2kπ时,即x=5π/3+kπ时,f(x)min=-1;
(3)x∈[-π/4,π/4],则2x-π/6∈[-2π/3,π/3],所以
f(x)min=f(-π/6)=sin(-π/2)=-1;
f(x)max=f(π/4])=sin(π/3)=二分之根号3
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