N(t) = 25t + 150 for 0< =t< 6, (200 + 80t)/(2 + 0.05t) for t> =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→[infinity]n(t). explain the meaning of limt→[infinity]n(t) in the context of the problem.

1600. This is a maximum limit of fish the pond can host.

Step-by-step explanation:

For , we must take the part of this function valid for high values, i.e., the second part, where .

Since we have two polynoms both in numerator and denominator, and both of them are of degree 1 (both linear), for high values of , the main part of each polynom shall be the linear part, neglecting lower degree parts (in this case, constant terms):

This means that the number of fish in a pond has an horizontal asymthote. In other words, there seems to be a natural limit for the number of fish that there will be in the pond as years pass. The maximum number of fish is actually 1600. With this function, no higher than this figure can be reached. This might imply limits in productivity.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!