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Math Help - What is the domain of this function? (I am so confused)

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    What is the domain of this function? (I am so confused)

    Hello.

    I would like to ask what is the domain of the following fucnction.

    y=(5x^x)^(x+1)

    Our teachers tell us that the domain of a power function (A^X where A is constant and X is a variable) is that the base must be positive and must differ from 1,so in the above example, it is logically that the solution of the following system of inequalities defines the domain:

    5x>0 and
    5x<>1 and
    5x^x>0 and
    5x^x<>1


    Am I right? And I must note that in our textbook the power function's domain(sorry if I misuse mathematical argo) A^X is that the base A must be positive but it does not say that it must differ from 1. Ohhhh! Please clarify this very obscure situation!
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    Quote Originally Posted by Val21 View Post
    Hello.

    I would like to ask what is the domain of the following fucnction.

    y=(5x^x)^(x+1)

    Our teachers tell us that the domain of a power function (A^X where A is constant and X is a variable) is that the base must be positive and must differ from 1,so in the above example, it is logically that the solution of the following system of inequalities defines the domain:

    5x>0 and
    5x<>1 and
    5x^x>0 and
    5x^x<>1


    Am I right? And I must note that in our textbook the power function's domain(sorry if I misuse mathematical argo) A^X is that the base A must be positive but it does not say that it must differ from 1. Ohhhh! Please clarify this very obscure situation!
    y=\left ( 5x^x \right )^{x+1} = 5^{x+1} \cdot x^{x(x+1)}

    Looking at these terms individually, we see that we have no restrictions on the function due to the first factor. The second factor gives us that x \geq 0, since the base must be non-negative, but note that if x = 0 the second factor is 0^0, which is undefined. So the domain is x > 0.

    -Dan
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    Must a base differ from 1?

    And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

    Thanks for replying
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    Quote Originally Posted by Val21 View Post
    And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

    Thanks for replying
    x can also be any integer, so it looks to me as though the domain is:

    <br />
\left[ \mathbb{R}_+ \cup \mathbb{Z} \right] \backslash \{0\} <br />

    The union of all positive reals, and non-zero integers.

    On second thoughts it looks as though this is also defined when x is a rational of
    the form -\frac{2n}{2m-1},\ n,\ m \in \mathbb{Z}_+ (negative rationals which in lowest form have an even numerator)

    So my latest bet is:

    <br />
\left[ \mathbb{R}_+ \cup \left( \mathbb{Z}\backslash \{0\} \right) \cup A \right]<br />

    where A=\{a: a=-2n/(2m-1),\ n,\ m \in \mathbb{Z}_+ \}.

    But somehow I don't think your teacher is looking for an answer of this nature.

    RonL
    Last edited by CaptainBlack; November 18th 2006 at 12:25 AM.
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    Quote Originally Posted by Val21 View Post
    And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

    Thanks for replying
    I don't see why 1 should be prohibited in the base, 0 yes as 0^0 is undefined, but not 1.

    When x=1, we have:

    y=(5x^x)^{x+1}=(5 \times 1^1)^2=5^2=25

    RonL
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