= Solution
Put $u=1+x$. Then $\dot x=-x(t-\tau)-x(t)x(t-\tau)$, so the <linearization> is $\dot x=-x(t-\tau)$. A mode $e^{\lambda t}$ gives the <characteristic equation of a delay differential equation>
$$
\boxed{\lambda=-e^{-\lambda\tau}.}
$$
For real $\lambda\geq0$, the left side is nonnegative and the right side negative. The two real graphs therefore never meet there, as drawn in the figure. \b[There is no nonoscillatory exponential instability.] If stability is lost, the unstable <eigenvalues> must be a nonreal conjugate pair, producing oscillatory growth; absence of a positive real root alone does not imply stability.
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