n00b Testing Latex

**beardedoneder** · June 28th 2009, 10:08 AM

Given: $f(x)=x+2, h(x)=\frac{1}{x-1}$
Find $\frac{[h(x)]^2}{f(x)}$ , and list domain restrictions
$\frac{[\frac{1}{x-1}]^2}{x+2} = \frac{\frac{1}{(x-1)(x-1)}}{\frac{x+2}{1}} = \frac{1}{(x-1)(x-1)} \bullet \frac{1}{x+2} =$
$\frac{1}{(x-1)(x-1)(x+2)}, x \neq \{1,-2\}$

**earboth** · June 28th 2009, 12:19 PM

Originally Posted by beardedoneder

Given: $f(x)=x+2, h(x)=\frac{1}{x-1}$
Find $\frac{[h(x)]^2}{f(x)}$ , and list domain restrictions
$\frac{[\frac{1}{x-1}]^2}{x+2} = \frac{\frac{1}{(x-1)(x-1)}}{\frac{x+2}{1}} = \frac{1}{(x-1)(x-1)} \bullet \frac{1}{x+2} =$
$\frac{1}{(x-1)(x-1)(x+2)}, x \neq \{1,-2\}$

Only one tip: Instead of \bullet us \cdot

$\frac{1}{(x-1)(x-1)} \cdot \frac{1}{x+2}$

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