Let's analyse the evolution of the curvature in the $\Lambda\text{CDM}$ model. If $\rho_R$, $\rho_M$, and $\rho_\Lambda$ are the densities of radiation, matter and dark energy, and
$$
\rho_c = \frac{3H^2}{8\pi G}
$$
is the critical density, then we can define
$$
\Omega_{R} = \frac{\rho_{R}}{\rho_{c}},\quad
\Omega_{M} = \frac{\rho_{M}}{\rho_{c}},\quad
\Omega_{\Lambda} = \frac{\rho_{\Lambda}}{\rho_{c}},
$$
and the quantity
$$
\Omega_{K} = 1 - \Omega_{R} - \Omega_{M} - \Omega_{\Lambda},
$$
which can serve as a measure of the curvature: if $\Omega_{K} = 0$ the universe is flat, if $\Omega_{K} < 0$ the curvature is positive, and if $\Omega_{K} > 0$ the curvature is negative. We can write these quantities in terms of their present-day values (indicated by subscripts "$0$") as follows:
$$
\rho_R = \rho_{R,0}\, a^{-4},\quad
\rho_M = \rho_{M,0}\, a^{-3},\quad
\rho_\Lambda = \rho_{\Lambda,0},
$$
where $a$ is the scale factor with present-day value $a=1$, so that
$$
\Omega_{R} = \Omega_{R,0}\frac{H_0^2}{H^2}a^{-4},\quad
\Omega_{M} = \Omega_{M,0}\frac{H_0^2}{H^2}a^{-3},\quad
\Omega_{\Lambda} = \Omega_{\Lambda,0}\frac{H_0^2}{H^2}.
$$
From the Friedmann equations, we also find (see this post for details) that
$$
H^2 = H_0^2\left(\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}\right),
$$
so that
$$
\Omega_{K}(a) = \frac{\Omega_{K,0}\,a^{-2}}{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}.
$$
From this we learn the following:
- If $\Omega_{K,0}=0$, then $\Omega_{K}\equiv 0$. That is, if the universe is exactly flat today, it has always been flat and always will be.
- As $a\rightarrow\infty$, the term $\Omega_{\Lambda,0}$ dominates, so that $\Omega_{K}\rightarrow 0$. In other words, if $\Omega_{K,0}\ne 0$, the curvature of the universe will go to zero in the future under the influence of dark energy.
- As $a\rightarrow 0$, the term $\Omega_{R,0}$ dominates, and again $\Omega_{K}\rightarrow 0$. So in the distant past, the curvature of the universe was also very close to zero; this is known as the flatness problem, and one of the motivations for the existence of an inflationary epoch.
Since $|\Omega_{K}|$ vanishes in the past and in the future, it must have had a maximum value at some intermediate time if it is nonzero today. This maximum occurs when the derivative of $\Omega_{K}(a)$ is zero. After some algebra, this reduces to solving
$$
2\,\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} - 2\,\Omega_{\Lambda,0} = 0.
$$
Incidentally, this is also the moment at which the expansion of the universe transitioned from deceleration to acceleration (i.e. when $\ddot{a}=0$, see the previous link for details). Using the values $\Omega_{R,0}\approx 0$, $\Omega_{M,0}\approx 0.3$ and
$\Omega_{\Lambda,0}\approx 0.7$, we find the solution
$$
a_m \approx \left(\frac{\Omega_{M,0}}{2\,\Omega_{\Lambda,0}}\right)^{1/3} \approx 0.6,
$$
and the corresponding curvature
$$
\Omega_{K,m} \approx \frac{\Omega_{K,0}\,a_m^{-2}}{\Omega_{M,0}\,a_m^{-3} + \Omega_{K,0}\,a_m^{-2} + \Omega_{\Lambda,0}} \approx \frac{\Omega_{K,0}}{\Omega_{K,0} + (3/2)\,\Omega_{M,0}\,a_m^{-1}} \approx \frac{\Omega_{K,0}}{\Omega_{K,0} + 0.75}.
$$
Observations indicate that the present-day curvature is
$$-0.02 < \Omega_{K,0} < 0.02,$$
so that the minimum/maximum curvature would have been
$$-0.027 < \Omega_{K,m} < 0.026.$$
In other words, the curvature of the universe has always been small.
