0
$\begingroup$

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

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?

  • 0
    @AméricoTavares: Yes sir, doing that right now, thank you very much.2012-03-17

2 Answers 2

2

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 $\frac{\left( x+2\right) 5-6}{\left( x-2\right) \left( x+2\right) }=\frac{5-\frac{6}{x+2}}{x-2}$.

  • 0
    @KerimAtasoy: You are welcome.2012-03-17
2

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)