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**blertta** · December 7th 2008, 06:09 AM

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)

**CaptainBlack** · December 7th 2008, 11:45 PM

Originally Posted by blertta

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)

This looks incomprehensible to me.

CB

**blertta** · December 8th 2008, 12:14 AM

Originally Posted by CaptainBlack

This looks incomprehensible to me.

CB

Because my engish is 2 bad and I don't know how to write with math language,for ex. I don't know how to express square root so I wrote it like a base with exponent 1/2.
Hope u understood me!

**Jameson** · December 8th 2008, 12:15 AM

What does "commune" mean?

**blertta** · December 8th 2008, 12:22 AM

Originally Posted by Jameson

What does "commune" mean?

For example the "beggining commune" of x*1/2 is R+.
Are the values of x which the function has realization.
Hope u understood me?

**Jameson** · December 8th 2008, 12:27 AM

Originally Posted by blertta

For example the "beggining commune" of x*1/2 is R+.
Are the values of x which the function has realization.
Hope u understood me?

I'm not sure I do.

If $f(x)=\sqrt{x-1}$ , 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.

**Chop Suey** · December 8th 2008, 12:29 AM

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

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

Square roots are defined, like you said, for $\mathbb{R}^+\cup \{0\}$ . Therefore:
$\log_2{(x-2)}\geq 0 \implies x-2\geq 1 \implies x\geq 3$

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

2. $y=\frac{4}{2x+2}$
This expression is defined for $2x+2\neq 0 \implies x\neq -1$

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

**blertta** · December 8th 2008, 12:39 AM

Originally Posted by Jameson

I'm not sure I do.

If $f(x)=\sqrt{x-1}$ , 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.

1/2 is an exponent. Anyway,thank you

**blertta** · December 8th 2008, 12:45 AM

Originally Posted by Chop Suey

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

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

Square roots are defined, like you said, for $\mathbb{R}^+\cup \{0\}$ . Therefore:
$\log_2{(x-2)}\geq 0 \implies x-2\geq 1 \implies x\geq 3$

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

2. $y=\frac{4}{2x+2}$
This expression is defined for $2x+2\neq 0 \implies x\neq -1$

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

i think you understood me.
At the first, as I know x-2 must be bigger or equal 0 too.
at the second, the exercise asks not for the domain but for the end "commune".Do you understand me?

**Chop Suey** · December 8th 2008, 12:51 AM

Originally Posted by blertta

i think you understood me.
At the first, as I know x-2 must be bigger or equal 0 too.
at the second, the exercise asks not for the domain but for the end "commune".Do you understand me?

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 $\mathbb{R}\backslash\{0\}$

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

**Chop Suey** · December 8th 2008, 01:16 AM

Originally Posted by blertta

i think you understood me.
At the first, as I know x-2 must be bigger or equal 0 too. This is wrong

$A = \{x~|~ x\in \mathbb{R}~\&~x-2>0\}$ 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.

$B = \{x~|~ x\in \mathbb{R}~\&~x-2\geq 1\}$ 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 $B \subset A$

**blertta** · December 8th 2008, 05:47 AM

Originally Posted by Chop Suey

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 $\mathbb{R}\backslash\{0\}$

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

Thank you very much and Happy Ramadan's day if you are muslim.

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