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Finding the domain of a function (Sqr root)

  1. Sep 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Here is a picture of the question:

    1mdzb.png


    2. Relevant equations



    3. The attempt at a solution

    Here it what I did....

    factored it...so

    1 / (x-3)(x+17)

    domain: (-infinity, -3)(-3,17)(17, +infinity) or rather (-3,17)

    I know that is not correct. I am really stuck on this one!
     
  2. jcsd
  3. Sep 7, 2012 #2

    Mark44

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    There are two things to consider here:
    1) The denominator in the radical can't be zero, which happens when x = - 3 or x = 17
    2) The denominator in the radical can't be negative. You have three intervals. Pick a number in each interval, to see if you get a negative value for the denominator. If you do, that interval is not in the domain.
     
    Last edited: Sep 7, 2012
  4. Sep 7, 2012 #3
    So, is the domain just 17?
     
  5. Sep 7, 2012 #4

    Mark44

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    No, not at all. 17 is NOT in the domain, because f(17) is undefined. The domain is an interval or maybe two, of real numbers.

    Evaluate x2 + 14x - 51 at any number from each of these intervals:

    (-∞, -3)
    (-3, 17)
    (17, ∞)

    If you get a negative result, that interval is NOT in the domain.
    If you get a positive result, that interval IS in the domain.
     
    Last edited: Sep 7, 2012
  6. Sep 7, 2012 #5

    Oh right...

    in the (-3,17) interval there are some numbers that give me a pos number.
    Also in the interval (17, infinity)

    so is it those 2 intervals? I am slightly confused
     
    Last edited by a moderator: Sep 7, 2012
  7. Sep 7, 2012 #6

    SammyS

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    OK !

    Let's go back to the original problem.

    As Mark44 said, there are two things to consider.

    So,

    (1): [itex]x^2 + 14x - 51\ne 0 [/itex]

    and

    (2): [itex]x^2 + 14x - 51\ge 0 [/itex]

    Of course, that leaves you with [itex]x^2 + 14x - 51>0\ .[/itex]

    Can you solve that ?
     
    Last edited by a moderator: Sep 7, 2012
  8. Sep 7, 2012 #7
    To solve that would I try and isolate the x ?
     
    Last edited by a moderator: Sep 7, 2012
  9. Sep 7, 2012 #8

    SammyS

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    It's an inequality, not an equation.

    There are several ways to approach solving a quadratic inequality. I have no idea regarding which method(s) you may have previously learned.
     
  10. Sep 7, 2012 #9
    SammyS, I also have had trouble solving inequalities. I am just getting back into the grove of things, and this was something I never really mastered unfortunately.

    What is a good method? One that you use, or rather one you recommend a novice like my self?
     
  11. Sep 7, 2012 #10
    From what I remember, I would just make it = to 0 then factor it.

    Then I would draw a number line, and since it much be greater than 0, I would pick the one that is greater than 0

    ?
     
    Last edited by a moderator: Sep 7, 2012
  12. Sep 7, 2012 #11

    SammyS

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    I suppose I looked at a post with a wrong number.

    Yes, it should be [itex]x^2 + 14x - 51>0\ .[/itex]
     
  13. Sep 7, 2012 #12

    Mark44

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    That was my fault - I copied a number incorrectly. I have fixed it throughout this thread.
     
  14. Sep 7, 2012 #13

    Mark44

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    No, there aren't. Every number in this interval gives you a negative product from (x + 3)(x - 17).

    Think about it: if -3 < x < 17, then x + 3 will be > 0, but x - 17 will be < 0. This gives you a negative product.
    For the interval (17, ∞), all numbers produce a positive product.

    What's the other interval that produces a positive product?
     
  15. Sep 8, 2012 #14

    HallsofIvy

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    nukeman, you said, originally, that the function could be written
    [tex]\sqrt{\frac{1}{(x- 3)(x+ 17)}}[/tex]
    and everything you need to know is in that. In order to take the square root the fraction inside the square root must be positive (it is never 0) and, since the numerator is always positive (it is always "1"), the denominator must be positive. But a product of numbers, like (x- 3)(x+ 17), is positive if and only if the two factors have the same sign. That is, they must both be positive or both negative.

    So there are two possibilities:
    1) x- 3> 0 and x+ 17> 0
    or
    2) x- 2< 0 and x+ 17< 0.

    What values of x satisfy one of those two? (crucial point: -17< 3)
     
    Last edited by a moderator: Sep 9, 2012
  16. Sep 8, 2012 #15
    It would be this one wouldn't it? x- 3> 0 and x+ 17> 0

    Man, I got really confused on this one. SO SORRY, but can we start right at the start.

    WHEN I see a question like this, lets take this exact one, what is the first thing I do?
    I would factor it correct?
     
    Last edited: Sep 8, 2012
  17. Sep 8, 2012 #16
    g(x)=x2+14x-51
    Sketch the graph. It is a parabola
    Check for x-intercept where g(x)=0.
    Check the values of x where g(x)>0 and g(x)<0

    f(x)=√1/g(x)
    Domain of f(x) only if g(x)>0.
     
    Last edited: Sep 8, 2012
  18. Sep 9, 2012 #17
    I am still not 100% on the final answer ?
     
  19. Sep 9, 2012 #18
    Is this the correct answer?

    (-17, 3) (3, +infinity)
     
  20. Sep 9, 2012 #19

    SammyS

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    Let's see.

    Did you graph [itex]y=x^2 + 14x - 51[/itex] as has been suggested?

    Test some values of x ...

    x = 0 is an easy one in the interval (-17, 3). Plugging in x=0 gives, y = -51 .

    It turns out that any other value you might pick for x from (-17, 3) will give you a negative result for y.

    Have you graphed [itex]y=x^2 + 14x - 51\ ?[/itex]

    Pick a value for x from (3, +∞) ...

    x = 10 is fairly easy.

    If x = 10 , [itex]y=(10)^2 + 14(10) - 51=100+140-51=189[/itex]

    It turns out that any other value you might pick for x from (3, +∞) will give you a positive result for y.

    Have you graphed [itex]y=x^2 + 14x - 51\ ?[/itex]
     
  21. Sep 9, 2012 #20
    I dont understand the steps then :(

    I look at the question: [itex]y=x^2 + 14x - 51[/itex]

    So, I factor it! ----> Then I get from factoring: (x - 3) (x +17)

    When I get here. What do I do next?


     
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