I have a question in my homework that asks for me to find the steady states of $$\frac{dn}{dt}=n(n-a)(1-n)-fn$$ where $a<1$.
I already showed in a previous part of the question that there exist two nontrivial steady states if and only if $$f<\frac{(1-a)^2}{4}$$
Graphing $y=(n-a)(1-n)$ and $y=f$, the graph crosses the n-axis (x-axis) at $n=a$ and $n=1$. And we know that $n=0$ is another steady state. However, when plugging $n=1$ and $n=a$ into $\frac{dn}{dt}$ you do not get zero.
The inequality of $f$ is greatly throwing me off in finding the steady states.
Since this is a biological model, I will assume the initial condition satisfies $n(0) \ge 0$. Likewise, I will restrict myself to solutions where $n(t) \ge 0$ for all $t$. For ease of notation, write $f^*=(1-a)^2/4$, the value you have identified in your question.
A steady state occurs only when $dn/dt=0$. That means that $n=0$ or $$ (n-a)(1-n)=f. $$ So the trivial steady state $n(t)=0$ for all $t$ will always exist.
The left side of the equation above is a quadratic expression in $n$, as you have noted. Since $a < 1$ That expression is positive for $a < n < 1$. It has a maximum at $n = (1+a)/2$. This maximum value is $f^*$, as you have calculated.
If $f >f^*$, then $dn/dt < 0$ for all $n$, and the population will die out, no matter what the initial condition is.
If $f=f^*$, then $dn/dt = 0$ only when $n = (1+a)/2$. Thus if $n(0) > (1+a)/2$, the population will shrink to the unique steady state. And and if $n(0) < (1+a)/2$, the population will die out. If $n(0)=(1+a)/2$, then any negative perturbation will wipe out the population. Positive perturbations will result in a return to steady state.
If $ -a \le f < f^*$ and $f$, then there are two steady states with $ n > 0$. These occur where the parabola described by the above equation intersects the horizontal line with intercept $f$. Call these two steady states $n_1 < n_2$. The steady state at $n_1$ is not stable. If the initial condition $n(0) < n_1$, then the population dies off, and if $n(0) > n_1$, the population will grow to $n2_2$ That steady state is stable.
Finally, if $ f < -a$, then there's only one steady state. This is the one where the the parabola intersects horizontal line with intercept $f$ and $n > 0$. It is stable. Even the trivial steady state is unstable because any positive perturbation will result in a population explosion.