Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ f(x)=7 x e^{-0.5 x} $$. B. The function has two horizontal asymptotes. The top asymptote is $$ \square $$ and the bottom asymptote is $$ \square $$ . (Type equations.) C. There are no horizontal asymptotes. Find any vertical asymptotes of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. The function has one vertical asymptote, $$ \square $$ . (Type an equation.) B. The function has two vertical asymptotes. The leftmost asymptote is $$ \square $$ and the rightmost asymptote is $$ \square $$ . (Type equations.) C. There are no vertical asymptotes. 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 integers or decimals for any numbers in the expression. 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 $$ \square $$ . C. The function is decreasing on $$ \square $$ It is never increasing.

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**Horizontal Asymptote** Examine the limits of $$ f(x)=7x\,e^{-0.5x} $$ as $$x\to\infty$$ and $$x\to -\infty$$. 1. As $$x\to\infty$$: The exponential decays very rapidly so that $$ \lim_{x\to\infty}7x\,e^{-0.5x}=0. $$ Thus, there is a horizontal asymptote $$y=0$$ as $$x \to \infty$$. 2. As $$x\to -\infty$$: Notice that $$ e^{-0.5x}=e^{0.5|x|} $$ grows without bound as $$x\to -\infty$$ and multiplied by $$7x$$ (which is negative for $$x \to -\infty$$) gives $$ \lim_{x\to -\infty}7x\,e^{-0.5x}=-\infty. $$ Therefore, there is no horizontal asymptote as $$x\to -\infty$$. In summary, the function has one horizontal asymptote: $$ y=0 \quad \text{(as } x\to \infty\text{)}. $$ **Vertical Asymptote** Since $$f(x)=7x\,e^{-0.5x}$$ is defined for all real numbers (there is no division by zero or other discontinuity), there are no vertical asymptotes. Thus, the correct choice is: **C. There are no vertical asymptotes.** **Intervals of Increase and Decrease** To find where $$f(x)$$ is increasing or decreasing, compute the derivative using the product rule: $$ f(x)=7x\,e^{-0.5x}. $$ Differentiate: $$ f'(x)=7\,e^{-0.5x}+7x\,\left(-0.5\,e^{-0.5x}\right)=7e^{-0.5x}\left(1-0.5x\right). $$ Since $$7e^{-0.5x}>0$$ for all $$x$$, the sign of $$f'(x)$$ is determined by the factor $$1-0.5x$$. Set it equal to zero to find a critical point: $$ 1-0.5x=0\quad\Longrightarrow\quad x=2. $$ Now test the sign of the derivative: - For $$x<2$$: $$1-0.5x>0$$ so $$f'(x)>0$$. Hence, $$f$$ is increasing on $$(-\infty,2)$$. - For $$x>2$$: $$1-0.5x<0$$ so $$f'(x)<0$$. Hence, $$f$$ is decreasing on $$(2,\infty)$$. Thus, the correct choice is: **B. The function is increasing on $$(-\infty, 2)$$. It is decreasing on $$(2, \infty)$$.** **Graphing Summary** - **Horizontal asymptote:** $$y=0$$ as $$x\to\infty$$ (no horizontal asymptote as $$x\to -\infty$$). - **Vertical asymptote:** None. - **Monotonicity:** Increasing on $$(-\infty,2)$$ and decreasing on $$(2,\infty)$$. A sketch of the graph would show the curve rising from very negative values (as $$x\to -\infty$$) and crossing the horizontal line $$y=0$$ as $$x\to\infty$$, with a turning point at $$x=2$$. **Horizontal Asymptote:** - $$ y = 0 $$ as $$ x \to \infty $$ **Vertical Asymptote:** - None **Intervals of Increase and Decrease:** - Increasing on $$ (-\infty, 2) $$ - Decreasing on $$ (2, \infty) $$

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