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Math Help - testing

  1. #1
    Senior Member
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    testing

    g(x) = \left\{<br />
\begin{array}{c l}<br />
  x^2 + x + 1 & x \in (-1,0]\\<br />
  x^2 + x & x \in (0,1)\\<br />
  x^2 & x \not\in (-1,1)<br />
\end{array}<br />
\right.<br />

    g(x) = \left\{<br />
\begin{array}{c l}<br />
  0                 & x \in (-1,0]\\<br />
  x^2 + x & x \in (0,1)\\<br />
  x^2 & x \not\in (-1,1)<br />
\end{array}<br />
\right.<br />
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  2. #2
    MHF Contributor

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    Re: testing

    Quote Originally Posted by hmmmm View Post
    g(x) = \left\{<br />
\begin{array}{c l}<br />
  x^2 + x + 1 & x \in (-1,0]\\<br />
  x^2 + x & x \in (0,1)\\<br />
  x^2 & x \not\in (-1,1)<br />
\end{array}<br />
\right.<br />

    g(x) = \left\{<br />
\begin{array}{c l}<br />
  0                 & x \in (-1,0]\\<br />
  x^2 + x & x \in (0,1)\\<br />
  x^2 & x \not\in (-1,1)<br />
\end{array}<br />
\right.<br />
    Remove all line-feeds.
    [TEX]g(x) = \left\{\begin{array}{c l} x^2 + x + 1 & x \in (-1,0]\\ x^2 + x & x \in (0,1)\\ x^2 & x \not\in (-1,1)\end{array}\right.[/TEX]
    gives
    g(x) = \left\{\begin{array}{c l}  x^2 + x + 1 & x \in (-1,0]\\  x^2 + x & x \in (0,1)\\  x^2 & x \not\in (-1,1)\end{array}\right.
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  3. #3
    Super Member

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    Re: testing

    Hello, all!

    By accident I found a way to write piece-wise functions
    . . with a "brace" at the left end only.


    f(x) = \begin{Bmatrix} x+3 & x < 2 \\ 6-x & x \ge 2 \end{array}

    f(x) = \begin{Bmatrix} x+3 & x < 2 \\ 6-x & x \ge 2 \end{array}

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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Yuma, AZ, USA
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    Re: testing

    Wow so many different ways. I have to throw in my two cents

    f(x) = \begin{cases} x+3, \quad x < 2 \\ 6-x. \quad x \ge 2 \end{cases}

    f(x) = \begin{cases} x+3, \quad x < 2 \\ 6-x. \quad x \ge 2 \end{cases}
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