Question about inverse? the function f(x) = x^2*sin 1/x for x not equal to zero (0) . find inverse of the f(x) for x not equal zero

I'm pretty sure I already answered this question already, but I'll do it again.

A function `f` is inversible if and only if it is injective. Basically, it is a "one-to-one" function, meaning there exists no two elements in the domain of `f` that are mapped to the same element in the codomain of `f`.

Let `f:X->Y`. Below is an example of an injective function.

![https://en.wikipedia.org/wiki/Injective_function]()

A function is inversible only if it is injective due to the definition of a function.

A function exists if there is no element `a` in its domain such that `f(a)` is multivalued. As such, if `f` is a non-injective function then there must exist `a` and `b` so that `f(a)=f(b)=alpha`. The inverse of `f`, `f^-1`, will map `alpha` to both `a` and `b`, thus making it not a function.

In our particular case, `f(x) = x^2sin(1/x)`.

Let `delta=1/pi` then `epsilon=-1/pi`:

`f(delta) = (color(red)(1/pi))^2sin(1/color(red)(1/pi))=(1/pi)^2sinpi=0`

`f(epsilon) = (color(red)(-1/pi))^2sin(1/color(red)(-1/pi)) = (1/pi)^2sin(-pi)=0`

As such, `f` is not injective and thus doesn't have an inverse.