To find the range of the function $f(x) = \frac{2}{x^2 - 16}$, we need to analyze the behavior of the function.

Step 1: Identify the domain

The function is undefined where the denominator equals zero. Set the denominator equal to zero and solve for $x$:

$x^2 - 16 = 0$

$x^2 = 16$

$x = \pm 4$

Thus, the function is undefined at $x = 4$ and $x = -4$. These are vertical asymptotes.

Step 2: Behavior of the function

Now, let's analyze the behavior of $f(x)$ as $x$ approaches the values $4$ and $-4$, as well as for large $x$.

• As $x \to 4^{+}$ or $x \to 4^{-}$, $x^2 - 16 \to 0^+$, which means the function $f(x) \to \infty$.
• As $x \to -4^{+}$ or $x \to -4^{-}$, $x^2 - 16 \to 0^+$, and again, $f(x) \to \infty$.
• As $x \to \pm \infty$, the denominator $x^2 - 16$ grows larger and larger, causing $f(x) \to 0$.

Step 3: Analyze the range

For the function to approach $0$ as $x \to \pm \infty$, the value $f(x)$ can get arbitrarily close to 0, but never actually equal 0.

Thus, the function has a horizontal asymptote at $y = 0$, and it can take all positive values because as $x$ approaches 4 or -4, the function heads toward positive infinity.

Conclusion: Range of $f(x)$

The range of the function is all positive real numbers:

$\boxed{(0, \infty)}$

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Would you like more details or further clarification on this problem?

Related Questions:

1. How would the range change if the function was $f(x) = \frac{3}{x^2 - 16}$?
2. What happens to the graph of the function as $x$ approaches 0?
3. How do you find the domain of rational functions?
4. How do horizontal asymptotes affect the range of a function?
5. What is the behavior of functions with vertical asymptotes?

Tip: When working with rational functions, always start by finding the domain and identifying the asymptotes to get a clear understanding of the function's behavior.