Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ f(x)=\frac{8 x}{x^{2}-4} $$ Find the intervals where $$ f(x) $$ is increasing or decreasing. Select the correct choice below and fill in the answer box(es) to complete your choice. (Type your answer in interval notation. Use a comma to separate answers as needed.) A. The function is increasing on $$ \square $$ It is never decreasing. B. The function is increasing on $$ \square $$ It is decreasing on $$ \square $$ I. C. The function is decreasing on $$ \square $$ It is never increasing. Find the location of any local extrema of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. There is a local minimum at $$ x= $$ $$ \square $$ There is no local maximum. (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There is a local maximum at $$ x= $$ $$ \square $$ and there is a local minimum at $$ x= $$ $$ \square $$ (Type integers or decimals. Use a comma to separate answers as needed.) C. There is a local maximum at $$ \mathrm{x}= $$ $$ \square $$ There is no local minimum. (Type an integer or a decimal. Use a comma to separate answers as needed.)

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**Step 1. Identify the Domain and Asymptotes**

1. The function is 
$$
f(x)=\frac{8x}{x^2-4}.
$$
The denominator factors as 
$$
x^2-4=(x-2)(x+2).
$$
Thus the function is undefined when $$x-2=0$$ or $$x+2=0$$; that is, when 
$$
x=2 \quad \text{or} \quad x=-2.
$$
The domain is 
$$
(-\infty,-2) \cup (-2,2) \cup (2,\infty).
$$

2. **Vertical asymptotes:** At $$x=-2$$ and $$x=2$$.

3. **Horizontal asymptote:** Since the degree of the numerator (1) is less than the degree of the denominator (2), 
$$
\lim_{x\to\pm\infty} f(x)=0.
$$
So there is a horizontal asymptote at $$y=0$$.

4. **Intercept:** Setting the numerator equal to 0 gives the $$x$$-intercept:
$$
8x=0\quad \Rightarrow \quad x=0.
$$
Since $$x=0$$ is in the domain, the graph passes through $$(0,0)$$.

---

**Step 2. Analyze the Increasing/Decreasing Behavior**

To determine the intervals where $$f(x)$$ is increasing or decreasing, compute the derivative $$f'(x)$$.

Using the quotient rule:
$$
f'(x)=\frac{(x^2-4)\cdot 8 - 8x\cdot (2x)}{(x^2-4)^2}.
$$

Simplify the numerator:
$$
8(x^2-4)-16x^2 = 8x^2-32-16x^2 = -8x^2-32.
$$
Factor the numerator:
$$
-8x^2-32 = -8\left(x^2+4\right).
$$
Thus,
$$
f'(x)=\frac{-8\left(x^2+4\right)}{(x^2-4)^2}.
$$

**Sign Analysis:**

- The factor $$x^2+4$$ is always positive for all real $$x$$.
- The constant $$-8$$ is negative.
- The denominator $$(x^2-4)^2$$ is always positive (except at $$x=\pm2$$ where the function is undefined).

Therefore, for all $$x$$ in the domain,
$$
f'(x)<0.
$$
This implies that $$f(x)$$ is strictly decreasing on each interval of its domain.

---

**Step 3. Local Extrema**

Since $$f'(x)<0$$ for all $$x$$ in the domain (and there is no point where $$f'(x)=0$$), there are no critical points that could yield local extrema.  
Thus, there are no local maxima or local minima.

---

**Step 4. Final Answers**

- **Intervals of Increasing/Decreasing:**  
  The function is decreasing on 
  $$
  (-\infty,-2) \cup (-2,2) \cup (2,\infty).
  $$
  It is never increasing.

- **Local Extrema:**  
  There are no local extrema.

The correct choices consistent with our findings are:

- For the increasing/decreasing part, select **Choice C** and fill in the interval:
  $$
  (-\infty,-2) \cup (-2,2) \cup (2,\infty).
  $$
- For the local extrema part, note that since no local extrema exist, neither option A, B, nor C (which all assert the existence of an extremum) applies. The appropriate statement is: "There is neither a local maximum nor a local minimum."
The function $$ f(x) = \frac{8x}{x^2 - 4} $$ is strictly decreasing on $$ (-\infty, -2) $$, $$ (-2, 2) $$, and $$ (2, \infty) $$. There are no local maxima or minima.

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