# Find the domain and range of the function $f(x) = \frac{x}{x^2 - 16}$

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

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

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?

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

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.

• That's really insightful. Thanks, that's what I'm looking for. Rules I can know and apply. @N.F.Taussig – Bucephalus Oct 14 '18 at 13:08

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 $$\mathbb{R}$$.

• Thankyou also, @TheSimpliFire . This is helpful too and answers my question and not just solves the problem. – Bucephalus Oct 14 '18 at 13:10
• @Bucephalus You're welcome :) – TheSimpliFire Oct 14 '18 at 13:12

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 $$\mathbb R$$

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