Does the existence of a superstable fixed point imply a $-\infty$ Lyapunov exponent?

Question

This is the Lyapunov exponent as a function of $r$ for the logistic map ($x_{n+1}=f(x_n)=r(x_{n}-x_{n}^2)$)

The big dips are centered around points where $f'(x)=0$ for some $x$ in the trajectory used to calculate the exponent $\lambda$, and therefore for those values $\lambda=-\infty$. These are precisely the $r$ values where there exists a superstable fixed point/cycle, which makes sense as $f'(x^*)=0$ is a necessary condition for superstability (where $x^*$ is some fixed point). This naturally led me to think that perhaps the existence of a superstable fixed point/cycle is a sufficient condition for any 1D map to have $\lambda = -\infty$

However, this is now a Lyapunov graph for $x_{n+1}=g(x_n)=r(x_{n}^2-x_{n}^3)$

For which $0$ is a superstable fixed point for all values of $r$. For $r<4$ we indeed see that $\lambda = -\infty$ but for some values of $r$ there is a real $\lambda$, for some even positive.

So what's the difference? Are these superstable fixed points different somehow? Also, the bluriness of the line in the second image results from most of the $r$ values in between still being $-\infty$ - why is this so?

Notes:

I used the standard algorithm to generate the $\lambda$ values. Namely, for each value of $r$ I draw uniformly some $x_0\in [0, 1]$, and calculate $\lambda = \left(\frac{1}{n}\sum_{i=1}^{n}\ln\left|f'\left(x_{i}\right)\right|\right)$ for $n=10^4$

The maximal $r$ value, $6.75$, was chosen to ensure the interval $[0,1]$ maps onto itself, as in the logistic map.

Answer 1

As Evgeny helpfully points out, the second system in the question exhibits bistability for $r>4$.

If we plot the attractor of each $x_0\in [0,1]$ (by iterating $g$ for 1000 iterations to get rid of the transients and then plotting the next 100 points), for $r<4$ this looks like (Note: The equation in the title of these figures is wrong, didn't notice at the time. Should be $r(x^2-x^3)$)

So 0 is a [super]stable fixed point, and its basin is all of $[0,1]$

However, for $r$ values greater than 4 and less than ~6.2 (where the Lyapunov exponents are still negative), a typical attraction map looks like So some (a little less than half) of the starting points are attracted to 0, but if we have $x_0$ in the middle of the range (approx. 0.25 to 0.95), the trajectory gets attracted to a 2-cycle.

Finally, for $r$ values greater than ~5.8, there appears a strange attractor, and hence the Lyapunov exponent is positive for some starting points $x_0$

Bonus: The full orbit diagram for the map