Show that `f: R->R` defined by `f(x)=(x-1)(x-2)(x-3)` is surjective but not injective.

Explanation

We have
` f: R to R, " where " f(x) =(x-1)(x-2)(x-3)`
Clearly `f(1)=f(2)=f(3)=0.`
So, f(x) is many-one.
Also, when x approaches to infinity, f(x) approaches to infinity and when x approaches to negative infinity, f(x) approaches to negative infinity.
Since, polynomial function continuously exists for all real x, given function f(x) takes all real values.
Hence, range of the f(x) is R.
Therefore , f(x) is onto.