f(x) = |x + 1| for ≤ 0  
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No, it is not one-to-one for x≤0.

When -1<x≤0, 0<f(x)≤1, but when -2<x≤-1, 0≤f(x)<1.

Therefore, we have the same value for f(x) for many different values of x. For example, if x=-½ or -1½, f(x)=½. One-to-one requires a unique mapping from x to f(x), and vice versa. f(x)=½ in this example maps back to two different values of x.


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