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If a>0, then x is in the interval [1, +∞), and f (x) = x+2+a/x > 0 also holds.
Generally speaking, a & gt-3
2. Similarly, if 2 ≦ x2 < x 1, then f(x 1)-f(x2) is substituted for simplification, so that f (x1)-f (x2) = (x1-x2). And f(x) is increasing function in the interval [2, +∞), then f (2) =11¢ a is constant, so a.
3. Let 2 ≦ x2 < x 1 ≦ 5, and you can get f (x1)-f (x2) = (x2-x1)/[(x1-kloc-0). 0, it can be seen that this function is a subtraction function, so f (x)