Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of \( \mathrm{f}(\mathrm{x})=7 \mathrm{xe}-0.5 \mathrm{x} \) C. The function is decreasing on \( \quad \). It is never increasing. Find the location of any local extrema of \( \mathrm{f}(\mathrm{x}) \). 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 \( \mathrm{x}= \) (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There is a local maximum at \( \mathrm{x}= \) and there is a local minimum at \( \mathrm{x}= \) (Type integers or decimals. Use a comma to separate answers as needed.) C. There is a local maximum at \( x=2 \). 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. Find the intervals where \( \mathrm{f}(\mathrm{x}) \) is concave upward or downward. 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 concave upward on \( \quad \). It is never concave downward. B. The function is concave downward on \( \square \). It is never concave upward. C. The function is concave upward on \( \square \). It is concave downward on \( \square \). C.

**Step 1. Find the first derivative and determine increasing/decreasing intervals.** The function is \[ f(x)=7x e^{-0.5x}. \] Differentiate using the product rule: \[ f'(x)=7e^{-0.5x}+7x\left(-0.5e^{-0.5x}\right)=7e^{-0.5x}\left(1-0.5x\right). \] Set \( f'(x)=0 \): \[ 7e^{-0.5x}\left(1-0.5x\right)=0. \] Since \( 7e^{-0.5x}>0 \) for all \( x \), it follows that \[ 1-0.5x=0 \quad\Longrightarrow\quad x=2. \] For \( x<2 \), \( 1-0.5x>0 \) so \( f'(x)>0 \) (increasing). For \( x>2 \), \( 1-0.5x<0 \) so \( f'(x)<0 \) (decreasing). Thus, the function is increasing on \((-\infty,2)\) and decreasing on \((2,\infty)\). **Step 2. Identify the local extrema.** Since the function changes from increasing to decreasing at \( x=2 \), there is a local maximum at \( x=2 \). There is no local minimum. *Correct choice for extrema:* C. There is a local maximum at \( x=2 \). There is no local minimum. **Step 3. Find the second derivative and determine intervals of concavity.** Differentiate the first derivative \[ f'(x)=7e^{-0.5x}(1-0.5x) \] using the product rule. Let \[ u=e^{-0.5x}, \qquad v=1-0.5x. \] Then \[ u'=-0.5e^{-0.5x}, \qquad v'=-0.5. \] Thus, \[ f''(x)=7\Big[u'v+uv'\Big]=7\left[-0.5e^{-0.5x}(1-0.5x)-0.5e^{-0.5x}\right]. \] Factor \( -0.5e^{-0.5x} \): \[ f''(x)=7\left[-0.5e^{-0.5x}\Big((1-0.5x)+1\Big)\right] =7\left[-0.5e^{-0.5x}(2-0.5x)\right]. \] This simplifies to \[ f''(x)=-\frac{7}{2} e^{-0.5x}(2-0.5x). \] An alternative form is achieved by factoring \( 2-0.5x=0.5(4-x) \): \[ f''(x)=-\frac{7}{2} e^{-0.5x}\cdot0.5(4-x) =-\frac{7}{4}e^{-0.5x}(4-x). \] Set \( f''(x)=0 \): \[ -\frac{7}{4}e^{-0.5x}(4-x)=0. \] Since \( -\frac{7}{4}e^{-0.5x}\neq0 \), we have \[ 4-x=0 \quad\Longrightarrow\quad x=4. \] Test the sign of \( f''(x) \): - For \( x<4 \): \( (4-x)>0 \) so \( f''(x)<0 \) (concave downward). - For \( x>4 \): \( (4-x)<0 \) so \( f''(x)>0 \) (concave upward). Thus, the function is concave downward on \((-\infty, 4)\) and concave upward on \((4, \infty)\). *Correct choice for concavity:* C. The function is concave upward on \((4,\infty)\) and concave downward on \((-\infty,4)\). **Step 4. Graph Summary of \( f(x)=7xe^{-0.5x} \)** - The function is increasing on \((-\infty,2)\) and decreasing on \((2,\infty)\) with a local maximum at \( x=2 \). - It is concave downward on \((-\infty,4)\) and concave upward on \((4,\infty)\). - The transition in concavity at \( x=4 \) is an inflection point. - Since \( f(0)=0 \) and \( f(2)=14/e \approx 5.15 \), the curve rises from the origin to the maximum and then decreases gradually while eventually leveling off as \( x\to\infty \). These analyses form the basis for sketching the graph.