Question

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 number.
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.

Answer 1

Answer:

See Below

Step-by-step explanation:

The function is 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 find the limit of the function as t goes to infinity. This means what is the max value of fish in the pond given times goes to infinity (on an on).

We will take the 2nd part of the equation since t falls into that range, t is infinity, which is definitely greater than 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 means the maximum number of fish at this pond is 1600, no matter how long it goes on.

b)

A function is continuous at a point if we have the limit and the functional value at that point same.

Functional value at t = 8 is (we use 2nd part of equation):

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

We do have a value and limit also goes to this as t approaches 8.

So, function is continuous at t = 8

c)

We want to find is there a "time" when the number of fishes in the pond is 250, during t from 0 to 6. We plug in 250 into N(t) and try to find t. Make sure to use the 1st part of the piece-wise function. Shown below:

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

The time is 4 years when the number of fishes in the pond is 250