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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: 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.

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 $t = 8$.

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 $N(t) = 250$.

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.