I recently searched around SE, and found:
The Attempt at a Solution
The answer is interesting.
"A function given that $$f(x)=f''(x)+f'(x)g(x)$$ could be an exponential function, sine, cosine , quadratic polynomial or $$f\equiv0$$. So we can say that the function is a continuous function $\in C^2$.
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
1. $$f(x_1)<0$$ (obviously contradiction)
2. $$f(x_1)=0$$ (it's impossible because this imply that f(x) = 0)
Analog for the other case"
But how does $$f''(x_1) > 0$$ show that $$f(x_1) < 0$$? I dont understand the contradiction for #1??