= Solution
Set $Y=U^{q/2}$ and $k=2b/q$. The proposed profile with $a,b>0$ gives
$$
U'=-kU(1-Y),\qquad U''=k^2U(1-Y)[1-(1+q/2)Y].
$$
Substituting into $U''+cU'+U(1-U^q)=0$ and dividing in the interior by $U(1-Y)$ yields
$$
k^2[1-(1+q/2)Y]-ck+1+Y=0.
$$
Equating the constant and $Y$ coefficients gives
$$
\boxed{k=\sqrt{\frac2{q+2}},\qquad b=\frac q{\sqrt{2(q+2)}},\qquad c=\frac{q+4}{\sqrt{2(q+2)}}.}
$$
Hence the <exact generalized Fisher front> is
$$
\boxed{U(z)=(1+ae^{qz/\sqrt{2(q+2)}})^{-2/q},\qquad a>0.}
$$
The free positive $a$ translates the front. It connects one on the left to zero on the right and is strictly decreasing. At $q=1$, its speed is $5/\sqrt6$.
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