Practicing Success
Show f: R → R defined by $f(x) = (x -1)(x-2)(x-3)$ is surjective but not injective. |
onto bijective one-one none of these |
onto |
We have f: R → 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. |