求函数f(x)=sinx+cosx在[-(π/2),π/2]上的最值
2个回答

f(x)=√2（sinxcosπ/4+cosxsinπ/4)=)=√2sin(x+π/4)

解法一：x∈(-π/2,π/2)则(x+π/4)∈(-π/4,3π/4)

令Y=x+π/4,则g(Y)=√2sinY又sinY,在Y∈[-π/4,π/2]单调递增,在Y∈[π/2,3π/4]单调递减,故

f(x)max=g(Y)max=g(π/2)=√2,f(x)min=g(Y)min=g(-π/4)=-1

解法二：f'(x)=√2cos(x+π/4)=0解得x=π/4

则有f(-π/2)=-1,f(π/2)=1,f(π/4)=√2

易知f(x)max=f(π/4)=√2,f(x)min=f(-π/2)=-1

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