Summarize the pettinent information obtained by applying the graphing strategy and sketch the graph of $$ \mathrm{f}(\mathrm{x})=7 \mathrm{x} e^{-0.5 \mathrm{x}} $$ A. The function has one vertical asymptote, (Type an equation.) B. The function has two vertical asymptotes. The leftmost asymptote is and the rightmost asymptote is (Type equations.) C. There are no vertical asymptotes. Find the intervals where $$ \mathrm{f}(\mathrm{x}) $$ 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 inlegers or decimals for any numbers in the expression. Use a comma to separate answers as needed.) A. The function is increasing on B. It is never decreasing. B. The function is increasing on $$ \square $$. It is decreasing on C. The function is decreasing on $$ \square $$. It is never increasing.

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To determine the intervals where the function $$ f(x) = 7xe^{-0.5x} $$ is increasing or decreasing, we need to find the critical points by taking the derivative of the function and setting it equal to zero. 1. Find the derivative of the function: $$ f'(x) = 7e^{-0.5x} - 3.5xe^{-0.5x} $$ 2. Set the derivative equal to zero to find the critical points: $$ 7e^{-0.5x} - 3.5xe^{-0.5x} = 0 $$ 3. Solve for $$ x $$ to find the critical points. After finding the critical points, we can determine the intervals where the function is increasing or decreasing by testing the sign of the derivative in each interval. Let's proceed with finding the critical points. Solve the equation by following steps: - step0: Solve for $$x$$: $$7e^{-0.5x}-3.5xe^{-0.5x}=0$$ - step1: Simplify: $$7e^{-\frac{1}{2}x}-3.5xe^{-\frac{1}{2}x}=0$$ - step2: Factor the expression: $$\frac{7}{2}\left(2-x\right)\left(e^{x}\right)^{-\frac{1}{2}}=0$$ - step3: Rewrite the expression: $$\frac{7\left(2-x\right)}{2\left(e^{x}\right)^{\frac{1}{2}}}=0$$ - step4: Cross multiply: $$7\left(2-x\right)=2\left(e^{x}\right)^{\frac{1}{2}}\times 0$$ - step5: Simplify the equation: $$7\left(2-x\right)=0$$ - step6: Rewrite the expression: $$2-x=0$$ - step7: Move the constant to the right side: $$-x=0-2$$ - step8: Remove 0: $$-x=-2$$ - step9: Change the signs: $$x=2$$ The critical point is $$ x = 2 $$. Now, we need to determine the intervals where the function $$ f(x) = 7xe^{-0.5x} $$ is increasing or decreasing by testing the sign of the derivative in each interval. Let's test the intervals around the critical point $$ x = 2 $$ to determine where the function is increasing or decreasing. Solve the inequality by following steps: - step0: Solve for $$x$$: $$7e^{-0.5x}-3.5xe^{-0.5x}>0$$ - step1: Simplify: $$7e^{-\frac{1}{2}x}-3.5xe^{-\frac{1}{2}x}>0$$ - step2: Factor the expression: $$\frac{7}{2}\left(2-x\right)\left(e^{x}\right)^{-\frac{1}{2}}>0$$ - step3: Expand the expression: $$\left(7-\frac{7}{2}x\right)\left(e^{x}\right)^{-\frac{1}{2}}>0$$ - step4: Rearrange the terms: $$\frac{14-7x}{2\left(e^{x}\right)^{\frac{1}{2}}}>0$$ - step5: Separate into possible cases: $$\begin{align}&\left\{ \begin{array}{l}14-7x>0\2\left(e^{x}\right)^{\frac{1}{2}}>0\end{array}\right.\&\left\{ \begin{array}{l}14-7x<0\2\left(e^{x}\right)^{\frac{1}{2}}<0\end{array}\right.\end{align}$$ - step6: Solve the inequality: $$\begin{align}&\left\{ \begin{array}{l}x<2\x \in \mathbb{R}\end{array}\right.\&\left\{ \begin{array}{l}x>2\x \notin \mathbb{R}\end{array}\right.\end{align}$$ - step7: Find the intersection: $$\begin{align}&x<2\&x \notin \mathbb{R}\end{align}$$ - step8: Find the union: $$x<2$$ The function is increasing on the interval $$ x < 2 $$. Now, let's test the intervals around the critical point $$ x = 2 $$ to determine where the function is decreasing. Solve the inequality by following steps: - step0: Solve for $$x$$: $$7e^{-0.5x}-3.5xe^{-0.5x}<0$$ - step1: Simplify: $$7e^{-\frac{1}{2}x}-3.5xe^{-\frac{1}{2}x}<0$$ - step2: Factor the expression: $$\frac{7}{2}\left(2-x\right)\left(e^{x}\right)^{-\frac{1}{2}}<0$$ - step3: Expand the expression: $$\left(7-\frac{7}{2}x\right)\left(e^{x}\right)^{-\frac{1}{2}}<0$$ - step4: Rearrange the terms: $$\frac{14-7x}{2\left(e^{x}\right)^{\frac{1}{2}}}<0$$ - step5: Separate into possible cases: $$\begin{align}&\left\{ \begin{array}{l}14-7x>0\2\left(e^{x}\right)^{\frac{1}{2}}<0\end{array}\right.\&\left\{ \begin{array}{l}14-7x<0\2\left(e^{x}\right)^{\frac{1}{2}}>0\end{array}\right.\end{align}$$ - step6: Solve the inequality: $$\begin{align}&\left\{ \begin{array}{l}x<2\x \notin \mathbb{R}\end{array}\right.\&\left\{ \begin{array}{l}x>2\x \in \mathbb{R}\end{array}\right.\end{align}$$ - step7: Find the intersection: $$\begin{align}&x \notin \mathbb{R}\&x>2\end{align}$$ - step8: Find the union: $$x>2$$ The function is decreasing on the interval $$ x > 2 $$. Therefore, the function $$ f(x) = 7xe^{-0.5x} $$ is increasing on the interval $$ x < 2 $$ and decreasing on the interval $$ x > 2 $$. The function $$ f(x) = 7xe^{-0.5x} $$ is increasing on $$ x < 2 $$ and decreasing on $$ x > 2 $$.

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