Question

已知函数f(X)=(sin2x+cos2x+1)/2cosx,求函数值域

Answer 1

先化简
f(x)
=(sin2x+cos2x+1)/2cosx
=(2sinxcosx+2(cosx)^2-1+1)/(2cosx)
=sinx+cosx
=√2(√2/2sinx+√2/2cosx)
=√2(sinxcosπ/4+sinπ/4cosx)
=√2sin(x+π/4)
-1<=sin(x+π/4)<=1
所以-√2<=f(x)<=√2
又因为cosx≠0，即x≠kπ
sin(x+π/4)≠±√2/2
f(x)≠±1

所以-√2<=f(x)<=√2，且f(x)≠±1

Answer 2

先化简
f(x)=(sin2x+cos2x+1)/2cosx=(2sinxcosx+2(cosx)^2-1+1)/(2cosx)
=sinx+cosx
易知定义域为{x|cosx不等于0}
所以其值域为[-根号2,-1]U[1,根号2]

Answer 3

f(X)=(sin2x+cos2x+1)/2cosx=(2sinxcosx+2(cosx)^2)/2cosx
=sinx+cosx=√2sin(x+π/4)