# Working out the interval in which the algae becomes extinct (how to get the interval)

I have a birth rate that is

$$b(p) = \frac{p^2}{p^2 + 3}$$

and a death rate that is

$$d(p) = \frac{p}{4}.$$

I therefore have a reproduction rate as $r = b - d$. In order for my algae to become extinct I need $b < d$. Rearranging and solving this gives me a formula which I get to be

$$p^2 - 4p + 3 = (p - 3)(p - 1) > 0.$$

Now I'm confused where the interval comes from. The answers say the interval is

$$p \in (0,1) \cup (3,+ \infty).$$

How do they get that? Because when I solve my polynomial, I get $p > 3$ and $p > 1$ and so I thought it is just when $p > 1$. How have they got this interval like that?

Am I right in saying that if I graph it, everything between $1$ and $3$ is in the negative bit of the $y$ axis and so that's not possible as that would imply a negative reproduction rate? And you can't have anything in the negative $x$ axis as that would be time (I think) and obviously you can't have negative time.

Is that correct?

If you did not eliminate one factor, you would have seen it immediately:

$$\dfrac{p^2}{p^2+3} < \dfrac{p}{4}$$

$$p(p-3)(p-1) > 0$$

This gives:

$0 < p < 1$

$p > 3$

• Again, would you notice this from the graph? I can't see it straight away by just looking at the values. Unless you say: The cubic is positive, so it starts by going up (through $p = 0$. There is a turning point then goes down through $p = 1$. Not it is in the $-y$ bit of the axis. It turns and goes back through $p = 3$ is positive again. – Kaish Mar 28 '13 at 19:36
• You can sketch the graph yes, or if you don't know how to, you can also put some test values. Say, you can use $p = -1$, $p = 0.5$, $p = 2$ and $p = 4$ to check where $p(p-3)(p-1)$ is indeed positive. – Jerry Mar 28 '13 at 19:51

$(p-3)(p-1)>0$ if and only if $p-3$ and $p-1$ are both non-zero and have the same sign. This happens when $p>3$ (both positive) or $p<1$ (both negative). We certainly can't have $p<0$ (that would mean a negative population), and if the population is $0$ then the algae isn't dying out, it's already dead! Hence, the algae is dying out when $0<p<1$ or $p>3$, meaning $p\in(0,1)\cup(3,+\infty).$

Also, I think you should have $r=b-d$. You can have a negative reproduction rate (in fact, that's what you want).

• Yeah sorry, that $r = d - b$ was a typo. – Kaish Mar 28 '13 at 19:30