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Math Help - piecewise function challenge

  1. #1
    Member integral's Avatar
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    piecewise function challenge

    <br />
f(x) = \left\{<br />
\begin{array}{lr}<br />
x=3,\, \, if\, \, .5<x<1.325 \\<br />
y=.5,\, \, if\, \, -3<y<-2.5<br />
\end{array}<br />
\right.<br />
    <br />
f(x) = \left\{<br />
\begin{array}{lr}<br /> <br />
x=-2,\, \, if\, \, .5<x<0.994 \\<br /> <br />
\end{array}<br />
\right.<br />
    <br />
\underbrace{f(x) = \left\{<br />
\begin{array}{lr}<br />
y=-x, \,\, if \,\, -1<x<-.5\\<br />
y=x+2,\, \, if\, \, -1.5<x<-1\\<br />
y=x+1,\,\, if \,\, -0.5<x<0\\<br />
y=-x+1 \,\, if \,\, 0<x<.5<br />
\end{array}<br />
\right.}<br />
    As:
    <br />
 \Delta \left\{<br />
\begin{array}{lr}<br />
y=x,\, \, if\, \, -1<x<1 \\<br />
y=-x,\, \, if\, \, -1<x<-1<br />
\end{array}<br />
\right. \to 0<br />

    for: y=7x^4
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by integral View Post
    <br />
f(x) = \left\{<br />
\begin{array}{lr}<br />
x=3,\, \, if\, \, .5<x<1.325 \\<br />
y=.5,\, \, if\, \, -3<y<-2.5<br />
\end{array}<br />
\right.<br />
    <br />
f(x) = \left\{<br />
\begin{array}{lr}<br /> <br />
x=-2,\, \, if\, \, .5<x<0.994 \\<br /> <br />
\end{array}<br />
\right.<br />
    <br />
\underbrace{f(x) = \left\{<br />
\begin{array}{lr}<br />
y=-x, \,\, if \,\, -1<x<-.5\\<br />
y=x+2,\, \, if\, \, -1.5<x<-1\\<br />
y=x+1,\,\, if \,\, -0.5<x<0\\<br />
y=-x+1 \,\, if \,\, 0<x<.5<br />
\end{array}<br />
\right.}<br />
    As:
    <br />
 \Delta \left\{<br />
\begin{array}{lr}<br />
y=x,\, \, if\, \, -1<x<1 \\<br />
y=-x,\, \, if\, \, -1<x<-1<br />
\end{array}<br />
\right. \to 0<br />

    for: y=7x^4
    Do you know the solution and how to solve this? If not it does not belong in this forum.

    Also the notation needs clarification.

    CB
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  3. #3
    Member integral's Avatar
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    Yes I have the answer
    And the notation is just find, what is wrong is the way you are looking at it.
    That is why it is in the challenges and puzzles section.


    -- \int
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by integral View Post
    Yes I have the answer
    And the notation is just find, what is wrong is the way you are looking at it.
    That is why it is in the challenges and puzzles section.


    -- \int
    1. First you don't know how I am looking at it. If I find your notation obscure then others will also. It is your job to make the statement of the problem clear.

    2. You are mixing x and y on the RHS of some of the function definitions. There is no requirement for y ever to appear on the right in fact it appears undefined leaving us to assume what it means. You also leave us to assume what you want \Delta to mean.

    3. You also do not make it clear if you intend three functions or in some obscure way you expect the reader to abstract a single function from the apparent three definitions.

    4. What is the horizontal brace doing?

    5. What you have posted is in fact not a question at all (it does not contain a statement equivalent to something like find, show, prove, demonstrate etc)

    6. Probably not important but I asked "do you know ... how to solve this?" You only answer that you "have the answer" not that you know how to solve it. But I will let that go and just assume that you do not see the distinction.

    CB
    Last edited by CaptainBlack; December 29th 2009 at 01:42 AM. Reason: Added item 6.
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  5. #5
    Member integral's Avatar
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    ok then Mr.unimaginative
    Half of the problem is thinking about it, then you will see that it all makes perfect mathematical sense... You need not make one assumption.
    But if you cant make heads or tails of it, I guess you can delete it.
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