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求下列函数的值域1)sinx+cosx;2)sin^x-cosx+1;3)y=cosx/(2cosx+1)
人气:391 ℃ 时间:2019-12-08 07:38:59
解答
sinx+cosx = (根号2)sin(x+π/4) 值域[-根号2,根号2]sin^x-cosx+1 = 1-cos^x+1 = -(cosx+1/2)^+9/4值域[0,9/4]y = cosx/(2cosx+1) 所以cosx = y/(2y-1)在-1到1之间解得值域(-无穷,1/3]∪[1,+无穷)...
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