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Math Help - Domain

  1. #1
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    Domain

    Sorry I had to write it out, I don't know how to put it on my computer and I don't have a scanner.

    Find the domain:
    f(x) = cube root of (t + 4)
    f(x) = fourth root of (x squared + 3x)
    f(x) = x - 5 / (square root of (x squared - 9))
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by superpike View Post
    Sorry I had to write it out, I don't know how to put it on my computer and I don't have a scanner.
    It's ok, we can easily read it. You put the parentheses correctly

    Find the domain:
    f(x) = cube root of (x + 4)
    f(x) = fourth root of (x squared + 3x)
    f(x) = x - 5 / (square root of (x squared - 9))
    Remember the domain is all the values x can have and still the function is defined at this point !

    What is under an even-th root should be positive. For odd-th root, it's ok. That is to say you don't care about the sign of x+4 in the cube (3) root (and hence the domain is...), but you'll care about the sign of x+3x under the fourth root, or the sign of x-9 under the square (2) root.

    When you have a quotient, the denominator can't be 0. So find the values of x for which it is 0. Then remoove them from the possible values of x.

    Now please, give it a try, or ask questions if explanations weren't clear enough.
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  3. #3
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    1) all x values for which x is greater than or equal to 4
    2) all real numbers greater than or equal to 0 except 0 or -3
    3) all real numbers except -3 and 3

    Am I right?
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  4. #4
    Moo
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    Quote Originally Posted by superpike View Post
    1) all x values for which x is greater than or equal to 4
    Hmm... I told you it didn't matter for the cube root. All values of x will fit in !
    And t+4>0 ---> t> -4, not t>4
    2) all real numbers greater than or equal to 0 except 0 or -3
    This is not a quotient.

    \sqrt[4]{x^2+3x}=\sqrt[4]{x(x+3)}
    x(x+3) has to be positive, that is to say if x is negative, then x+3 is negative. If x is positive, x+3 is positive.
    This gives 2 possibilities. But I notice that it is quite difficult to give you only tips if you've never done it before... I hoped you would know

    \begin{array}{c|ccc} & -\infty ~~~~~~ & -3 ~~~~~~ & 0 ~~~~~~ + \infty \\ \hline<br />
x & ~~~~~~ -  & \vline ~~~~~~ + & \vline ~~~~~~ + \\ \hline<br />
x+3 & ~~~~~~ - & \vline ~~~~~~ - & \vline ~~~~~~ + \\ \hline \hline<br />
x(x+3) & ~~~~~~ + & \vline ~~~~~~ - & \vline ~~~~~~ +<br />
 \end{array}

    Try to understand this table please.


    3) all real numbers except -3 and 3
    Try the reasoning above.
    And indeed, -3 and 3 are not possible values

    Am I right?
    Unfortunately no... But it will come to the Truth lol
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  5. #5
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    So then 1 is all real numbers.

    I do not understand the chart at all, sorry. Is that the infinity sign? What's its relevance?
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  6. #6
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    The chart tells you in what intervals the quantities x, x+3, and x(x+3) will be positive or negative.
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  7. #7
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    I understand that part of the chart (at least I think), but I don't know how Moo generated the top half or what s/he is exactly doing or even what it means.
    I looked at Moo's first post closer and realized that x = radical negative three if I make f(x) = 0... does that help at all?


    I'm sorry, my teacher never taught me how to do this...
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  8. #8
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    Quote Originally Posted by superpike View Post
    I understand that part of the chart (at least I think), but I don't know how Moo generated the top half or what s/he is exactly doing or even what it means.
    I looked at Moo's first post closer and realized that x = radical negative three if I make f(x) = 0... does that help at all?


    I'm sorry, my teacher never taught me how to do this...
    Basically, the first part of making this chart is to find out where x and x + 3 are equal to zero. We know that x is zero at x = 0 and x + 3 is zero at x = -3. Now, for all values other than x = 0 and x = 3, the product of x(x+3) will be either positive or negative, because both of the factors will be nonzero. All the chart says is that for all values of x < -3, both x and x + 3 will be negative, which will make the product positive. For -3 < x < 0, x + 3 will be positive but x will be negative so the product will be negative. And for all values of x > 0, both x and x+3 will be positive so their product will be positive.
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  9. #9
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    So does that make the domain -3 < x < 0, because the value when inserted in has to be positive as it is in the chart?
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  10. #10
    Super Member 11rdc11's Avatar
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    Not quite, notice that using Moo's chart in the interval you chose

     x = -

    x + 3 = +

    So (+)(-) = -
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  11. #11
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Moo View Post
    Hmm... I told you it didn't matter for the cube root. All values of x will fit in !
    And t+4>0 ---> t> -4, not t>4

    This is not a quotient.

    \sqrt[4]{x^2+3x}=\sqrt[4]{x(x+3)}
    x(x+3) has to be positive, that is to say if x is negative, then x+3 is negative. If x is positive, x+3 is positive.
    This gives 2 possibilities. But I notice that it is quite difficult to give you only tips if you've never done it before... I hoped you would know

    \begin{array}{c|ccc} & -\infty ~~~~~~ & -3 ~~~~~~ & 0 ~~~~~~ + \infty \\ \hline<br />
x & ~~~~~~ - & \vline ~~~~~~ {\color{red}-} & \vline ~~~~~~ + \\ \hline<br />
x+3 & ~~~~~~ - & \vline ~~~~~~ {\color{red}+} & \vline ~~~~~~ + \\ \hline \hline<br />
x(x+3) & ~~~~~~ + & \vline ~~~~~~ - & \vline ~~~~~~ +<br />
\end{array}

    Try to understand this table please.



    Try the reasoning above.
    And indeed, -3 and 3 are not possible values


    Unfortunately no... But it will come to the Truth lol
    Made a correction for you Moo
    Last edited by 11rdc11; September 15th 2008 at 06:28 PM.
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