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Domain of a function

  1. Sep 14, 2003 #1
    ok guys, I need some help. I have been working on this problem for half an hour and im stuck. Every answer I have came up with has been wrong. Here is the problem, find the domain of this function:

    F(x)=square root of x^2-49



    Thanks for your help.
     
  2. jcsd
  3. Sep 14, 2003 #2
    On the assumption that you're working with real numbers, work out where f(x) stops being a real number.
     
  4. Sep 14, 2003 #3
    like what number could I plug in there?
     
  5. Sep 14, 2003 #4
    Think about when √x doesn't exist for a real number x. √4 = 2, √0 = 0, √(-1) = ...?
     
  6. Sep 14, 2003 #5
    ok, so the number is a nonreal number when it is a negative. I don't understand what you are trying to say.
     
  7. Sep 14, 2003 #6
    OK, you have f(x) = √(x2-49).

    Since f is a function from the reals to the reals (assumedly), we need (x2-49) to be non-negative, or f(x) will not be a real number. Specifically, find the range of numbers x where (x2-49) is non-negative.
     
  8. Sep 14, 2003 #7
    I somewhat understand that. I'm just really lost with finding domains of a function. My teacher is german with a hardcore accent and he doesn't explain math very well.
     
  9. Sep 14, 2003 #8
    I can sympathise with your situation. My Linear Algebra teacher had a thick German accent. Hearing him say such things as "eigenvector" was quite intimidating.

    To find the domain, try to find where the function "breaks down". As practise, find the domain of f(x) = 1/(x-1) for x a real number. Think about what value of x would make f(x) make no sense.
     
  10. Sep 15, 2003 #9
    I have no idea. I'm totally lost. This is due in less than an hour. Please help me out. Thanks
     
  11. Sep 15, 2003 #10
    This may have come too late, but anyway.

    The solution is the domain of x is |x| > 7.

    Why? Well, we have f(7) = f(-7) = 0. If we set x = 7, and then decrease x slightly, f(x) turns into an imaginary number, and since we're working with reals, this is of no use to us. Similarly for x = -7, if we increase x slightly, we again get an imaginary number. Since (x2-49) is continuous, we know values of f(x) between x = -7 and x = 7 will also give us imaginary numbers. We need x to be greater than 7, or less than -7. We can express this as |x| > 7.

    With respect to your teacher, isn't there another one you could ask? It sometimes helps a lot to get more than one explanation for something, particularly if you can't understand the first...
     
  12. Sep 15, 2003 #11
    Btw, Lonewolf meant |x| [>=] 7 (the absolute value of x must be equal to or greater than 7), since at x = 7 or x = -7, the function value is [squ]0 = 0.

    Another way of saying "|x| [>=] 7" is "x [<=] -7 or 7 [>=] 7"

    Btw, I'll try to give an explanation of a function's natural domain. If it is unclear or wrong please point me out.

    --- natural domain of a function ---

    [note that we are discussing a function that maps a real number to another real number]

    The domain of a function can be stated explicitly, such as

    f(x) = 1, x >0

    The domain of the function above is all real numbers greater than 0. With that restriction, f(0), f(-2), f(-3.43), etc are undefined.

    If a function has no domain stated explicitly, then it is understood that the domain consists of all real number x for which f(x) is defined. This implicit domain is called the natural domain of f(x).

    Let's look at the function f(x) = 1/x

    for x = 1, f(x) = 1/1 = 1 (defined). For x = 5, f(x) = 1/5 (defined also). It turns out that only at x = 0 is f(x) undefined. That is because 1/0 is undefined. So the domain of the function f(x) = 1/x is all x values other than 0.

    How about the function f(x)=[squ]x?

    At x equal or greater than 0, f(x) is defined. At x less than 0, f(x) is undefined. For example, [squ](-1) is undefined (because there are no real number y that satisfies the property y^2 = -1). That means the natural domain of the function f(x) = [squ]x is all x equal to or greater than 0.

    Now, let's go back to your problem. Your function is f(x)=[squ](x^2-49). Let's introduce a new variable b, where b = x^2-49. We can then rewrite your function as f(x) = [squ]b. We already know from the previous explanation that in order to make f(x) defined, b must be equal to or greater than 0. So we write

    b [>=] 0

    But b = x^2-49, so we can write

    x^2-49 [>=] 0

    By factoring, we get

    (x+7)(x-7) [>=] 0

    At x = 7 or x = -7, the value of b = (x+7)(x-7) will be zero (do you know why?). Since [squ]0 is defined, f(x) is defined at these x = 7 and x = -7. At values of x between 7 and -7, the value of b will be less than 0 (verify this). That means f(x) is undefined for -7 < x < 7. For x < -7 or x > 7, b is positive (verify this also) and that means f(x) is defined.

    Thus we can conclude that the natural domain of f(x) consists of all x such that
    x is equal to -7, or
    x is smaller than -7, or
    x is equal to 7, or
    x is greater than 7
    We can write this as |x| [>=] 7.

    Does every part of my explanation makes sense to you?
     
    Last edited: Sep 15, 2003
  13. Sep 15, 2003 #12
    Yes, I did mean that. Thanks.
     
  14. Sep 15, 2003 #13
    Any number that you put in for X must result in a number greater than or equal to 0 if you square that number and subtract 49. The numbers that will NOT do this are the numbers that have an absolute value less than 7.
     
    Last edited: Sep 15, 2003
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