= Solution
For $\lambda>1$, the new fixed point in $[0,1]$ is
$$
x^*=\sqrt{1-\frac1\lambda}.
$$
The origin has multiplier $F_x(0,\lambda)=\lambda>1$ and is therefore unstable. At the new fixed point,
$$
F_x(x^*,\lambda)
=\lambda\left(1-3\left(1-\frac1\lambda\right)\right)
=3-2\lambda.
$$
Hence
$$
|F_x(x^*,\lambda)|<1
\quad\Longleftrightarrow\quad
1<\lambda<2.
$$
The point $x^*$ is stable in that interval and unstable for $\lambda>2$. At $\lambda=2$ its multiplier is $-1$, the threshold for a <period-doubling bifurcation>.
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