Finding Equilibria

An equilibrium (or fixed point) of $\frac{dy}{dt} = f(y)$ is a value $y^*$ where the derivative is zero:

$f(y^*) = 0$

At an equilibrium, the system does not change. Equilibria are the long-term destinations (or starting points) of solutions.

Examples

System	Equilibria
$\frac{dy}{dt} = -y$	$y^* = 0$
$\frac{dy}{dt} = y(1 - y)$	$y^* = 0$ and $y^* = 1$
$\frac{dy}{dt} = \sin(y)$	$y^* = n\pi$ for all integers $n$

Numerical Root Finding: Bisection

To find equilibria numerically, we look for sign changes in $f$ over an interval. Between any two points where $f$ changes sign, there must be a zero (Intermediate Value Theorem).

For each sign change, we narrow down the root using bisection:

lo, hi = x0, x1
for _ in range(50):
    mid = (lo + hi) / 2
    if f(lo) * f(mid) <= 0:
        hi = mid
    else:
        lo = mid

50 bisection steps gives precision $\approx \text{interval} / 2^{50} \approx 10^{-15}$ .

Your Task

Implement find_equilibria(f, a, b) that finds all zeros of $f$ in $[a, b]$ using sign changes and bisection. Return a list of equilibrium values rounded to 6 decimal places.

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