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I seriously need help in this functions question! Chop chop!

  1. Mar 3, 2005 #1
    hey guys, I'm new here but whatever, who cares.
    ANYWAYS, I have this math question on hybrid functions.
    I desperately need help for this cause I'll be having an important test next week.
    HELP!!

    OK anyways, in hybrid function,
    example:

    this question asked me to form the function by giving me a hybrid functions graph.
    So i managed to find f(x) but I had a problem understanding the domains.

    2x-1 , x<0
    f(x)= 3x²-2x , 0x≤2
    4 , x>2​
    ok my question is.. for the domain, how do you know if you put ≤ instead of < ? HOw come it is not x0 and 0<x≤2 instead?

    PS: well if you don't really know how to explain, or don't really understand what im trying to ask, then, it'll be helpful if you could provide me a link to a website I can go to for help on hybrid functions.

    Thanks!!!!! (=
     
    Last edited: Mar 3, 2005
  2. jcsd
  3. Mar 3, 2005 #2

    James R

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    At x=0, the expression 2x-1 has value -1.
    At x=0, the expression 3x^2 - 2x has value 0.

    For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.

    See?

    By the way, on a graph, where there is a discontinuity it is usual to put a closed or open dot to indicate the value at the discontinuity. So, for the function above, the 2x-1 line would have an open dot at the point (0,-1), while the 3x^2 - 2x curve would have a closed dot or circle at the point (0,0).
     
    Last edited: Mar 3, 2005
  4. Mar 3, 2005 #3
    ok if u say that, this question comes from my textbook

    Sketch the graph of the function

    -2x-2 ,x<0
    f(x)= x-2 , 0≤x<2
    3x-6 , x≥2

    how come at x=0, both expressions -2x-2 and x-2 has the same value at -2?
     
  5. Mar 3, 2005 #4

    James R

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    Since in that case the function is continuous at x=0, you could just as easily write:

    -2x-2 ,x0
    f(x)= x-2 , 0≤x<2
    3x-6 , x≥2

    or even

    -2x-2 ,x0
    f(x)= x-2 , 0<x<2
    3x-6 , x≥2

    It would make no difference in this case.
     
  6. Mar 3, 2005 #5

    dextercioby

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    You can ask the same question for the point x=2...:wink:It means that the function is continuous in the 2 points...I don't know if you know what that means,but it's useful to know that it's not essential for this problem...So basically,u have to draw 3 lines and that's it.Can u do that...?

    Daniel.
     
  7. Mar 3, 2005 #6

    erm, how did u know that
    For the composite function, you need to decide which value you want at x=0. It seems from above that you have chosen the function to have the value 0 at x=0.
     
  8. Mar 3, 2005 #7
    dextercioby: I seriously have zilch idea on what you're trying to say???
     
  9. Mar 3, 2005 #8

    dextercioby

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    Perfect,at least can u draw tha graph...?

    Daniel.
     
  10. Mar 3, 2005 #9
    I can but it's not the drawing of the graph that im having a problem with, it's the... detestable signs. ARGHH!!
     
  11. Mar 3, 2005 #10

    dextercioby

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    What do you mean signs...?It's just + and -...There's nothing complicated about it.Basically u'll have to plot the function on each interval.

    Daniel.
     
  12. Mar 3, 2005 #11

    James R

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    lilsheltie:

    Let me give you a simple example.

    Suppose we have

    [itex]f(x) = 3,\qquad x \geq 1[/itex]
    [itex]f(x) = 7, \qquad x < 1[/itex]

    What is the value of f(x) at x=1? Well, which interval contains x=1? Obviously, the interval [itex]x < 1[/itex] doesn't include x=1, so that part of the function definition doesn't apply. On the other hand, the other part of the function does include x=1. Therefore, for this function f(1) = 3, and f(1) definitely doesn't equal 7.

    Now compare the function in your original question, at the point x=0. Which interval contains x=0?
     
  13. Mar 4, 2005 #12

    ok. I didn't know that.
     
  14. Mar 4, 2005 #13
    Huh? Ok i've given you quite a few examples so im pretty much confused.
    Anyways, I kinda get the picture now! (=
     
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