testing

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October 19th 2011, 03:06 AM
hmmmm

testing

$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. $

$g(x) = \left\{ \begin{array}{c l} 0 & x \in (-1,0]\\ x^2 + x & x \in (0,1)\\ x^2 & x \not\in (-1,1) \end{array} \right. $
October 19th 2011, 03:24 AM
Plato

Re: testing

Quote:

Originally Posted by hmmmm

$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. $

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

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
$g(x) = \left\{\begin{array}{c l} x^2 + x + 1 & x \in (-1,0]\\ x^2 + x & x \in (0,1)\\ x^2 & x \not\in (-1,1)\end{array}\right.$
October 20th 2011, 11:57 AM
Soroban

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}

$f(x) = \begin{Bmatrix} x+3 & x < 2 \\ 6-x & x \ge 2 \end{array}$
October 20th 2011, 12:01 PM
TheEmptySet

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}