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### Given Information:
- The number of fish in a pond at time \( t \) years is modeled by the function \( N(t) \).
- The function \( N(t) \) is piecewise defined as follows:
  \[
  N(t) = \begin{cases} 
  f(t) & 0 \leq t < 6, \\
  25t + 150 & 6 \leq t < 8, \\
  \frac{200 + 80t}{2 + 0.05t} & t \geq 8.
  \end{cases}
  \]
- Additionally, \( f \) is a continuous function with \( f(0) = 80 \).

### Part (a): 
**Find \( \lim_{t \to \infty} N(t) \). Explain the meaning of this limit in the context of the problem.**

#### Solution:
For large \( t \) (as \( t \to \infty \)), we are concerned with the third piece of the piecewise function:
\[
N(t) = \frac{200 + 80t}{2 + 0.05t}, \quad t \geq 8.
\]
To find \( \lim_{t \to \infty} N(t) \), we divide the numerator and denominator by \( t \):
\[
N(t) = \frac{200 + 80t}{2 + 0.05t} = \frac{80t(1 + \frac{200}{80t})}{t(0.05 + \frac{2}{t})} = \frac{80(1 + \frac{200}{80t})}{0.05 + \frac{2}{t}}.
\]
As \( t \to \infty \), the terms involving \( \frac{1}{t} \) go to 0. Therefore, we have:
\[
\lim_{t \to \infty} N(t) = \frac{80}{0.05} = 1600.
\]
**Meaning of the limit**: The limit represents the long-term population of fish in the pond. As time progresses indefinitely, the number of fish stabilizes around 1600 fish.

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### Part (b):
**Is the function \( N(t) \) continuous at \( t = 8 \)? Justify your answer.**

#### Solution:
To check the continuity of \( N(t) \) at \( t = 8 \), we need to verify the following conditions:
1. \( \lim_{t \to 8^-} N(t) \) exists.
2. \( \lim_{t \to 8^+} N(t) \) exists.
3. \( \lim_{t \to 8^-} N(t) = \lim_{t \to 8^+} N(t) = N(8) \).

- For \( t = 8^- \), \( N(t) = 25t + 150 \). So:
  \[
  \lim_{t \to 8^-} N(t) = 25(8) + 150 = 350.
  \]
- For \( t = 8^+ \), \( N(t) = \frac{200 + 80t}{2 + 0.05t} \). Substituting \( t = 8 \):
  \[
  N(8) = \frac{200 + 80(8)}{2 + 0.05(8)} = \frac{840}{2.4} = 350.
  \]

Since both limits exist and are equal, and \( N(8) = 350 \), the function \( N(t) \) is continuous at \( t = 8 \).

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### Part (c):
**The function \( N(t) \) is continuous at \( t = 6 \). Is there a time \( t \), for \( 0 \leq t \leq 6 \), at which \( N(t) = 250 \)? Justify your answer.**

#### Solution:
For \( 0 \leq t < 6 \), the function \( N(t) = f(t) \) is continuous, but we do not know the exact form of \( f(t) \). However, by the Intermediate Value Theorem (IVT), if \( f(t) \) is continuous and takes on values at both ends of the interval, we can determine whether \( N(t) = 250 \).

Since \( f(0) = 80 \) and \( N(6) = 25(6) + 150 = 300 \), and \( 250 \) lies between 80 and 300, by the IVT, there must be some \( t \in [0, 6] \) such that \( N(t) = 250 \).

Thus, the answer is **yes**, there is a time \( t \) where \( N(t) = 250 \).

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Would you like more detailed explanations or have any questions about these steps? Here are five related questions to explore further:

1. How would the limit change in part (a) if the coefficients in the third part of \( N(t) \) were different?
2. Can you determine the exact form of \( f(t) \) from the given data?
3. Why is the Intermediate Value Theorem essential in part (c)?
4. How would the continuity at \( t = 6 \) be affected if the function for \( t = 6 \) was slightly altered?
5. Could there be discontinuities for \( t < 6 \) in this problem?

**Tip:** Always check the continuity at transition points in a piecewise function by evaluating the limits from both sides and the function's actual value at those points.

The number of fish in a pond at time t years is modeled by the function N(t), defined as a piecewise function with three different expressions for different intervals of t. The question involves finding limits, checking continuity, and applying the Intermediate Value Theorem.

Explore a piecewise function modeling fish population in a pond. Learn to calculate limits as t approaches infinity, check continuity at specific points, and apply the Intermediate Value Theorem to determine specific values. Perfect for AP Calculus students tackling limits and continuity.

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