Find the beginning commune

y=((log2(x-2) ))*1/2 (the first 2 is the base of log)

Find the end (valuation)commune

y=4/(x*2+2)

Results 1 to 12 of 12

- Dec 7th 2008, 06:09 AM #1

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- Dec 7th 2008, 11:45 PM #2

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- Dec 8th 2008, 12:14 AM #3

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- Dec 8th 2008, 12:15 AM #4

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- Dec 8th 2008, 12:22 AM #5

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- Dec 8th 2008, 12:27 AM #6

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I'm not sure I do.

If $\displaystyle f(x)=\sqrt{x-1}$, is the commune of f(x) x>1? If this is what you mean then commune I take to mean domain of x. I fear though this is not exactly what you mean. Try looking around online for different vocabulary to use. I'm sorry I don't quite get you. Maybe someone else will and I'm missing something.

- Dec 8th 2008, 12:29 AM #7

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Ah, I think you're referring to the domain.

1. $\displaystyle y=\sqrt{\log_2{(x-2)}}$

Square roots are defined, like you said, for $\displaystyle \mathbb{R}^+\cup \{0\}$. Therefore:

$\displaystyle \log_2{(x-2)}\geq 0 \implies x-2\geq 1 \implies x\geq 3$

Therefore, domain set is $\displaystyle [3, \infty)$

2. $\displaystyle y=\frac{4}{2x+2}$

This expression is defined for $\displaystyle 2x+2\neq 0 \implies x\neq -1$

Therefore, domain set is $\displaystyle \mathbb{R} \backslash \{-1\}$

- Dec 8th 2008, 12:39 AM #8

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- Dec 8th 2008, 12:45 AM #9

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- Dec 8th 2008, 12:51 AM #10

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Oh, then I think you're referring to the range. Determining the range may be a bit tricky so we refer to a graph.

I attached the graph of the second expression.

As you can see, the y values covered are $\displaystyle \mathbb{R}\backslash\{0\}$

I hope I understood you correctly. You may want to refer to these terms as domain ("beginning commune") and range ("end commune").

- Dec 8th 2008, 01:16 AM #11

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$\displaystyle A = \{x~|~ x\in \mathbb{R}~\&~x-2>0\}$ is the set for which the logarithm is defined. But not all of the range is defined for the square root. Therefore, A is not the domain set.

$\displaystyle B = \{x~|~ x\in \mathbb{R}~\&~x-2\geq 1\}$ however, is the set for which the logarithm are greater than or equal to zero. This is the domain set for the expression. Note that $\displaystyle B \subset A$

- Dec 8th 2008, 05:47 AM #12

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