Let f:R defined on f(x)=|x|-1/|x|+1, then f is ----nether one-one nor onto. How??

As given f(x)=(|x|-1)/(|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=1.5 both f has same image in co-domain so it is not one-one function. Now, definition of onto function is that if each element in the co-domain have at least one preimage in the domain. Here I had taken co-domain of f as R. Method to find if function is onto is that first find its inverse function and then find domain of inverse function. If domain of inverse function is equal to co-domain of given function then given function is onto function. Here (f^(-1))(x) = ((1-x)/(x-1))+{(1-x)/(x-1)}. {} is fractional part. Now our inverse function is not defined for x=1 so our function is not onto. Thank you. I am Ankit of class X.