Summarize the pertinent information oblained by applying the graphing strategy and sketch the graph of $$ f(x)=7 x e^{-05 x} $$ B. The domain is all real $$ x $$. Find the $$ x $$-intercepts of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The $$ x $$-intercept(s) is/are at $$ x=0 $$ (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There are no $$ x $$-intercepts Find the $$ y $$-intercepts of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The $$ y $$-intercept(s) is/are at $$ y= $$ $$ \square $$ . (Type an integer or a decimal. Use a comma to separate answers as needed ) B. There are no $$ y $$-intercepts.

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**1. Domain**

The domain of $$ f(x)=7xe^{-0.5x} $$ is all real numbers, i.e., 
$$
(-\infty,\infty).
$$

**2. $$ x $$-Intercept(s)**

To determine the $$ x $$-intercepts, set $$ f(x)=0 $$:
$$
7xe^{-0.5x}=0.
$$
Since the exponential factor $$ e^{-0.5x} $$ is never zero, we have:
$$
7x=0 \quad\Longrightarrow\quad x=0.
$$
Thus, the $$ x $$-intercept is at:
$$
x=0.
$$

**3. $$ y $$-Intercept(s)**

To determine the $$ y $$-intercept, evaluate $$ f(0) $$:
$$
f(0)=7\cdot0\cdot e^{-0.5\cdot 0}=0.
$$
Thus, the $$ y $$-intercept is at:
$$
y=0.
$$

**4. Additional Graphing Information**

- **Critical Point:**  
  Find the derivative using the product rule:
  $$
  f(x)=7xe^{-0.5x} \quad\Longrightarrow\quad f'(x)=7e^{-0.5x}+7x\left(-0.5e^{-0.5x}\right).
  $$
  Factor out $$ 7e^{-0.5x} $$:
  $$
  f'(x)=7e^{-0.5x}\left(1-0.5x\right).
  $$
  Setting $$ f'(x)=0 $$ (note $$ 7e^{-0.5x}\neq0 $$):
  $$
  1-0.5x=0 \quad\Longrightarrow\quad x=2.
  $$
  The function has a critical point at $$ x=2 $$.  
  Calculate the $$ y $$-value:
  $$
  f(2)=7\cdot2\cdot e^{-0.5\cdot2}=14e^{-1}=\frac{14}{e}.
  $$
  This is a relative maximum at $$ \left(2,\frac{14}{e}\right) $$.

- **End Behavior:**  
  - As $$ x\to\infty $$, the exponential decay dominates so:
    $$
    \lim_{x\to\infty} f(x)=0^+.
    $$
  - As $$ x\to -\infty $$, note that:
    $$
    e^{-0.5x}=e^{0.5|x|} \to +\infty,
    $$
    and since $$ x $$ is large and negative, we obtain:
    $$
    \lim_{x\to -\infty} f(x)=-\infty.
    $$
    
**5. Sketching the Graph**

When sketching the graph of $$ f(x)=7xe^{-0.5x} $$:

- Plot the intercepts: the graph passes through the origin $$(0,0)$$.
- At $$ x=2 $$, plot the maximum point $$\left(2,\frac{14}{e}\right)$$.
- For $$ x>2 $$, the graph decreases approaching $$ y=0 $$ as $$ x\to\infty $$.
- For $$ x<0 $$, the graph comes from $$ -\infty $$ as $$ x\to -\infty $$ and increases to meet the origin.

**Answers:**

- $$ x $$-intercepts: $$ x=0 $$.

- $$ y $$-intercepts: $$ y=0 $$.
- $$ x $$-intercept: $$ x=0 $$. - $$ y $$-intercept: $$ y=0 $$.

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