How Can We Restrict the Function f(x)=x^3-x to Create a Bijective Function g?

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To create a bijective function g from f(x) = x^3 - x, the domain must be restricted to intervals where f is either non-increasing or non-decreasing, avoiding intervals that include more than one root. The roots of f are found at x = -1, 0, and 1, which can guide the selection of these intervals. A proposed mapping defines g such that g(-1) = -∞, g(0) = 0, and g(1) = ∞, establishing a one-to-one correspondence. While multiple intervals can yield a valid g, the key is ensuring that the chosen interval maintains the function's injectivity. Thus, careful selection of the domain is essential for achieving a bijective function.
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Homework Statement


Let f: R->R and f(x)=x3-x. By restricting the domain and range of f appropriately, obtain from f a bijective function g.

Homework Equations


x3-x=(x+1)(x-1)x
g(x): R->R

The Attempt at a Solution


we can find roots from the polynomial form (x+1)(x-1)x and restrict the domain and range by avoiding intervals including more than one root. There are several ways to obtain g, because we are free to choose an interval, but what I'm interested in is this.

if we say g:R->R such that
x=-1 -> g(x)=-∞
x=0 -> g(x)=0
x=1 -> g(x)=∞

we can define a 1 to 1 map because R is an infinite set, then can we say this is also an answer? Because it is bijective to R.
 
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Although they do not say so explicitly, one can infer that they want the function g to match f on the shared domain.
Since f is not injective, some of the domain of f is going to have to be removed.

BTW, avoiding intervals including more than one root is not necessary or sufficient. What you need is an interval on which f is purely non-increasing or purely non-decreasing.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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