Example 4. For what value of $\mathrm{k}$ will the roots of the equation $(k+1) x^{2}$ $-2 k x+1=0$ both lie in the interval $(0,1)$?
Solving this problem essentially means that the parabola
$$
\begin{array}{l}
f(x)=x^{2}-\frac{2 k}{k+1} x+\frac{1}{k+1} \\
=\left(x-\frac{k}{k+1}\right)^{2}-\frac{k^{2}-k-1}{(k+1)^{2}}
\end{array}
$$

intersects the $x$-axis at two points that are both between $0$ and $1$ (as shown in Figure 3).
From Figure 3, it is not difficult to deduce:
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(0)>0, \\
00
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
k>0, \\
k \geqslant \frac{\sqrt{5}+1}{2} \text { or } k \leqslant \frac{1-\sqrt{5}}{2} \\
\frac{2-k}{k+1}>0
\end{array}\right. \\
\Rightarrow k \in\left[\frac{\sqrt{5}+1}{2}, 2\right] .
\end{array}
$$