已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.f(x)=2sin(2x+π/6)+1记t=2x+π/6,t∈[π/6,2π+π/6]f(t)=2sin(t)+1f(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]
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![已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.f(x)=2sin(2x+π/6)+1记t=2x+π/6,t∈[π/6,2π+π/6]f(t)=2sin(t)+1f(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]](/uploads/image/z/10724884-52-4.jpg?t=%E5%B7%B2%E7%9F%A5%2C%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%3D2sin%EF%BC%882x%2B%CF%80%2F6%EF%BC%89%2B1%E6%B1%82%E5%AE%83%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E.f%28x%29%3D2sin%282x%2B%CF%80%2F6%29%2B1%E8%AE%B0t%3D2x%2B%CF%80%2F6%2Ct%E2%88%88%5B%CF%80%2F6%2C2%CF%80%2B%CF%80%2F6%5Df%28t%29%3D2sin%28t%29%2B1f%28t%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E4%B8%BA%EF%BC%9At%E2%88%88%5B%CF%80%2F6%2C%CF%80%2F2%5D%E6%88%96t%E2%88%88%5B3%CF%80%2F2%2C2%CF%80%2B%CF%80%2F6%5D%E6%AD%A4%E6%97%B6%EF%BC%9Ax%E2%88%88%5B0%2C%CF%80%2F6%5D%E6%88%96x%E2%88%88%5B2%CF%80%2F3%2C%CF%80%5D)
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已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.f(x)=2sin(2x+π/6)+1记t=2x+π/6,t∈[π/6,2π+π/6]f(t)=2sin(t)+1f(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]
已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.
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
已知,函数f(x)=2sin(2x+π/6)+1求它在[0,π]上的单调递增.f(x)=2sin(2x+π/6)+1记t=2x+π/6,t∈[π/6,2π+π/6]f(t)=2sin(t)+1f(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]
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
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