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Math Help - [SOLVED] Domain and Range of this Function

  1. #1
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    Cool [SOLVED] Domain and Range of this Function

    f(x)=\frac{6}{2x+3} For x is greater than or equal to Zero

    How can I find the range of this function? I would normally use the Completing Square Method, but I am not sure how to apply that method in such functions. Can I use it here if so then how?

    Help will be appreciated!

    Thanks!
    Last edited by unstopabl3; May 7th 2010 at 06:35 AM. Reason: Edited the Latex
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  2. #2
    Senior Member Dinkydoe's Avatar
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    You mean the range of f(x) for x> 0 since f is not defined in x=0.

    Have you thought of determining f'(x) ?

    If you check carefully you see that the derivative is negative for all x>0 wich means that f is a strictly decreasing function. (not very surprising is it?).

    And f(x)> 0 for all x.

    f 's given domain is the open interval (0,\infty) and since as x\to 0
    we see that f(x) \to \infty.

    What can you conclude from this?
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  3. #3
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    I didn't understand all of that but I've understood most of it.

    The correct answer is f>0 and f is less than or equal to 2.

    But I am still not sure how to get to this answer.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    Indeed, the open interval (3,\infty) is the range

    What exactly do you mean by f is less then or equal to 2?
    It's not, as far as I know. (we just determined that 3<f(x)<\infty for all x>0)

    Btw, is f(x)= \frac{6}{2}x+3 or \frac{6}{2x}+3. It's quite different
    I assumed the second case.

    (made a typo, hope you see the refreshed version ;p)

    Look carefully what happens when x\to\infty
    Last edited by Dinkydoe; May 7th 2010 at 06:17 AM.
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  5. #5
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    Well here's the question itself.



    As the range for a function is the domain of it's inverse, that's why I was trying to figure out the range of the function f to write it down as the domain of it's inverse.

    But the correct answer is f>0 and f is less than or equal to 2. ( Domain of f –1: 0 < x ≤ 2 )

    Any idea how they got it?
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  6. #6
    Senior Member Dinkydoe's Avatar
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    haha ok, so actually it's even a different function. It's f(x)=\frac{6}{2x+3}. Be careful next time with your notation, not to leave anything to the imagination of the reader

    But yes, now I have a good idea how they got there.

    as x\to\infty we see that the denominator becomes very large, while the numerator stays constant 6. This means that f(x) \to 0 as x becomes very large. (but never really hits the zero)

    While f(0)= \frac{6}{2\cdot 0 + 3} = 2

    We can not immediately conclude that 0<f(x)\leq 2 but we're almost there. If we notice that f'(x)< 0 for all x\geq 0 (check this for yourself) we see that f is strictly decreasing and \lim_{x\to\infty}f(x)=0 so f can not take any other values then in the half open interval (0,2].
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  7. #7
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    I've got the concept behind it. Thanks!
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