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1.y=sin(1/2x+π/6),x属于[0,π/3] 2.y=-cos(3x-π/3),x属于[-π/3,π/3]
人气:181 ℃ 时间:2020-05-10 19:32:24
解答
1
y=sin(1/2x+π/6)
x属于[0,π/3]
π/6≤1/2x+π/6≤π/3
1/2 ≤ sin(1/2x+π/6) ≤ 根号3/2
值域【1/2,根号3/2】
2
y=-cos(3x-π/3)
x属于[-π/3,π/3]
-π-π/3 ≤ 3x-π/3≤ π- π/3
(3x-π/3)的变化范围涵盖了从--π-π/3 到π- π/3正好2π的区间
∴y=-cos(3x-π/3)值域为【-1,1】
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