# testing

• Oct 19th 2011, 03:06 AM
hmmmm
testing
$\displaystyle g(x) = \left\{ \begin{array}{c l} x^2 + x + 1 & x \in (-1,0]\\ x^2 + x & x \in (0,1)\\ x^2 & x \not\in (-1,1) \end{array} \right.$

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

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

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

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

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

• Oct 20th 2011, 12:01 PM
TheEmptySet
Re: testing
Wow so many different ways. I have to throw in my two cents

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

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