$f(x)=\frac{5x+4}{x^2-4}\; \rightarrow x=\;?\;$
Well, I got this from that equation:
$\frac{5x+4}{x^2-2^2}=\frac{(x+2)*5-6}{(x-2)(x+2)}=\frac{-1}{x-2}=\frac{1}{-x+2} \; \rightarrow x=\mathbb{R}-\left\{2,-2\right\}$
So, my solution approaching for this equation is right? Maybe you could share yours as well?
Domain of "f(x)"
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0What are you trying to do? You start with a function, then treats is an *equation*. – 2012-03-17
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0I'm just trying to learn here, not showing off... So, if any help or idea available, it might be very nice and kind...? – 2012-03-17
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1Do you want to find the [domain](http://en.wikipedia.org/wiki/Domain_of_a_function) of the function $f(x)$? – 2012-03-17
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0@AméricoTavares: I think yes. Yes, I do, sir. Thank you very much for the "wiki" clue... :) – 2012-03-17
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0Where is $f(x)$ defined? – 2012-03-17
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0How did you get $\frac{(x+2)*5-6}{(x-2)(x+2)}=\frac{-1}{x-2}$?? – 2012-03-17
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0Please change the title and the text accordingly. – 2012-03-17
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0Also, "concluding" from $\frac{1}{-x+2}$ that $x$ cannot be $2$ or $-2$ is just wrong. – 2012-03-17
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0@AméricoTavares: It asks for the maxiumum available range for the domain set. – 2012-03-17
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0@AméricoTavares: The title, yes, I got it already... I'm trying to change it now... – 2012-03-17
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0@TMM: I did it to solve the equation... If you think there is something wrong there, please, just go ahead and add your any available correction or idea. That's what I'm looking for... – 2012-03-17
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1@Kerim: In general, $(a \cdot b) - c \neq a \cdot (b - c)$, which you did apply to the numerator in the second equality on the second line. Also $1/(-x+2)$ is perfectly well-defined for $x = -2$, hence my second comment. – 2012-03-17
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0@TMM: I think I understand you... But, doesn't "(((x+2)*5)-6)/((x-2)*(x+2)) equal (-1)/(x-2) ? Where I'm doing wrong, show me please...? – 2012-03-17
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0@KerimAtasoy: see the PS in my answer. – 2012-03-17
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0@AméricoTavares: Yes sir, doing that right now, thank you very much. – 2012-03-17
2 Answers
Hint: The function $f(x)$ is defined for every $x\in \mathbb{R} $ such that the denominator $x^{2}-4\neq 0$.
PS. Note that $\frac{\left( x+2\right) 5-6}{\left( x-2\right) \left( x+2\right) }\neq \frac{-1}{x-2}$, because if you divide both the numerator and the denominator by $x+2$ you get $\frac{\left( x+2\right) 5-6}{\left( x-2\right) \left( x+2\right) }=\frac{5-\frac{6}{x+2}}{x-2}$.
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0I see it now... THANK YOU VERY MUCH, SIR... :) – 2012-03-17
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0@KerimAtasoy: You are welcome. – 2012-03-17
A problem like this is partly an exercise in learning to parse equations and partly an exercise in organization.
The expression (not equation!) $\frac{5x+4}{x^2-4}$ is a quotient expression: the "outermost" operation is division, and it can be rewritten as
- The quotient of "$5x+4$" divided by "$x^2-4$"
The domain of a quotient is everywhere where:
- The numerator is defined
- The denominator is defined, and nonzero
So you need to solve the subproblems:
- What is the domain where $5x+4$ is defined
- What is the domain where $x^2-4$ is defined and nonzero?
As an aside, the domain of $\frac{x-1}{x-1}$ is not all of $\mathbb{R}$: it is everywhere except $1$. As partial[1] functions:
$$ \frac{x-1}{x-1} \neq 1 $$
However, the continuous extension of $\frac{x-1}{x-1}$ is, in fact, $1$.
As an aside... be aware that there is an implicit hypothesis: $x$ is a variable ranging over the domain of all real numbers. Sometimes, you have variables that range over smaller domains: e.g. if $y$ is a variable ranging over positive real numbers, then the domain of $y-1$ is also merely the positive real numbers.
[1]: Really, you're working with partial functions, not functions. (If you were working with functions, there would be no questions about domain: if $x$ is a variable ranging over all reals, then either the domain of $f(x)$ is all of $\mathbb{R}$, or the problem is ill-posed. In particular, $1/x$ would be an ill-defined expression)