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$f(x)=\frac{5x+4}{x^2-4}\; \rightarrow x=\;?\;$

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?

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What are you trying to do? You start with a function, then treats is an equation. –  Fredrik Meyer Mar 17 '12 at 13:47
    
I'm just trying to learn here, not showing off... So, if any help or idea available, it might be very nice and kind...? –  Kerim Atasoy Mar 17 '12 at 13:52
1  
Do you want to find the domain of the function $f(x)$? –  Américo Tavares Mar 17 '12 at 13:53
    
@AméricoTavares: I think yes. Yes, I do, sir. Thank you very much for the "wiki" clue... :) –  Kerim Atasoy Mar 17 '12 at 13:59
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@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. –  TMM Mar 17 '12 at 14:29

2 Answers 2

up vote 1 down vote accepted

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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I see it now... THANK YOU VERY MUCH, SIR... :) –  Kerim Atasoy Mar 17 '12 at 14:43
    
@KerimAtasoy: You are welcome. –  Américo Tavares Mar 17 '12 at 14:44

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)

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