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Thread: function

  1. #1
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    function

    The function g is defined by g(x)=(x+1)/(x-2), x is not equal to k and m, find the values of k and m

    of course, I know that x is definitely not equal to 2, but how about the other one?
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: function

    That's it -- x cannot equal 2. Are you sure you wrote the function correctly?
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  3. #3
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    Re: function

    The other value would be $\displaystyle k+m-2$.
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    Re: function

    I suspect there is some aspect of this problem being ignored. Is this the exact and complete language of the problem?

    $a \in \mathbb R\ and\ a \ne 0 \implies \dfrac{1}{a} \in \mathbb R.$

    $\therefore a, b \in \mathbb R\ and\ a \ne 0 \implies b * \dfrac{1}{a} \equiv \dfrac{b}{a} \in \mathbb R.$

    So if the only restriction on g(x) is that it is a real-valued function, x = 2 is the only number necessarily outside its domain. Of course the definition of g may exclude any other real number.
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    Re: function

    Perhaps the "other value" is for $\displaystyle g(x)$. While $\displaystyle x \neq 2$, it is fairly straightforward to show that $\displaystyle g(x) \neq 1$ for any value of $\displaystyle x$. In other words, $\displaystyle g(x)$ has an inverse function for any value except at $\displaystyle g(x)=1$.
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  6. #6
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    Re: function

    Quote Originally Posted by SlipEternal View Post
    Perhaps the "other value" is for $\displaystyle g(x)$. While $\displaystyle x \neq 2$, it is fairly straightforward to show that $\displaystyle g(x) \neq 1$ for any value of $\displaystyle x$. In other words, $\displaystyle g(x)$ has an inverse function for any value except at $\displaystyle g(x)=1$.
    That's a very clever guess.
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    Re: function

    Quote Originally Posted by JeffM View Post
    I suspect there is some aspect of this problem being ignored. Is this the exact and complete language of the problem?

    $a \in \mathbb R\ and\ a \ne 0 \implies \dfrac{1}{a} \in \mathbb R.$

    $\therefore a, b \in \mathbb R\ and\ a \ne 0 \implies b * \dfrac{1}{a} \equiv \dfrac{b}{a} \in \mathbb R.$

    So if the only restriction on g(x) is that it is a real-valued function, x = 2 is the only number necessarily outside its domain. Of course the definition of g may exclude any other real number.
    yeah, that's it. The answer given is k=2, m=5. I can't understand
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    Re: function

    Quote Originally Posted by Trefoil2727 View Post
    yeah, that's it. The answer given is k=2, m=5. I can't understand
    Given the answer, I can come up with a question that would have that as its answer. What values of $\displaystyle x$ are not in the domain of $\displaystyle (g\circ g)$? That is the function $\displaystyle g$ composed with itself.

    $\displaystyle (g\circ g)(x) = g(g(x)) = \dfrac{g(x)+1}{g(x)-2} = \dfrac{\dfrac{x+1}{x-2}+1}{\dfrac{x+1}{x-2}-2}$

    From here, you can simplify:

    $\displaystyle \dfrac{\dfrac{x+1}{x-2}+1}{\dfrac{x+1}{x-2}-2}\cdot \dfrac{x-2}{x-2} = \dfrac{x+1+x-2}{x+1-2(x-2)} = \dfrac{2x-1}{5-x}$

    Obviously, $\displaystyle x\neq 5$ for this. But, since $\displaystyle (g\circ g)(x)$ is only defined when $\displaystyle g(x)$ is defined, we also have $\displaystyle x \neq 2$.
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  9. #9
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    Re: function

    Clearly $\displaystyle f(x)= \frac{x+1}{x- 2}$ is undefined when x= 2. To determine a value that f(x) cannot be equal to, do the division: $\displaystyle \frac{x+1}{x- 2}= 1+ \frac{3}{x- 2}$. Since the fraction is never 0, f(x) is never 1.
    Last edited by HallsofIvy; Jun 24th 2014 at 04:03 PM.
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