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Domain of a Radial Function

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the domain of the function.

    h(x) = [tex]\sqrt{4 - x}[/tex] + [tex]\sqrt{x^2 - 1}[/tex]

    2. Relevant equations
    a2 - b2 = (a - b) (a + b)

    I think that's all...

    3. The attempt at a solution

    For starters, I know we can't have a negative number in the radical.

    I looked at the 2nd radical first and broke it up...

    [tex]\sqrt{x^2 - 1}[/tex] = [tex]\sqrt{(x + 1) (x - 1)}[/tex]



    Then I looked at the first radical...

    [tex]\sqrt{4 - x}[/tex]

    [tex]\sqrt{-x + 4}[/tex]

    [tex]\sqrt{- (x - 4)}[/tex]


    So I'm left with...

    = [tex]\sqrt{- (x - 4)}[/tex] + [tex]\sqrt{(x + 1) (x - 1)}[/tex]

    Then found the zero's, but are those necessary with a radical question? I know we can't have zero as a denominator but a zero in a radical is fine.

    x = 1
    x = -1
    x = 4

    Now I feel stuck though...I think it's the addition of the two radicals that's throwing me off. If there was just one then it would probably be simpler.
     
  2. jcsd
  3. Jan 16, 2010 #2

    vela

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    You're making things hard on yourself. ;)

    You said, "For starters, I know we can't have a negative number in the radical." That's all you need. You want both quantities in the radical to be non-negative, so you have:

    [tex]4-x \ge 0[/tex]

    and

    [tex]x^2-1 \ge 0[/tex]

    Find the conditions on [itex]x[/itex] to satisfy each inequality. The domain of the function are those values of [itex]x[/itex] which satisfy both inequalities simultaneously.
     
  4. Jan 16, 2010 #3
    It's a habit. :P

    Okay...

    4 - x [tex]\geq[/tex] 0

    4 [tex]\geq[/tex] x

    x2 - 1 [tex]\geq[/tex] 0

    x2 [tex]\geq[/tex] 1

    x [tex]\geq[/tex] [tex]\sqrt{1}[/tex] or just 1

    So, now...

    1 [tex]\leq[/tex] x [tex]\leq[/tex] 4

    D: [1,4] ???
     
  5. Jan 16, 2010 #4

    vela

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    So far so good.

    Not quite. Remember that [itex]\sqrt{x^2} = |x|[/itex]. In other words, you missed a bunch of negative solutions, like [itex]x=-2[/itex] since [itex](-2)^2=4\geq 1[/itex].

    That's one piece. You need to fix your mistake above to find the rest of the domain.
     
  6. Jan 16, 2010 #5
    I don't know...

    The answer is supposed to be (- infinity, -1] U [1,4], but I'm not making sense of that. Anything squared, positive or negative, will end up being positive. Meaning that -22 (from your example) will be positive and therefore part of the domain. But I'm not sure I even understand what I'm saying. If I were presented with a similar question, then I'd be back to square one again.

    Would it be possible for you to elaborate a little bit? Let's assume that I'm drawing a complete blank here...:redface:
     
  7. Jan 16, 2010 #6

    vela

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    You have to find all the solutions to [itex]x^2\geq 1[/itex]. The positive answers [itex]x\geq 1[/itex] are the ones you normally think of, but there are negative ones too, namely [itex]x\leq -1[/itex].

    It's similar to when you solve [itex]x^2=1[/itex]. There are two solutions, +1 and -1. In your case, instead of the equals sign, you have an inequality instead. You just have to be careful you get the directions right.
     
  8. Jan 16, 2010 #7

    Mentallic

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    It's easiest to understand by looking at a graph.
    Take the function [itex]f(x)=x^2-1[/itex]. Now, to find where [itex]x^2-1\geq0[/itex] just look at where the function f(x) is above the x-axis. This occurs at [itex]x\geq 1[/itex] AND [itex]x\leq -1[/itex].

    Also, you can test this by just using random values of x:

    test [itex]x^2-1\geq 0[/itex]

    x=2, [itex]2^2-1=4\geq 0[/itex] therefore true.

    x=-3 [itex](-3)^2-1=9-1=8\geq 0[/itex] therefore also true. This is why you missed the entire interval of (-[itex]\infty[/itex],-1]
     
  9. Jan 17, 2010 #8
    Okay, it's clicking. Making the graph helped a fair bit. Would you always recommend looking at a graph if you're stuck? Or do you find that to be too time consuming?

    Once I have f(x) = x2 - 1 graphed, would it make sense to put f(x) = 4 - x on the same graph? If they intersect with each other, is that...okay? Is rewriting f(x) = 4 - x as f(x) = -x + 4 acceptable?

    "In your case, instead of the equals sign, you have an inequality instead. You just have to be careful you get the directions right."

    I believe I did get the directions incorrect as I had both:

    x [tex]\geq[/tex] 1

    and

    x [tex]\geq[/tex] -1

    That led some confusion.

    Sorry for all the questions. Math generally goes in one ear and then out the other. I don't use it enough for it to stick.
     
  10. Jan 17, 2010 #9

    vela

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    Graph away! Graphing can be incredibly helpful, even when you're not stuck, to understand what's going on and to give you another way of looking at a problem that you might have overlooked.

    In this case, I don't think putting them on the same graph would buy you anything.
     
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