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Math Help - Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

  1. #1
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    Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

    Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

    So I'm able to do the first two steps
    f(g(sinx))
    f(1/sin)

    But I'm not sure what the last step does. Normally something enclosed inside ||, such as |x| means the magnitude of x, right? So what does that operator mean in the context of this problem? How does y=1/sinx differ from y=|1/sinx|? Obviously, this can be written as y=|csc|.

    b) Describe what each composition does in the order that it is applied. Again, I'm fine until the last function, where I don't know what the operator does.

    c) sketch the curve. My curve looks like a csc function, is this correct? Please help
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by stacey View Post
    Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?

    So I'm able to do the first two steps
    f(g(sinx))
    f(1/sin)

    But I'm not sure what the last step does. Normally something enclosed inside ||, such as |x| means the magnitude of x, right? So what does that operator mean in the context of this problem? How does y=1/sinx differ from y=|1/sinx|? Obviously, this can be written as y=|csc|.

    b) Describe what each composition does in the order that it is applied. Again, I'm fine until the last function, where I don't know what the operator does.

    c) sketch the curve. My curve looks like a csc function, is this correct? Please help

    h(g(f(x)))=h(g(sin(x)))=h(1/sin(x))=|1/sin(x)|=1/|sin(x)|=|cosec(x)|

    There is a difference between cosec(x) and |cosex(x)|.

    Start by drawing y=cosec(x) and try to understand what the operation | | doing to cosec(x).

    (Maybe you should draw the function f(x)=x tnd afterwards to draw g(x)=|x|, it will give you a clue)
    Last edited by Also sprach Zarathustra; May 27th 2011 at 05:52 PM.
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  3. #3
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    Given f(x) = sin(x) , g(x) = 1/x , h(x) = |x|, state h(g(f(x)))?
    Let's start with the inside, and work our way out. g(f(x)) is the same as g(sin(x)), which is the same as 1/sin(x). Our final result is then h(1/sin(x)). When you subsitute the x in |x| with 1/sin(x), what do you get?
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  4. #4
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    Hello, stacey!

    \text{Given: }\:f(x)\,=\,\sin(x),\;\;g(x)\,=\,\tfrac{1}{x},\;\;  h(x) \,=\, |x|

    \text{Find: }\:h(g(f(x)))

    But I'm not sure what the last step does.
    Normally something enclosed inside ||, such as |x| means the magnitude of x, right?
    So what does that operator mean in the context of this problem?
    How does y\,=\,\tfrac{1}{\sin x} differ from y\,=\,\left|\tfrac{1}{\sin x}\right|?
    Obviously, this can be written as y \,=\,|\csc x| . Yes!

    b) Describe what each composition does in the order that it is applied.
    Again, I'm fine until the last function, where I don't know what the operator does.

    Evidently, you're not familiar with "absolute values".
    (I'm not sure how that happened to you.)

    |x| means: take the positive value of x.

    . . \begin{array}{ccc}\vdots && \vdots \\ |\text{-}2| &=& 2 \\ |\text{-}1| &=& 1 \\ |0| &=& 0 \\ |1| &=& 1 \\ |2| &=& 2 \\ \vdots && \vdots\end{array}

    When graphing, any portion below the x-axis is reflected upward above the x-axis.




    c) sketch the curve.
    My curve looks like a csc function. .Is this correct?

    Not quite . . . It is the cosecant graph, but everything is "made positive".

    It looks something like this:

    . . \begin{array}{c} | \\ \cup \cup | \cup \cup \\ | \\ --+-- \\ | \end{array}

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