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Function question. Is this correct?

  1. Nov 21, 2012 #1
    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
     
  2. jcsd
  3. Nov 21, 2012 #2

    Mark44

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    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.
     
  4. Nov 21, 2012 #3
    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..
     
  5. Nov 21, 2012 #4

    Mark44

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    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.
     
  6. Nov 21, 2012 #5
    I only sketched a graph,
     
  7. Nov 21, 2012 #6

    Dick

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    Ok, then what's a value in R that x^2-4 can never equal?
     
  8. Nov 21, 2012 #7

    haruspex

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    Sure, but to be surjective on ℝ the inverse must have a solution for all real y. Does it?
     
  9. Nov 22, 2012 #8
    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..
     
  10. Nov 22, 2012 #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?
     
  11. Nov 22, 2012 #10
    Can't 5 make it <0?
     
  12. Nov 22, 2012 #11

    Dick

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    If you mean there is no real value of x such that 4-x^2=5, that would be correct.
     
  13. Nov 22, 2012 #12
    But aren't negative numbers be.. real?
     
  14. Nov 22, 2012 #13

    haruspex

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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?
     
  15. Nov 22, 2012 #14
    If x=-4 that could make f(x) < 0
     
  16. Nov 22, 2012 #15

    haruspex

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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.
     
  17. Nov 22, 2012 #16
    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..
     
  18. Nov 22, 2012 #17
    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?
     
  19. Nov 22, 2012 #18
    Umm that Y is > or equal to 4 no matter number you use?
     
  20. Nov 22, 2012 #19
    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.)
     
  21. Nov 22, 2012 #20
    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?
     
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