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Math Help - Composite functions

  1. #1
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    Composite functions

    If h(x)=\frac{9-5x}{1-x}, x\neq1

    The question requires the function h^{2}(x)

    I've worked out h^{2}(x), which is given by
    h^{2}(x)=\frac{4x-9}{x-2}

    Obviously h^{2} is undefined when x=2.

    My question is whether h^{2} is defined when x=1

    Even though the expression \frac{4x-9}{x-2} is defined when x=1, h^{2} is not defined at x=1 since h is not defined at x=1, am I correct?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by acc100jt View Post
    If h(x)=\frac{9-5x}{1-x}, x\neq1

    The question requires the function h^{2}(x)

    I've worked out h^{2}(x), which is given by
    h^{2}(x)=\frac{4x-9}{x-2}

    Obviously h^{2} is undefined when x=2.

    My question is whether h^{2} is defined when x=1

    Even though the expression \frac{4x-9}{x-2} is defined when x=1, h^{2} is not defined at x=1 since h is not defined at x=1, am I correct?
    I do not understand the notation that you have employed. What do you mean by h^2(x) ?
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  3. #3
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    compose h with itself, i.e.
    h^{2}(x)=h(h(x))
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by acc100jt View Post
    compose h with itself, i.e.
    h^{2}(x)=h(h(x))
    Oh. Well, then no, a restriction at x=1 must be made because of the way we have defined h^2(x).

    We can't get passed h(x) through x=1. This means we'd never get to h^2(x). This is subtle, but it is the case when dealing with functions.
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    Oh. Well, then no, a restriction at x=1 must be made because of the way we have defined h^2(x).
    You mean, "Well, then yes" since he was asking if it was correct that the function was NOT defined at x= 1!

    We can't get passed h(x) through x=1. This means we'd never get to h^2(x). This is subtle, but it is the case when dealing with functions.
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