- #1

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## Homework Statement

h:x → 4-x

^{2}, x E ℝ

show that it is not surjective(not onto ℝ)

## 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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- Thread starter lionely
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- #1

- 576

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h:x → 4-x

show that it is not surjective(not onto ℝ)

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

- #2

Mark44

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If it fails the## Homework Statement

h:x → 4-x^{2}, x E ℝ

show that it is not surjective(not onto ℝ)

## The Attempt at a Solution

Since the line tests fail.

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.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

- #3

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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..

- #4

Mark44

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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.

- #5

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I only sketched a graph,

- #6

Dick

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I only sketched a graph,

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

- #7

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Sure, but to be surjective on ℝ the inverse must have a solution forx=±√(4-y) if y is like 3 x is a real number..

- #8

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

Umm I'm not sure..

- #9

HallsofIvy

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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?## Homework Statement

h:x → 4-x^{2}, x E ℝ

show that it is not surjective(not onto ℝ)

## 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

- #10

- 576

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Can't 5 make it <0?

- #11

Dick

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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.

- #12

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But aren't negative numbers be.. real?

- #13

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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?Can't 5 make it <0?

- #14

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

- #15

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

- #16

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- #17

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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?

- #18

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

- #19

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

What is your reasoning? Substitute some values for [itex]x[/itex] into [itex]f[/itex] and see what you get, or what you

- #20

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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?

- #21

Mark44

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The line above should be x = ±sqrt(4-y)But solving it alebraically isn't it this??

y= 4-x^2

x^2= 4-y

x= sqrt(4-y)

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

As you said earlier(or at least, alluded to), the range of the function y = 4-x

- #22

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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 [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

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