To examine the horizontal asymptotes for the functions $b) \ f(x) = 5 + x^2 e^x$ and $c) \ f(x) = x + e^{-2x+1}$, let's proceed:

(b) $f(x) = 5 + x^2 e^x$:

1. As $x \to \infty$:

   • $e^x \to \infty$ and $x^2 \to \infty$, so $x^2 e^x \to \infty$.
   • The function $f(x) \to \infty$.
   • Hence, no horizontal asymptote exists for $x \to \infty$.

2. As $x \to -\infty$:

   • $e^x \to 0^+$ (since $e^x$ approaches zero for $x \to -\infty$), and $x^2 > 0$, so $x^2 e^x \to 0$.
   • Therefore, $f(x) \to 5 + 0 = 5$.
   • Horizontal asymptote: $y = 5$ as $x \to -\infty$.

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

1. As $x \to \infty$:

   • In the term $e^{-2x+1}$, the exponent $-2x+1 \to -\infty$, so $e^{-2x+1} \to 0^+$.
   • The function $f(x) \approx x + 0 = x \to \infty$.
   • Hence, no horizontal asymptote exists for $x \to \infty$.

2. As $x \to -\infty$:

   • The exponent $-2x+1 \to \infty$, so $e^{-2x+1} \to \infty$.
   • The term $x \to -\infty$, but $e^{-2x+1}$ dominates. Therefore, $f(x) \to \infty$.
   • No horizontal asymptote exists for $x \to -\infty$.

Summary of Results:

• For $b) f(x) = 5 + x^2 e^x$, horizontal asymptote: $y = 5$ as $x \to -\infty$.
• For $c) f(x) = x + e^{-2x+1}$, no horizontal asymptote exists.

Would you like further explanations or visualizations for these results?

Related Questions:

1. What is the definition of a horizontal asymptote, and why does $f(x)$ approach it as $x \to \infty$ or $x \to -\infty$?
2. How can we determine horizontal asymptotes using limits algebraically?
3. Why does $e^x \to \infty$ dominate polynomial growth such as $x^2$?
4. Could $c) f(x) = x + e^{-2x+1}$ have other asymptotes (e.g., oblique)?
5. How does $e^{-2x+1}$ decay compared to other exponential terms like $e^{-x}$?

Tip: Always compute the limit of $f(x)$ as $x \to \pm\infty$ to identify horizontal asymptotes clearly.