Function question. Is this correct?

lionely

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

5hassay

What is your reasoning? Substitute some values for [itex]x[/itex] into [itex]f[/itex] and see what you get, or what you can't get. Does your answer change? Even better, use graphing software to visualize [itex]f[/itex], 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.)

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?

Mark44

The line above should be x = ±sqrt(4-y)

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.

5hassay

Do you mean letting [itex]y=5[/itex]? If so, then yes, you get [itex]\pm \sqrt{-1}[/itex], which is not a real number, and therefore there does not exist any [itex]x[/itex] in the domain of [itex]f[/itex] such that [itex]f(x)=5[/itex]. Consequently, the range of [itex]f[/itex] does not include [itex]5[/itex], for example. That is enough to show that surjectivity is not held by [itex]f[/itex], as the set of real numbers (the domain) does not equal the set of real numbers missing [itex]5[/itex] (the codomain) (of course, it missed more than just [itex]5[/itex]).

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

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