= Solution
Put $x(t)=1+u(t)$. Retaining terms linear in $u$ gives
$$
u'(t)=\alpha\bigl[u(t)-2u(t-T)\bigr].
$$
The exponential ansatz $u=e^{\lambda t}$ gives the <characteristic equation of a delay differential equation>
$$
\lambda=\alpha\left(1-2e^{-\lambda T}\right).
$$
At $T=0$ its root is $\lambda=-\alpha$, and the remaining delay roots lie far in the left half-plane, so the equilibrium is stable for sufficiently small $T$.
A stability change can first occur when $\lambda=i\omega$. Separating real and imaginary parts gives
$$
\cos(\omega T)=\frac12,
\qquad
\omega=2\alpha\sin(\omega T).
$$
For the first positive crossing, $\omega T=\pi/3$ and $\omega=\sqrt3\alpha$. Thus
$$
\boxed{T_c=\frac{\pi}{3\sqrt3\alpha}.}
$$
All positive-frequency crossings have $\omega T=\pi/3+2\pi k$, so there is no earlier imaginary-axis crossing. Implicit differentiation of
$$
F(\lambda,T)=\lambda-\alpha+2\alpha e^{-\lambda T}=0
$$
gives, with $q=\alpha T_c$,
$$
\operatorname{Re}\frac{d\lambda}{dT}
=\frac{3\alpha^2}{(1-q)^2+3q^2}>0,
$$
so the conjugate pair crosses into the right half-plane as $T$ increases. Therefore the <stability threshold for delayed quadratic crowding> is
$$
\boxed{x=1\text{ is stable for }0\leq T<T_c\text{ and loses stability at }T=T_c.}
$$
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