Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ \mathrm{f}(\mathrm{x})=5+2 e^{-0.2 \mathrm{x}} $$. 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 a comma to separate answers as needed.) A. The function is concave upward on . It is never concave downward. B. The function is concave downward on It is never concave upward. C. The function is concave upward on Find the location of any inflection points of $$ \mathrm{f}(\mathrm{x}) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice A. There is an inflection point at $$ \mathrm{x}= $$ (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There are no inflection points.

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$$
\textbf{Step 1: Analyze the function and sketch key features}
$$

The function is 
$$
\mathrm{f}(x)=5+2e^{-0.2x}.
$$
Key observations include:
- The domain of $$\mathrm{f}$$ is all real numbers, $$(-\infty, \infty)$$.
- As $$x \to \infty$$, note that $$e^{-0.2x} \to 0$$ so $$\mathrm{f}(x) \to 5$$. Hence, the horizontal asymptote is $$y=5$$.
- As $$x \to -\infty$$, $$e^{-0.2x} \to \infty$$ and thus $$\mathrm{f}(x) \to \infty$$.
- At $$x=0$$, $$\mathrm{f}(0)=5+2e^0=7$$.

Thus, the graph starts high for negative $$x$$, decreases monotonically (as will be seen from the derivative), and approaches $$y=5$$ from above as $$x$$ increases.

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\textbf{Step 2: Compute the first derivative for monotonicity}
$$

Differentiate $$\mathrm{f}(x)$$:
$$
\mathrm{f}'(x)=\frac{d}{dx}\left(5+2e^{-0.2x}\right)=2 \cdot \left(-0.2\right)e^{-0.2x}=-0.4e^{-0.2x}.
$$
Since $$e^{-0.2x}>0$$ for all $$x$$, we have $$\mathrm{f}'(x)<0$$ for all $$x$$. Thus, the function is strictly decreasing.

$$
\textbf{Step 3: Compute the second derivative to determine concavity}
$$

Differentiate $$\mathrm{f}'(x)$$:
$$
\mathrm{f}''(x)=\frac{d}{dx}\left(-0.4e^{-0.2x}\right) = -0.4\cdot\left(-0.2\right)e^{-0.2x}=0.08e^{-0.2x}.
$$
Since $$e^{-0.2x}>0$$ for every $$x$$, it follows that
$$
\mathrm{f}''(x) > 0 \quad \text{for all } x.
$$
Thus, the function is concave upward on the entire domain $$(-\infty, \infty)$$.

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\textbf{Step 4: Identify potential inflection points}
$$

Inflection points occur where the concavity changes, i.e., where $$\mathrm{f}''(x)=0$$ or is undefined and changes sign. Since $$\mathrm{f}''(x)=0.08e^{-0.2x}$$ is never zero or undefined, there are no inflection points.

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\textbf{Final Answers:}
$$

$$
\textbf{Concavity:} \quad \text{The function is concave upward on } (-\infty, \infty).\quad \text{It is never concave downward.}
$$

$$
\textbf{Inflection Point:} \quad \text{There are no inflection points.}
$$
The function $$ \mathrm{f}(x) = 5 + 2e^{-0.2x} $$ is concave upward on $$ (-\infty, \infty) $$ and has no inflection points.

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