Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ f(x)=7 x e^{-0.5 x} $$ 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 C. The function is increasing on $$ (-\infty, 2) $$. It is decreasing on $$ (2, \infty) $$. Thecreasing. 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 $$ x=\square $$. There is no local minimum. (Type an integer or a decimal. Use a comma to separate answers as needed.) D. There are no local extrema.

Question

UpStudy AI · Accepted Answer

The function is given by $$ f(x)=7x\,e^{-0.5x}. $$ 1. **Vertical Asymptotes:** Since $$e^{-0.5x}$$ is defined for every real number and the only factor that could produce a zero in the denominator is absent, there are no vertical asymptotes. 2. **Intervals of Increase and Decrease:** We find the derivative using the product rule: $$ f'(x)=\frac{d}{dx}\Big(7x\,e^{-0.5x}\Big)=7e^{-0.5x}+7x\left(-0.5e^{-0.5x}\right)=7e^{-0.5x}\Big(1-0.5x\Big). $$ Since $$7e^{-0.5x}>0$$ for all $$x$$, the sign of $$f'(x)$$ depends on $$1-0.5x$$: - $$f'(x)>0$$ when $$1-0.5x>0$$ $$\Rightarrow x<2$$. - $$f'(x)<0$$ when $$1-0.5x<0$$ $$\Rightarrow x>2$$. Thus, the function is increasing on $$ (-\infty,2) $$ and decreasing on $$ (2,\infty). $$ 3. **Local Extrema:** Setting $$f'(x)=0$$: $$ 7e^{-0.5x}\Big(1-0.5x\Big)=0\quad \Rightarrow \quad 1-0.5x=0, $$ which gives $$x=2$$. Since $$f'(x)$$ changes from positive to negative at $$x=2$$, there is a local maximum at this point. There is no local minimum. 4. **Additional Graphing Information:** - **$$x$$-intercept:** $$f(0)=7\cdot0\cdot e^{0}=0$$. - **End behavior:** As $$x\to\infty$$, the exponential decay dominates, and $$ \lim_{x\to\infty}7x\,e^{-0.5x}=0. $$ As $$x\to -\infty$$, $$e^{-0.5x}$$ grows without bound while $$x$$ is negative, so $$ \lim_{x\to -\infty}7x\,e^{-0.5x}=-\infty. $$ - **Inflection Point (for concavity change):** Compute the second derivative. Starting from $$ f'(x)=7e^{-0.5x}\Big(1-0.5x\Big), $$ differentiate again: $$ f''(x)=7e^{-0.5x}\left(-0.5\Big(1-0.5x\Big) + \Big(-0.5\Big)\right)=7e^{-0.5x}\Big(0.25x-1\Big). $$ Setting $$f''(x)=0$$ gives: $$ 0.25x-1=0 \quad \Rightarrow \quad x=4. $$ For $$x<4$$, $$0.25x-1<0$$ (concave down); for $$x>4$$, $$0.25x-1>0$$ (concave up). Thus, there is an inflection point at $$x=4$$. 5. **Answer Selections:** - **Intervals of Increase/Decrease:** The function is increasing on $$ (-\infty,2) $$ and decreasing on $$ (2,\infty). $$ - **Local Extrema:** There is a local maximum at $$ x=2 $$ and no local minimum. The summary of the pertinent information is as follows: - No vertical asymptotes. - Increasing on $$(-\infty, 2)$$ and decreasing on $$(2, \infty)$$. - A local maximum at $$x=2$$ (with $$f(2)=7\cdot2\,e^{-1}$$ or approximately $$14/e$$). - An inflection point at $$x=4$$ where the concavity changes. - As $$x\to\infty$$, $$f(x)\to 0$$; as $$x\to -\infty$$, $$f(x)\to -\infty$$. Thus, the correct choices are: C. The function is increasing on $$ (-\infty, 2) $$ and decreasing on $$ (2, \infty) $$. C. There is a local maximum at $$ x=2$$. There is no local minimum. The function $$ f(x) = 7x e^{-0.5x} $$ has no vertical asymptotes. It is increasing on $$ (-\infty, 2) $$ and decreasing on $$ (2, \infty) $$. There is a local maximum at $$ x = 2 $$ and no local minimum.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Calculus Questions