Function question. Is this correct?

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.

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

Mark44

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)?
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.

lionely

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..

Mark44

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.

lionely

I only sketched a graph,

Dick

Ok, then what's a value in R that x^2-4 can never equal?

haruspex

Sure, but to be surjective on ℝ the inverse must have a solution for all real y. Does it?

lionely

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..

HallsofIvy

Almost correct. First, as Mark44 said, it should be "[itex]\pm[/itex]" and surely you know that "4- y" is NOT "4 times -y"! [itex]x= \pm\sqrt{4- y}[/tex]. Now, can you find a value of y so that 4- y< 0?

lionely

Can't 5 make it <0?

Dick

If you mean there is no real value of x such that 4-x^2=5, that would be correct.

lionely

But aren't negative numbers be.. real?

haruspex

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?

lionely

If x=-4 that could make f(x) < 0

haruspex

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.

lionely

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..

5hassay

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[/itex].

Here, [itex]\mathbb{R}[/itex] is both the domain and codomain of [itex]f[/itex].

Surjectivity is the property that the image of the domain of [itex]f[/itex], which is defined and denoted to be [itex]f[\mathbb{R}]=\{f(x) : x \in \mathbb{R}\}[/itex], equals the codomain of [itex]f[/itex].

Thus, we want to see if we can generate all the real numbers with [itex]f[/itex].

Analytically, this function is a parabola starting at [itex](0,4)[/itex] and opening down. What does this imply, then?

Also, a algebraic argument can provide a solution. Suppose [itex]y \in \mathbb{R}[/itex] is some value in the codomain of [itex]f[/itex]. Furthermore, suppose that there exists some value [itex]x \in \mathbb{R}[/itex] in the domain of [itex]f[/itex] such that [itex]f(x)=4-x^2=y[/itex]. If you solve for [itex]y[/itex], what do you discover?