Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real n

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2 months ago

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Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f(x) is a real n

umber.
N(t)=⎧⎩⎨⎪⎪⎪⎪f(t)25t+150200+80t2+0.05tfor0≤t<6for6≤t<8fort≥8

The number of fish in a pond at time t years is modeled by the function N defined above, where f is a continuous function such that f(0)=80.

(a) Find limt→∞N(t). Explain the meaning of limt→∞N(t) in the context of the problem.

(b) Is the function N continuous at t=8 ? Justify your answer.

(c) The function N is continuous at t=6 . Is there a time t , for 0≤t≤6 , at which N(t)=250 ? Justify your answer.

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-02-16T23:42:34+03:00

Answer:

Refer below

Step-by-step breakdown:

The function behaves as a piecewise function defined as:

$N(t)=\left \{ {{25t+150} \ 0 \leq t \leq 6 \atop {\frac{200+80t}{2+0.05t} \ t \geq 8}} \right.$

a)

We need to evaluate the limit of the function as t approaches infinity. This means determining the maximum number of fish present in the pond as time extends indefinitely.

We consider the second segment of the equation since t fits into that range, whereby t is infinite, clearly exceeding 8.

$\lim_{t \to \infty} \frac{200+80t}{2+0.05t} \\\lim_{n \to \infty} \frac{80t}{0.05t}\\ \lim_{n \to \infty} \frac{80}{0.05} =1600$

This indicates that the maximum fish population in this pond is 1600, regardless of the time.

b)

A function is considered continuous at a specific point if the limit and the function value at that point are the same.

The function value at t = 8, according to the second part of the equation, is:

$\frac{200+80t}{2+0.05t}\\\frac{200+80(8)}{2+0.05(8)}\\=350$

We observe that a value exists, and the limit approaches this as t nears 8.

<pthus>the function maintains continuity at t = 8

c)

We seek to determine if there exists a "time" during t from 0 to 6 when the fish count in the pond reaches 250. Substituting 250 into N(t) allows us to solve for t using the first portion of the piecewise function as shown below:

$N(t)=25t+150\\250=25t+150\\25t=250-150\\25t=100\\t=4$

The time is 4 years when the fish count in the pond becomes 250

</pthus>