Let f: R → R is defined by f(x) = |x| / (|x| + 1). Then f is:

As given f(x) = (|x| / (|x| + 1)). We know that the definition if if distinct elements in the domain of a function f have distinct images in the co-domain, then f is said to be one-one. Here for x = 1 and x = -1, both f has the same image in the co-domain so it is not a one-one function. Now, the definition of an onto function is that if each element in the co-domain has at least one preimage in the domain. Here I had taken the co-domain of f as R. The method to find if a function is onto is that first find its inverse function and then find the domain of the inverse function. If the domain of the inverse function is equal to the co-domain of the given function then the given function is an onto function. Here (f⁻¹)(x) = (1 - x) / (x - 1). is the fractional part. Now our inverse function is not defined for x = 1 so our function is not onto.