Let $f,g: [a,b] → R$, two differential functions and suppose $f(a) = f(b) = 0$.
If $W(f,g): [a,b] → R$ and $W(f,g)(x) = f(x)g'(x) - g(x)f'(x)$ doesn't equal 0 for all $x$ in $[a,b]$, show that a $c$ in $[a,b]$ must exist such that $g(c) = 0$.
I thought about using the Rolle Theorem to say that there exists a $k$ in [a,b] such that $f'(k) = 0$ and then use the Intermediate Value Theorem to show that g(a) and g(b) are the opposite signs therefore a c such that g(c)=0 must exist, but I get lost at that last part.
Can anyone help me? Thank you!
If possible, take $g\neq 0 $ for all $x\in [a,b] $
Then we can easily define the differentiable function $\frac{f}{g} $ on $[a,b]$.
Clearly, $(\frac{f}{g})(a) = 0 = (\frac{f}{g})(b) $.
Then by Rolle's theorem, there exist at least one $c\in (a,b) $ such that $(\frac{f}{g})' (c) = \frac{gf'-fg'}{g^2}=\frac{-W(f,g)}{g^2} = 0 \implies W(f,g)=0 $ , and this is a contradiction .
So, there exist $c\in (a,b) $ such that $g(c)=0$
Edit: Here, we shouldn't use the closed interval $[a,b]$, for this statement, $\text{there exist $c\in (a,b) $ such that g(c)=0 } $.
Since at $a,b$, if either of any $g(a)=0$ or $g(b)=0$ true, Then , $W(f,g)$ becomes $0$, which makes the fact " $f,g$ linearly independent " wrong.
So, we use $(a,b)$ instead of $[a,b]$, for the conclusion of the proof.