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

Question

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?

Answer 1

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:

$0 < p < 1$

$p > 3$

Answer 2

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