若函数f(x+1)的定义域是[-2,3],则y=f(2x-1)的定义域是?解析是.因为函数y=f(x+1)的定义域为x∈[-2,3],即-1≤x+1≤4,所以函数f(x)的定义域为[-1,4].由f(x)与f(2x-1)的关系可得-1≤2x-1≤4,可是为
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![若函数f(x+1)的定义域是[-2,3],则y=f(2x-1)的定义域是?解析是.因为函数y=f(x+1)的定义域为x∈[-2,3],即-1≤x+1≤4,所以函数f(x)的定义域为[-1,4].由f(x)与f(2x-1)的关系可得-1≤2x-1≤4,可是为](/uploads/image/z/12010540-4-0.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%2B1%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF%5B-2%2C3%5D%2C%E5%88%99y%3Df%282x-1%29%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E6%98%AF%3F%E8%A7%A3%E6%9E%90%E6%98%AF.%E5%9B%A0%E4%B8%BA%E5%87%BD%E6%95%B0y%3Df%EF%BC%88x%2B1%EF%BC%89%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAx%E2%88%88%5B-2%2C3%5D%2C%E5%8D%B3-1%E2%89%A4x%2B1%E2%89%A44%2C%E6%89%80%E4%BB%A5%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BA%5B-1%2C4%5D%EF%BC%8E%E7%94%B1f%EF%BC%88x%EF%BC%89%E4%B8%8Ef%EF%BC%882x-1%EF%BC%89%E7%9A%84%E5%85%B3%E7%B3%BB%E5%8F%AF%E5%BE%97-1%E2%89%A42x-1%E2%89%A44%2C%E5%8F%AF%E6%98%AF%E4%B8%BA)
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