(1) y=x²-2x,x∈{0,1,2,3} 值域为:y∈{0,-1,3}
(2)y=x²-4x+6,x∈[1,5)
=(x-2)²+2
所以可得当x=2有最小值为:y=2
当x=5时有最大值为:y=11
所以可得其值域为:y∈[2,11)
求下列函数的值域 (1) y=x²-2x,x∈{0,1,2,3} (2)y=x²-4x+6,x∈[]
求下列函数的值域 (1) y=x²-2x,x∈{0,1,2,3} (2)y=x²-4x+6,x∈[]
求下列函数的值域
(1) y=x²-2x,x∈{0,1,2,3}
(2)y=x²-4x+6,x∈[1,5)
求下列函数的值域
(1) y=x²-2x,x∈{0,1,2,3}
(2)y=x²-4x+6,x∈[1,5)
数学人气:727 ℃时间:2019-09-09 11:38:35
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