Solution
= Solution
Again put $x=1+u$. Then
$$
u'(t)=\alpha\bigl(1+u(t-T)\bigr)(-u(t))
=-\alpha u(t)-\alpha u(t-T)u(t).
$$
The linearized equation is therefore
$$
u'=-\alpha u,
$$
whose sole characteristic exponent is $-\alpha<0$ and is independent of $T$. By <delay-independent stability from an undelayed linearization>,
$$
\boxed{x(t)=1\text{ is locally asymptotically stable for every }T\geq0.}
$$