Let's solve 14 b and 14 c step by step.

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Problem 14 b:

Function: $f(x) = 5 + x^2 e^{-x}$

Step 1: Behavior as $x \to \infty$:

• The term $e^{-x} \to 0$ as $x \to \infty$, since the exponential decay dominates.
• Thus, $x^2 e^{-x} \to 0$, because $x^2$ grows slower than $e^x$.
• Therefore: $f(x) \to 5 \quad \text{as } x \to \infty.$

Step 2: Behavior as $x \to -\infty$:

• As $x \to -\infty$, $e^{-x} \to \infty$ (exponential growth).
• Thus, $x^2 e^{-x} \to \infty$, since $x^2$ is always positive and grows large.
• Therefore: $f(x) \to \infty \quad \text{as } x \to -\infty.$

Horizontal Asymptote:

For $x \to \infty$, the horizontal asymptote is: $y = 5.$

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Problem 14 c:

Function: $f(x) = x + e^{-2x+1}$

Step 1: Behavior as $x \to \infty$:

• The term $e^{-2x+1} = e^{-2x} \cdot e^1 \to 0$, since $e^{-2x} \to 0$ as $x \to \infty$.
• Thus, $f(x) \to x + 0 = x$.
• $f(x) \to \infty \quad \text{as } x \to \infty$.

Step 2: Behavior as $x \to -\infty$:

• As $x \to -\infty$, $e^{-2x+1} \to \infty$, since $-2x \to \infty$.
• Therefore: $f(x) \to -\infty \quad \text{as } x \to -\infty.$

Horizontal Asymptote:

There is no horizontal asymptote, as the function does not settle to a constant value in either direction.

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Final Results:

• b) $y = 5$ is the horizontal asymptote as $x \to \infty$.
• c) No horizontal asymptote.

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Would you like a deeper explanation or visualization of these results?

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Follow-Up Questions:

1. What is the graphical behavior of $x^2 e^{-x}$ near zero and infinity?
2. Why does $e^{-x}$ dominate polynomial terms as $x \to \infty$?
3. How do transformations like $-2x+1$ in exponentials affect their growth or decay?
4. Can a function have different horizontal asymptotes in different directions?
5. What happens if a logarithmic term replaces $e^{-x}$ in similar functions?

Tip: For exponential functions, always focus on how the exponential term grows or decays—it usually determines the behavior of the entire function.