I have been given to prove that $f:(- \infty, -1] \rightarrow (- \infty, 3]$ defined by $f(x) = x^3-3x+2$ is not onto.
Now I found that the function is increasing in its given domain, and it's range is $(- \infty, 4]$
The definition of an onto function is that "if $f:A \rightarrow B$ then every element of B must have a pre image in A"
Here, that implies, every element of $(- \infty, 3]$ must have a pre image in the domain right? Isn't that what's happening? Then why isn't this an onto function exactly?
Best Answer
The question is wrong. $f$ is not a function from $(-\infty, -1]$ into $(-\infty, 3]$ since it takes values outside $(-\infty, 3]$.
Perhaps they wanted to ask if $(-\infty, 3]$ is contained in the range of $f$ when the codomain of $f$ is taken as $\mathbb R$. In that case the answer is YES.