How to solve this derivative of f proof? A function $f$ satisfies:
$$f''(x) + f'(x)g(x) - f(x) = 0$$ for some function $g$. Prove that if $f$ is $0$ at two points, then $f$ is $0$ on the interval between them. 
Can someone verify my proof?
Scratchwork:
So let $I = [a, b]$ and $f(a) = f(b) = 0$. $g(x)$ is some function, doesn't matter. I will use the second derivative test idea: 
Proof
Suppose $f(x) > 0 \space \forall x \in (a, b)$
$$\because f(a) = f(b) = 0 \space \exists x_1 \in (a, b) \implies f'(x_1) = 0 \tag1$$
$$f''(x_1) + \overbrace{f'(x_1)g(x_1)}^{0} - f(x_1) = 0$$
$$ \therefore f''(x_1) = f(x_1) > 0$$
I just need help to reach a contradiction please?? 
 A: Since $f$ is continuous on the compact set $[a,b]$ it attains its maximum at some $x_1\in[a,b]$. We assume that $f(x_1)>0$. Then $x_1\in(a,b)$ and $x_1$ is a local maximum, which implies $f'(x_1)=0$ and therefore $f''(x_1)=f(x_1)>0$. But then $x_1$ is a strict (local) minimum, a contradiction. Thus, $f(x_1)\leq0$ and hence $f(x)\leq0$ for all $x\in[a,b]$. Now, repeat the proof with "minimum" instead of "maximum" to show that $f(x)\geq0$ for all $x\in[a,b]$.
A: The right negation is that $f(x)\ge0$ in $(a,b)$ and exist a point c | $f(c)>0$.
You have that $f''(x_1)\ge0$ (the function in that point is convex) so in that point you have a minima so there are two case


*

*$f(x_1)<0$ (obviously contradiction)

*$f(x_1)=0$ (it's impossible because this imply that $f(x)=0$  $ \forall x \in (a,b) $)


Analog for the other case
A: To make what you wrote more watertight:
Starting from your assumptions, we have that $f$ is continouous, hence there are $x_1,x_2\in[a,b]$ such that $f(x_1)=\inf_{x\in[a,b]}f(x)$ and $f(x_2)=\sup_{x\in[a,b]}f(x)$. 
If $x_1\in(a,b)$, then we have a local minimum, that is $f'(x_1)=0$ and $f''(x_1)\ge 0$. Then 
$$f(x_1)=f''(x_1)+f'(x_1)g(x_1)=f''(x_1)\ge 0.$$
Of course also $f(x_1)\ge 0$ if $x_1\in\{a,b\}$, thus definitely $$\inf_{x\in[a,b]}f(x)\ge0.$$
By the same argument with $x_2$, we obtain
$$\sup_{x\in[a,b]}f(x)\le0.$$
Hence for arbitrary $x\in[a,b]$
$$0\le\inf_{x\in[a,b]}f(x)\le f(x)\le \sup_{x\in[a,b]}f(x)\le 0.$$
