Question

求函数f(x)=sinx+cosx在{-2分之π,2分之π}上的最大值和最小值

Answer 1

解：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

Answer 2

f(x)=sinx+cosx=根号2sin(x+π/4)
x∈[-π/2,π/2]
故x+π/4∈[-π/4,3π/4]
故f(x)∈[-根号2,1]