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$$f(x) = \frac{x}{x^2-16}$$

$$f(x) = \frac{x}{(x-4)(x+4)}$$

I can see that the domain is $\{x|x\neq \pm 4\}$

I'm not sure what to do for the range though.

$$y = \frac{x}{(x-4)(x+4)}$$

$$x = y(x-4)(x+4)$$

From here, to me it looks like there can be no value of $y$ that will create an unreal number $x$, so I would say the range of $f(x)$ is any real number.

But I'm not sure I did this the mathematically correct way? Is there a more correct way to verify the range of this function or is this the way it's done?

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  • $\begingroup$ Why is $-4$ included in the domain? You are correct that the range is the set of all real numbers. $\endgroup$ – N. F. Taussig Oct 14 '18 at 12:57
  • $\begingroup$ aahh, yes, you're right. Let me correct that. $\endgroup$ – Bucephalus Oct 14 '18 at 12:59
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Your domain is correct.

For the range, since $$y = \frac{x}{x^2 - 16}$$ we obtain \begin{align*} y(x^2 - 16) & = x\\ yx^2 - 16y & = 0\\ yx^2 - x - 16y & = 0 \end{align*} For $y$ to be in the range, this quadratic equation must have real roots. Hence, we require that the discriminant be nonnegative. Since the discriminant is $$\Delta = b^2 - 4ac = (-1)^2 - 4y(-16y) = 1 + 16y^2$$ the discriminant is positive for every real value of $y$. Hence, the range is the set of all real numbers.

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    $\begingroup$ That's really insightful. Thanks, that's what I'm looking for. Rules I can know and apply. @N.F.Taussig $\endgroup$ – Bucephalus Oct 14 '18 at 13:08
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The range is the set of all values that $y$ can take.

Since $y=\dfrac x{(x+4)(x-4)}$, over $(-4,4)$, $y$ is continuous.

Now when $x=-4+\epsilon$, $y\to\dfrac{-4+\epsilon}{\epsilon(-8+\epsilon)}=\dfrac{4-\epsilon}{\epsilon(8-\epsilon)}$ and as $\epsilon\to0$, we see that $y\to\infty$.

And when $x=4-\epsilon$, $y\to\dfrac{4-\epsilon}{-\epsilon(8-\epsilon)}$ and as $\epsilon\to0$, we see that $y\to-\infty$.

By the Mean Value Theorem, there exists a value $t\in\mathbb{R}$ such that $t=\dfrac x{x^2-16}$, so the range is the whole of $\mathbb{R}$.

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    $\begingroup$ Thankyou also, @TheSimpliFire . This is helpful too and answers my question and not just solves the problem. $\endgroup$ – Bucephalus Oct 14 '18 at 13:10
  • $\begingroup$ @Bucephalus You're welcome :) $\endgroup$ – TheSimpliFire Oct 14 '18 at 13:12
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there's two asymptotes ---> $x=4$ and $x=-4$ so the domain is: $$x \in \mathbb R \setminus \text{{-4,4}}$$

but for all other $x \in \mathbb R$, f(x) is exists, so the range is $\mathbb R$

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The domain is given by $$x\ne \pm 4$$ and the range is given by all real numbers.

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