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I am dealing with (below) and r > 0

$$

N_{*} = \frac{rN_*}{(1 + aN_*)^b}

$$

So the steady states are $N_* = 0$ and $N_* = \frac{\sqrt

$$

N_{*} = \frac{rN_*}{(1 + aN_*)^b}

$$

So the steady states are $N_* = 0$ and $N_* = \frac{\sqrt

**{r} - 1}{a}$.**

Let $f(x) = \frac{rx}{(1 + ax)^b}$.

Then $f'(x) = (ax + 1)^{-b}\left(\frac{br}{ax + 1} + (1 - b)r\right)$.

Evaluating the derivative at $N_*$, we obtain

$$

f'(0) = r \quad \text{and} \quad

f'\left(\frac{\sqrtLet $f(x) = \frac{rx}{(1 + ax)^b}$.

Then $f'(x) = (ax + 1)^{-b}\left(\frac{br}{ax + 1} + (1 - b)r\right)$.

Evaluating the derivative at $N_*$, we obtain

$$

f'(0) = r \quad \text{and} \quad

f'\left(\frac{\sqrt

**{r} - 1}{a}\right) = \frac{r}{a}\left(\frac{b}{\sqrt****{a}} - b + 1\right).**

$$

For $N_* = 0$, $0 < f'(0) < 1$ for $0 < r < 1$.

For this range of $r$, $N_* = 0$ is stable monotonically.

For $r > 1$, $f'(0) > 1$ which is unstable monotonically, and at $r = 1$, we have a tangent bifurcation.

Now when I examine $N_* = \frac{\sqrt$$

For $N_* = 0$, $0 < f'(0) < 1$ for $0 < r < 1$.

For this range of $r$, $N_* = 0$ is stable monotonically.

For $r > 1$, $f'(0) > 1$ which is unstable monotonically, and at $r = 1$, we have a tangent bifurcation.

Now when I examine $N_* = \frac{\sqrt

**{r} - 1}{a}$, I need to find when $f'(N_*) = 1$ for the tangent bifurcation and when $f'(N_*) = -1$ for the pitchfork.**

However, I have this if the f'(N_*) = 1

$$

r = \frac{a}{\frac{b}{\sqrtHowever, I have this if the f'(N_*) = 1

$$

r = \frac{a}{\frac{b}{\sqrt

**{a}}-b+1}**

$$

So would this be the correct way to do this or does it need to be done be considering a and b?

For $N_* = \frac{\sqrt$$

So would this be the correct way to do this or does it need to be done be considering a and b?

For $N_* = \frac{\sqrt

**{r} - 1}{a}$, $0 < f'\left(\frac{\sqrt****{r} - 1}{a}\right) < 1$ for $0 < r < \frac{a}{\frac{b}{\sqrt****{a}} - b + 1}$ which is stable monotonically.**

If $r > \frac{a}{\frac{b}{\sqrtIf $r > \frac{a}{\frac{b}{\sqrt

**{a}} - b + 1}$, then $f'\left(\frac{\sqrt****{r} - 1}{a}\right) > 1$ which is unstable monotonically, and at $r = \frac{a}{\frac{b}{\sqrt****{a}} - b + 1}$, we have a tangent bifurcation.**

For $\frac{-a}{\frac{b}{\sqrtFor $\frac{-a}{\frac{b}{\sqrt

**{a}} - b + 1} < r < 0$, $-1 < f'\left(\frac{\sqrt****{r} - 1}{a}\right) < 0$ so we have a stable oscillations.**

When $r < \frac{-a}{\frac{b}{\sqrtWhen $r < \frac{-a}{\frac{b}{\sqrt

**{a}} - b + 1}$, there are unstable oscillations, and at $r = \frac{-a}{\frac{b}{\sqrt****{a}} - b + 1}$, we have a pitchfork bifurcation.**

I could I say this to contend for a and b:

This stability analysis is valid for all $a,b$ such that $r > 0$.I could I say this to contend for a and b:

This stability analysis is valid for all $a,b$ such that $r > 0$.

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