= Solution
Let $x^*(\lambda)=\sqrt{1-1/\lambda}$ and center the map at this parameter-dependent fixed point:
$$
G(X,\mu)=F(x^*(2+\mu)+X,2+\mu)-x^*(2+\mu).
$$
A <Taylor polynomial> in $X$, with the coefficients expanded at $(x,\lambda)=(1/\sqrt2,2)$, gives
$$
G(X,\mu)=-X+\alpha\mu X+\beta X^2+\gamma X^3+O(\mu^2)
$$
under the period-doubling scaling $X=O(\sqrt\mu)$. The coefficients are
$$
\alpha=\left.\frac{d}{d\lambda}F_x(x^*(\lambda),\lambda)\right|_{\lambda=2}=-2,
$$
$$
\beta=\frac12F_{xx}\left(\frac1{\sqrt2},2\right)=-3\sqrt2,
\qquad
\gamma=\frac16F_{xxx}\left(\frac1{\sqrt2},2\right)=-2.
$$
Here we used $F_x(x^*(\lambda),\lambda)=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.
Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ gives
$$
G(G(X,\mu),\mu)-X
=-2\alpha\mu X-2(\gamma+\beta^2)X^3
+o(\mu X+X^3).
$$
Besides the fixed solution $X=0$, the two leading solutions therefore satisfy
$$
X^2=-\frac{\alpha\mu}{\gamma+\beta^2}.
$$
Thus the period-two points born at $\lambda=2$ are
$$
\boxed{x_\pm=x^*\pm
\sqrt{\frac{-\alpha\mu}{\gamma+\beta^2}}}
$$
to leading order. With the coefficients above, the radicand is positive for $\mu>0$.
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