Do Wronskians have the intermediate value property? I wonder if the following is true:

Conjecture: Let $I \subset \Bbb R$ be an open interval and $f, g: I \to \Bbb R$ be differentiable functions. Then the Wronskian
$$
W(f,g) =\begin{vmatrix}f &g \\f' & g'\end{vmatrix} = f g' - f'g
$$
is a Darboux function.

A Darboux function is a real-valued function $f$ which has the “intermediate value property”: for any two values $a$ and $b$ in the domain of $f$, and any $y$ between $f(a)$ and $f(b)$, there is some $c$ between $a$ and $b$ with $f(c) = y$.
Motivation and thoughts:

*

*In Does there exists two differentiable functions $f, g$ on $I$ such that $W(f, g) (x) >0$ on $A$ and $W(f, g) <0$ on $I\setminus A$? it what proved that

If the Wronskian takes both positive and negative values on an interval, then it must be zero somewhere.

and my conjecture would be a natural generalization.
However, I do not yet see how the case of an arbitrary intermediate value $y$ can be reduced to the special case of $y = 0$ as the intermediate value.


*The conjecture is (trivially) true if both $f$ and $g$ are continuously differentiable, since then $W(f, g)$ is continuous. So the interesting case is that $f$ and $g$ are just assumed to be differentiable.


*Derivatives have the Darboux property, that covers the case that $f$ or $g$ is constant, e.g. $W(1, g) = g'$.


*Sums and products of Darboux functions are not necessarily Darboux functions (see for example The sum of Darboux is a Darboux function?). So even if all terms in $f g' - f'g$ have the intermediate value property, there is no immediate way to conclude the conjecture.
A (failed) proof attempt:
Assume that $w = W(f, g)$ does not take a value $y \in \Bbb R$, and consider the sets
$$
A = \{ x \in I \mid w(x) > y \} \, , \, B = \{ x \in I \mid w(x) < y \} \, .
$$
If we can show that both $A$ and $B$ are open then one of them must be empty (since $I$ is connected), and we are done.
If $f(x_0) \ne 0$ then we can define $h(x) = y \int_{x_0}^x f(t)^{-2} dt$ in a neighborhood of $x_0$, and
$$
 W(f, g) -y = f^2 \left( \frac gf - h\right)'
$$
shows that $W(f, g) -y$ does not change its sign near $x_0$, so that $x_0$ is an interior point of $A$ or of $B$.
A similar argument works if $g(x_0) \ne 0$. However, other than in my previous answer, one can not exclude the case $f(x_0) = g(x_0) = 0$. That is where my I am stuck in my current proof attempt.
 A: Wronskian of two differentiable functions may not have Darboux property. Below the conformation due to Józef Banaś; Wagdy Gomaa El-Sayed
Let $I=[0, 1]$  . Consider two functions $f, g $ defined by
$f(x) =\begin{cases}x^2 \sin(\frac{1}{x^4})&x\neq 0\\0 & \text{otherwise}\end{cases}$
$g(x) =\begin{cases}x^2 \cos(\frac{1}{x^4})&x\neq 0\\0 & \text{otherwise}\end{cases}$
For $x=0$ , $W(f, g) (0) =0$
For $x\neq 0$ ,
$\begin{align}W(f, g) (x) &=\begin{vmatrix} \sin(\frac{1}{x^4}) & \cos(\frac{1}{x^4})\\2x\sin(\frac{1}{x^4})-\frac{4}{x^3} \cos(\frac{1}{x^4})&2x\cos(\frac{1}{x^4})-\frac{4}{x^3} \sin(\frac{1}{x^4})
  \end{vmatrix}\\&=\frac{4}{x}\end{align}$
Hence $x\to W(f, g) (x) $ doesn't have the Darboux property.

$\color{red}{\textbf{  Theorem}}:$

Let $f, g: I \to\Bbb{R}$  be functions differentiable on the interval
$I$.  Assume that the set $ Z_f $(or $Z_g$) has no accumulation
points, and $Z_f \cap \overline{ Z_g} = \emptyset$   (or $
 \overline{Z_f} \cap{Z_g} = \emptyset$). Then $W(f,g)$ has the Darboux
property on $I$.

Source : BANAS, J.—EL-SAYED, W. G.: Darboux property of the Wronski determinant, Math. Slovaca $45 (1995), 57–61.$  (Online available here.)
