Question

Suppose the population of cells is `N(t) = N_0(2)^(t/4)`, and `N_0 = 3`, where `t` is in hours, then how do you answer the following questions?

a) What is the effect of `N_0 = 3` on the graph of `N(t)`?

b) How many cells are there after 1 day?

c) What is the domain and range of this situation?

Answer 1

See below

Explanation:

a) Assuming `N_0=1`, the graph would look like this: graph{2^(1/2x) [-10, 10, -5, 5]}

b) `N(24)=N_0xx2^(24/4)=64N_0`

c) The equation would become `N(t)=N_0 2^(t/2)` and the graph would be more compressed.

Answer 2

a) If the initial popluation is `3`, the function becomes:

`N(t) = 3(2)^(t/4)`

Which has a graph

graph{y = 3(2)^(x/4 [-14.24, 14.23, -7.12, 7.11]}

b) We have `t = 24`, therefore:

`N(24) = 3(2)^6 = 192`

There will be `192` cells after `24` hours.

c) The equation would become `N(t) = 3(2)^(t/2)`, which is to say that if `t = a` in the initial graph and `t = a` in the second graph, `N` will be larger in the second graph because there will be more bacteria since they reproduce more frequently.

In summary, the graph will be steeper, increasing at a quicker rate.

Here's the new graph:

graph{3(2)^(x/2) [-14.24, 14.23, -7.12, 7.11]}

d) Since time will always be positive, the domain will be `t >= 0`. The range for `N(t) = 3(2)^(t/4)` will be `N >= 3`, since the colony starts with `3` members and increases from there without bounds.

Hopefully this helps!