## Function question. Is this correct?

Umm that Y is > or equal to 4 no matter number you use?

 Quote by lionely Umm that Y is > or equal to 4 no matter number you use?
What is your reasoning? Substitute some values for $x$ into $f$ and see what you get, or what you can't get. Does your answer change? Even better, use graphing software to visualize $f$, and then everything should be pretty clear. (Example, search "wolframalpha" into a search engine and type "f(x) = 4 - x^2" into the search bar on the website. A graph over a restricted amount of points will be generated. Personally, I would recommend solving this algebraically, but analytically is totally fine, too.)
 But solving it alebraically isn't it this?? y= 4-x^2 x^2= 4-y x= sqrt(4-y) putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?

Mentor
 Quote by lionely But solving it alebraically isn't it this?? y= 4-x^2 x^2= 4-y x= sqrt(4-y)
The line above should be x = ±sqrt(4-y)

 Quote by lionely putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?
In other words, if y = 5, there is no real value of x for which 4 - x2 = 5.

As you said earlier(or at least, alluded to), the range of the function y = 4-x2 is the set {y | y ≤ 4}. This should be enough to convince anyone that this function is not onto the reals.

 Quote by lionely But solving it alebraically isn't it this?? y= 4-x^2 x^2= 4-y x= sqrt(4-y) putting in y=5 x would = sqrt(-1) so... it's not surjective as the codomain doesn't equal the range?
Do you mean letting $y=5$? If so, then yes, you get $\pm \sqrt{-1}$, which is not a real number, and therefore there does not exist any $x$ in the domain of $f$ such that $f(x)=5$. Consequently, the range of $f$ does not include $5$, for example. That is enough to show that surjectivity is not held by $f$, as the set of real numbers (the domain) does not equal the set of real numbers missing $5$ (the codomain) (of course, it missed more than just $5$).

Recall you said $f$ attains all numbers greater than or equal to $5$ only--do you see the error now?

Also, recall that taking the square root always requires one to put a plus or minus $\pm$ sign in front of the root. Why? Well, for any number $x$ such that the square root of it is defined, we have both $(\sqrt{x})^2 = x$ and $(-\sqrt{x})^2 = x$.