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Math Help - domain and range

  1. #1
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    Exclamation domain and range

    Can someone explain domain and range to me? I recall doing these problems this past year, but since all textbooks were turned in in May, I have no point of reference for my summer homework, and my memory is a little foggy.

    The first problem says find the domain of the function: f(x)= (square root of (x-5))/(x^2-6x+8)

    The other problem says find the domain and range of the funtion:
    f(x)=(4/5)l8-xl

    A refresher course in domain and range would be helpful.
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  2. #2
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    Quote Originally Posted by iheartthemusic29 View Post
    The first problem says find the domain of the function: f(x)= (square root of (x-5))/(x^2-6x+8)
    f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}

    What are exactly the conditions you must set so the function is well defined?
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  3. #3
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    Quote Originally Posted by iheartthemusic29 View Post
    The other problem says find the domain and range of the funtion:
    f(x)=(4/5)l8-xl
    f(x)=\frac45\left|8-x\right|

    How'd you find the domain?

    Having an absolute value function, what could you conclude about the range?
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    Quote Originally Posted by Krizalid View Post
    f(x)=\frac{\sqrt{x-5}}{x^2-6x+8}

    What are exactly the conditions you must set so the function is well defined?
    What do you mean? I gave all the information I have...after that, I know nothing.
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  5. #5
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    Sorry, I didn't read the first

    Well, in few words... domain means to find the values of x for which the function is well defined.

    For example, in your first problem, you can't have a negative root, so one condition would be setting x-5\ge0, you can set it equal to zero 'cause numerator can be zero.

    But the denominator can't be zero, so you must set x^2-6x+8\ne0 (got the idea?)

    About the range, are all possible y values that the function can take.

    Sorry for this useless explanation but I have more things to do
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  6. #6
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    Quote Originally Posted by iheartthemusic29 View Post
    The first problem says find the domain of the function: f(x)= \sqrt{\frac{x-5}{x^2-6x+8}}
    Domains are fairly easy. Typical things you want to do is check for any of the following:
    1) Places where you get a 0 under a fraction
    2) Where you would get a negative under a square root
    These are the big ones, and you have both in this problem.

    Ranges are tougher. ThePerfectHacker has a method for doing these (go figure that I couldn't get the search engine to find any of his examples). Probably the simplest way is simply to graph it and see what happens.

    In your case we get a zero in the denominator for
    x^2 - 6x + 8 = 0

    (x - 2)(x - 4) = 0

    So the points x = 2 and x = 4 are out of the domain.

    We also want to see where the fraction
    \frac{x-5}{x^2-6x+8} = \frac{x - 5}{(x - 2)(x - 4)}
    is negative. We search for where the numerator and denominator are 0. This is at x = 5, x = 2, and x = 4. (These are called "critical points.") Now split up the real number line and ask the question:
    ( -\infty, 2 ): \frac{x - 5}{(x - 2)(x - 4)} < 0

    ( 2, 4 ): \frac{x - 5}{(x - 2)(x - 4)} > 0

    ( 4, 5 ): \frac{x - 5}{(x - 2)(x - 4)} < 0

    ( 5, \infty ): \frac{x - 5}{(x - 2)(x - 4)} > 0

    So we have the conditions on the domain: x cannot be 2 or 4, and cannot be in the intervals (-\infty, 2) nor (4, 5). The domain is everything else.

    So the domain of the function is (2, 4) \cup [5, \infty). (We include x = 5 because there is no reason we can't have a 0 under the square root.)

    For the range, look at the graph below. It looks to me like the range will be  [0, \infty).

    -Dan
    Attached Thumbnails Attached Thumbnails domain and range-function.jpg  
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