Function question. Is this correct?


by lionely
Tags: correct, function
lionely
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#1
Nov21-12, 08:03 PM
P: 509
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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Mark44
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#2
Nov21-12, 08:24 PM
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Quote Quote by lionely View Post
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 Quote by lionely View Post

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 Quote by lionely View Post

A root of a negative number is not possible so f(x) is not surjective onto R
lionely
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#3
Nov21-12, 08:52 PM
P: 509
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
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#4
Nov21-12, 09:36 PM
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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.
lionely
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#5
Nov21-12, 09:39 PM
P: 509
I only sketched a graph,
Dick
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#6
Nov21-12, 10:21 PM
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Quote Quote by lionely View Post
I only sketched a graph,
Ok, then what's a value in R that x^2-4 can never equal?
haruspex
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#7
Nov21-12, 10:23 PM
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Quote Quote by lionely View Post
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?
lionely
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#8
Nov22-12, 06:21 AM
P: 509
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
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#9
Nov22-12, 06:35 AM
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Quote Quote by lionely View Post
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
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
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#10
Nov22-12, 10:37 AM
P: 509
Can't 5 make it <0?
Dick
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Nov22-12, 10:59 AM
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Quote Quote by lionely View Post
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.
lionely
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#12
Nov22-12, 02:36 PM
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But aren't negative numbers be.. real?
haruspex
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Nov22-12, 03:51 PM
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Quote Quote by lionely View Post
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?
lionely
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#14
Nov22-12, 04:00 PM
P: 509
If x=-4 that could make f(x) < 0
haruspex
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#15
Nov22-12, 05:05 PM
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Quote Quote by lionely View Post
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.
lionely
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#16
Nov22-12, 05:25 PM
P: 509
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
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#17
Nov22-12, 06:31 PM
P: 82
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?
lionely
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#18
Nov22-12, 06:40 PM
P: 509
Umm that Y is > or equal to 4 no matter number you use?


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