Question

已知，函数f（x）=2sin（2x+π/6）+1求它在[0，π]上的单调递增。求大神

答案给
f(x)=2sin(2x+π/6)+1

记t=2x+π/6，t∈[π/6,2π+π/6]
f(t)=2sin(t)+1
f(t)的单调递增区间为：
t∈[π/6,π/2]或t∈[3π/2,2π+π/6]
此时：
x∈[0,π/6]或x∈[2π/3,π]

我就想知道明明是t∈[π/6,2π+π/6]是咋个变成
t∈[π/6,π/2]或t∈[3π/2,2π+π/6]的

Answer 1

t∈[π/6,2π+π/6]是t的取值范围
而t∈[π/6,π/2]或t∈[3π/2,2π+π/6]
是函数的单调增区间
∵sinα的单调递增区间为[-π/2+2kπ，π/2+2kπ]
∴k取0时，为[-π/2，π/2]得出t∈[π/6,π/2]
k取1时，为[3π/2，5π/2]得出t∈[3π/2,2π+π/6]
∴可得出t∈[π/6,π/2]或t∈[3π/2,2π+π/6]

追问

请问k取0时是为[-π/2，π/2]，那t咋个就从[π/6,2π+π/6]变成
t∈[π/6,π/2]的呢？我比较笨求讲解

追答

我白天不在，既然他给你讲解了就行了

Answer 2

只帮你理解答案，不从解题的角度，你先不要管X，只考虑t，f(t)=2sin(t)+1的单调递增区间为t∈[π/6,π/2]或t∈[3π/2,2π+π/6]，这个好理解吧？然后在考虑t=2x+π/6，t∈[π/6,π/2]或t∈[3π/2,2π+π/6]时，x∈[？]就行了

追问

额。我就是不懂f(t)=2sin(t)+1的单调递增区间为t∈[π/6,π/2]或t∈[3π/2,2π+π/6]是咋个从
t∈[π/6,2π+π/6]求出来的。。后面我都知道- -。。
呵呵我有点笨，求大神帮帮忙

追答

结合sin(t)函数图形，他的递增区间为t∈[-π/2+2kπ,π/2+2kπ]，当K=0时，t∈[-π/2,π/2]，但是由于x取值的限制t∈[π/6,2π+π/6]，也就是说t最小值是π/6，最大值是2π+π/6，t要从这个区间取值，所以单调递增区间为t∈[π/6,π/2]或t∈[3π/2,2π+π/6]，就是t∈[π/6,π/2]是递增的，然后t∈[π/2,3π/2]递减，到t∈[3π/2,2π+π/6]又是递增的了