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Thread: Composite functions

  1. #1
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    Composite functions

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

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

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

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

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

    Even though the expression $\displaystyle \frac{4x-9}{x-2}$ is defined when $\displaystyle x=1$, $\displaystyle h^{2}$ is not defined at $\displaystyle x=1$ since $\displaystyle h$ is not defined at $\displaystyle x=1$, 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 $\displaystyle h(x)=\frac{9-5x}{1-x}, x\neq1$

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

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

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

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

    Even though the expression $\displaystyle \frac{4x-9}{x-2}$ is defined when $\displaystyle x=1$, $\displaystyle h^{2}$ is not defined at $\displaystyle x=1$ since $\displaystyle h$ is not defined at $\displaystyle x=1$, am I correct?
    I do not understand the notation that you have employed. What do you mean by $\displaystyle h^2(x)$ ?
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  3. #3
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    compose h with itself, i.e.
    $\displaystyle h^{2}(x)=h(h(x))$
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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.
    $\displaystyle h^{2}(x)=h(h(x))$
    Oh. Well, then no, a restriction at x=1 must be made because of the way we have defined $\displaystyle h^2(x)$.

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