# Math Help - Who can?

1. ## Who can?

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)

2. 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

3. 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!

4. What does "commune" mean?

5. 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?

6. 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.

7. 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 $\mathbb{R} \backslash \{-1\}$

8. 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

9. 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?

10. 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").

11. 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 $B \subset A$

12. 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.