## Function question. Is this correct?

1. The problem statement, all variables and given/known data
h:x → 4-x2, x E ℝ

show that it is not surjective(not onto ℝ)

3. The attempt at a solution

Since the line tests fail.

y= 4-x^2

x= √(4-y) = 2√-y

A root of a negative number is not possible so f(x) is not surjective onto R
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 Quote by lionely 1. The problem statement, all variables and given/known data h:x → 4-x2, x E ℝ show that it is not surjective(not onto ℝ) 3. The attempt at a solution Since the line tests fail.
If it fails the horizontal line test (meaning that a horizontal line intersects two or more points on the graph), that means that the function is not one-to-one. What did you mean?

Do you understand the definition of "onto" (or surjective)?
 Quote by lionely y= 4-x^2 x= √(4-y) = 2√-y
What are you doing here? When you solve for x, you should get two values; namely, x = ±√(4 - y). Also, it is NOT true that √(4-y) = 2√-y.
 Quote by lionely A root of a negative number is not possible so f(x) is not surjective onto R
 I meant when I did the line tests it looked like the function was injective and surjective and hm... I'm not too sure about the 2nd part now umm x=±√(4-y) if y is like 3 x is a real number..

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## Function question. Is this correct?

For the function y = x2 - 4 to be onto the real numbers, it must be true that any choice of y is paired with some value of x.

Have you graphed this equation? That would probably give you a good idea about whether it is onto the reals. That wouldn't be proof, but it would get you thinking the right way.
 I only sketched a graph,

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 Quote by lionely I only sketched a graph,
Ok, then what's a value in R that x^2-4 can never equal?

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 Quote by lionely x=±√(4-y) if y is like 3 x is a real number..
Sure, but to be surjective on ℝ the inverse must have a solution for all real y. Does it?
 No y is not surjective for the positive Real numbers though.. "Ok, then what's a value in R that x^2-4 can never equal? " Umm I'm not sure..

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Almost correct. First, as Mark44 said, it should be "$\pm$" and surely you know that "4- y" is NOT "4 times -y"! $x= \pm\sqrt{4- y}[/tex]. Now, can you find a value of y so that 4- y< 0?  Can't 5 make it <0? Recognitions: Homework Help Science Advisor  Quote by lionely Can't 5 make it <0? If you mean there is no real value of x such that 4-x^2=5, that would be correct.  But aren't negative numbers be.. real? Recognitions: Homework Help Science Advisor  Quote by lionely Can't 5 make it <0? Of course, but HofI wasn't suggesting you could not. Halls just said, find such a value of y. Now, having found it, what value of x will be mapped to this y?  If x=-4 that could make f(x) < 0 Recognitions: Homework Help Science Advisor  Quote by lionely If x=-4 that could make f(x) < 0 You are not trying to make f(x) < 0. Go and look at Halls' post again. You were trying to make 4-y < 0. You correctly found that y=5 would do that. Now the question is whether you can find an x that makes f(x) = 5. If no such x exists then f is not surjective.  Oh well it's not surjective then because I can't think of any value... If this is the answer I'm sorry for being so difficult, I need MUCH more practice in functions..  Your notation for the function definition isn't correct, I think. I believe you meant [itex]f : \mathbb{R} \rightarrow \mathbb{R}, x \in \mathbb{R} \mapsto 4 - x^2$. Here, $\mathbb{R}$ is both the domain and codomain of $f$. Surjectivity is the property that the image of the domain of $f$, which is defined and denoted to be $f[\mathbb{R}]=\{f(x) : x \in \mathbb{R}\}$, equals the codomain of $f$. Thus, we want to see if we can generate all the real numbers with $f$. Analytically, this function is a parabola starting at $(0,4)$ and opening down. What does this imply, then? Also, a algebraic argument can provide a solution. Suppose $y \in \mathbb{R}$ is some value in the codomain of $f$. Furthermore, suppose that there exists some value $x \in \mathbb{R}$ in the domain of $f$ such that $f(x)=4-x^2=y$. If you solve for $y$, what do you discover?