# Proving this function is not surjective

I have the following question $h(x)=4-x^2, x\in \Bbb{R}$

From my understanding it is only surjective when for all values in the Range that is in the Co domain there exist a $x$ value in the domain for it, so the co-domain and range must be the same, not surjective I assume that a value exist in the co domain that doesn't exist in the range.

This is what I tried:

$-(\sqrt{y-4})=x$

when y = 3:

$-(\sqrt{1})=x$

Since the co-domain is restricted to Real numbers and that is an imaginary number can I safely say that it is not surjective ?

• h(1) = 3. But, try to find something that maps to 2. Nov 9, 2016 at 2:56
• @DougM but is that the mathematical way to solve it , trying values ? i dont think i am doing it properly. Nov 9, 2016 at 3:05
• You only need one example to show that it is not a subjection. If x is in the integers, there is no x that maps to 2, and you are done. That is the "mathematical" way to solve it. Nov 9, 2016 at 3:07
• @user2861799 May I know the co-domain is $\Bbb{R}$ or $\Bbb{Z}$? Nov 9, 2016 at 3:16

$h(x) \le 4$ for all $x \in \mathbb R$.
Therefore, there is no $x$ such that $h(x)=5$ for instance. Thus, $h$ is not surjective.