Question

求函数y=3sin(π/4-2x)的单调区间并确定其值域

详细过程

Answer 1

因为y=sinx的单调增区间为[-π/2+2kπ，π/2+2kπ]k属于整数，,-π-4kπ,-π-4kπ+π/4y=3sin(π/4-2x)的单调增区间为
[-3π/4-4kπ,-4kπ+5π/4],k属于整数
-1所以值域为[-3,3]
希望对你有帮助

Answer 2

sinx单调增区间 2kπ-π/2 ≤ x ≤ 2kπ+π/2
令2kπ-π/2 ≤ π/4-2x ≤ 2kπ+π/2，可求y的单调增区间。[kπ-π/8,kπ+3π/8]
sinx单调减区间 2kπ+π/2 ≤ x ≤ 2kπ+3π/2
令2kπ+π/2 ≤ π/4-2x≤ 2kπ+3π/2，可求单调减区间。[kπ-5π/8,kπ-π/8]
值域[-3,3]

Answer 3

画草图，得
增区间：[3π/8+kπ，7π/8+kπ]
减区间：[-π/8+kπ，3π/8+kπ]
k是正整数。
值域：[-3,3]

Answer 4

递增区间-π/8+kπ~3π/8+kπ
递减区间3π/8+kπ~7π/8+kπ
值域[-3,3]