TITLE: Show that $f$ does not change sign on some interval $(\beta,+\infty)$.
QUESTION [6 upvotes]: Let $a,b,f$ are continuous functions on some interval $(\alpha,+\infty)$ such that $a,b$ have constant sign on $(\alpha,+\infty)$ and $f$ is differentiable on $(\alpha,+\infty)$. Suppose $f'=af+b$. Show that $f$ does not change sign on some interval $(\beta,+\infty)$.
So I consider this as cases:
Case I: $a>0,b>0$
So I can rewrite $f=(f'-b)/a$. To the contrary assume that given $\beta \in \mathbb{R}$ there is $c_1,c_2>\beta $ such that $f(c_1)>0$ and $f(c_2)<0$. Then by intermediate value theorem there is $c\in(c_1,c_2)$ such that $f(c)=0$. So,$f'(c)=b(c)>0$. But after that I was stuck. Even a periodic function like sin function does have the property I got. So how do I prove the result?

REPLY [2 votes]: There exists a function $A$ with $A'=a$ because $a$ is continuous. Then $$(e^{-A} f)'=e^{-A}b.$$   So the function $g(x)= e^{-A(x)}f(x)$ is  monotonic. Thus,  either (1): $ \exists \beta\;( x>\beta\implies g(x)\ne 0)$, and hence $x>\beta \implies f(x)\ne 0$,  which implies that $f$ cannot change sign on $(\beta,\infty)$ because $f$ is continuous; or (2): $\exists \beta \;(x>\beta \implies g(x)=0)$ and hence $x>\beta\implies f(x)=0.$