Q. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(x)+2 f(x)>0$ for all $x \in \mathbb{R}$, where $f^{\prime}$ denotes the derivative of $f$ and $\mathbb{R}$ denotes the set of all real numbers. Then which one of the following statements is true?
A. $f(x)>0$, for all $x>0$ and $f(x)<0$, for all $x<0$
B. $f(x)<0$, for all $x \neq 0$
C. $f(x)>0$, for all $x \neq 0$
D. $f(x)<0$, for all $x>0$ and $f(x)>0$, for all $x<0$
It seems to me that option $A$ is true by taking the derivative of the function $g(x)=e^{2x}f(x)$, But I am not able to take counter examples for other options. Can we suggest some function $f(x)$ satisying the hypothesis.
Best Answer
From the inequality, we know that $f'(0)>0$ hence $f$ is strictly increasing at $0$. In particular, it must change sign and statements B and C are false. Also since it is increasing and $f(0)=0$, statement D is also false.
Suppose $f$ vanishes at $x_0 \ne 0$, if it does not change its sign, then $f'(x_0)+2f(x_0)=0$, hence a contradiction. Thus, if $f$ vanishes at such $x_0 \ne 0$, it must change sign.
Since $f(0)=0$ and strictly increasing, if $f$ changes sign there is a first positive $x_0$ where it happens and $f'(x_0)<0$. So $f'(x_0)+2f(x_0)<0$, contradiction. From the same argument, let $x_0$ the greatest negative root of $f$, then $f'(x_0)<0$ and $f'(x_0)+2f(x_0)<0$, also a contradiction.
As a conclusion, $f$ must vanish at $0$ only and is strictly increasing at this point. So statement A is true.