$$f(x) = \frac{x}{x^2-16}$$
$$f(x) = \frac{x}{(x-4)(x+4)}$$
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?
Best Answer
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 $\mathbb{R}$.