函数f:R→R满足下述条件:对所有实数x,有f(x+19)≤f(x)+19 和 f(x+94)≥f(x)+94.求证:对所有实数x,f(x+1)=f(x)+1.
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![函数f:R→R满足下述条件:对所有实数x,有f(x+19)≤f(x)+19 和 f(x+94)≥f(x)+94.求证:对所有实数x,f(x+1)=f(x)+1.](/uploads/image/z/10348211-11-1.jpg?t=%E5%87%BD%E6%95%B0f%EF%BC%9AR%E2%86%92R%E6%BB%A1%E8%B6%B3%E4%B8%8B%E8%BF%B0%E6%9D%A1%E4%BB%B6%EF%BC%9A%E5%AF%B9%E6%89%80%E6%9C%89%E5%AE%9E%E6%95%B0x%2C%E6%9C%89f%28x%2B19%29%E2%89%A4f%28x%29%2B19+%E5%92%8C+f%28x%2B94%29%E2%89%A5f%28x%29%2B94.%E6%B1%82%E8%AF%81%EF%BC%9A%E5%AF%B9%E6%89%80%E6%9C%89%E5%AE%9E%E6%95%B0x%2Cf%28x%2B1%29%3Df%28x%29%2B1.)
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