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)
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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