Show that $f$ does not change sign on some interval $(\beta,+\infty)$. 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?
 A: 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 $x>\beta\implies f(x)=0.$ 
A: You choose an arbitrary $\beta$ and then
assume that $f$ changes sign at least once in the interval
$(\beta, +\infty)$, hoping to derive a contradiction.
But in fact you can easily find an example of functions $a$, $b$, and $f$ that meet all the criteria of your theorem, such that $f$ changes sign once
in the interval $(\alpha, +\infty)$;
and you can then modify these functions to make that change of sign
occur as far "to the right" (as far from $\alpha$) as you like.
Therefore, if all you do is choose a $\beta$ and assume that
$f$ changes sign somewhere in $(\beta, +\infty)$,
how do you know that you aren't just looking at one of those
examples of $f$ that change sign once in $(\alpha, +\infty)$,
with the change in sign occurring so far to the right that it is
also to the right of your chosen $\beta$?
Let's re-examine the theorem's conclusion.
It allows $f$ to change sign, but
it says that if $f$ does change sign,
if you go out far enough to the right,
eventually you will pass the last sign change.
After that sign change, there will be no others.
One way to see why a theorem must be so is to attempt to construct a
counterexample and see what goes wrong.
I tried to make such a counterexample; no trouble finding $a$ and $b$
that had constant signs and satisfied $f' = af + b$ in the neighborhood
of a sign change in $f$;
but when I tried to extend this to the right, through a few sign changes
of $f$ (remember, the counterexample to the theorem must have
infinitely many of these), I quickly ran into a roadblock.
Try it and see.
If you can figure out why this does not work, you should be able to
prove the theorem.
As long as you can demonstrate an upper bound on the number of sign changes
of $f$, you can always place $\beta$ past the last sign change,
and then $f$ does not change sign on the interval $(\beta, +\infty)$.
So is there a maximum number of times $f$ can change sign? If so, how
many times is that?
