= Solution
{c}
For every real $t$, $f(t)-t=1/(1+e^t)>0$, so \b[there is no <fixed point>]. Direct <differentiation> gives
$$
f'(t)=1-\frac{e^t}{(1+e^t)^2}.
$$
Writing $u=e^t>0$, the inequality $(1+u)^2\geq4u$ shows
$$
\boxed{\frac34\leq f'(t)<1.}
$$
In particular $0<f'(t)<1$, but $f'(t)\to1$ as $t\to\pm\infty$. There is no common <contraction constant> smaller than one. Strict pointwise bounds on the <derivative> therefore do not suffice for the <contraction mapping theorem>.
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