= Solution
Extend the given functions continuously to zero by $w_1(0)=w_2(0)=0$. This is forced by their stated differentiability. Their derivatives at zero are both zero, since the corresponding difference quotients tend to zero.
For $x<0$, $w_2=w_2'=0$, and for $x>0$, $w_1=w_1'=0$. At zero all four quantities vanish. Thus the <Wronskian> is
$$
\boxed{W(w_1,w_2)(x)=w_1w_2'-w_1'w_2=0\quad\text{for all }x.}
$$
Nevertheless, if the constant <linear combination> $Aw_1+Bw_2$ vanishes identically, evaluation at $x=-1$ gives $-A=0$, and evaluation at $x=1$ gives $B=0$. Therefore \b[the two functions are <linearly independent>].
This illustrates why a <vanishing Wronskian does not imply linear dependence> for arbitrary <differentiable functions>. Here both functions are continuously differentiable, but neither has a second derivative at zero. The common regular differential equation hypothesis in the next part is absent.
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