Let $f(x)$ be any continuous function, then is it true that $$Z^{+}\left(\alpha f(x)+(x+\beta)f'(x)\right)\leq Z^{+}\left(f'(x)\right)$$ 


where $\alpha>1$ is a real number and $\beta$ is any positive integer. $Z^+$ denotes the number of positive zeros.


Note: If $f(x)$ is a polynomial then it's easy to see that the above inequality holds.


Any help or small hint will be really appreciated. Thanks.



Change in the number of positive zeros of a continuous function
No. Let $a:=\alpha$ and $b:=\beta$. If e.g. $f(x)=(x+b)^{-a}$, then $a f(x) + (x + b) f'(x)=0$ for all real $x\ge0$, so that the left-hand side (lhs) of your inequality is infinite, whereas the right-hand side (rhs) is $0$.




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If you now insist that the lhs be a finite number, we can modify the above example as follows: let $a=2$, $b=1$, and
$$f(x):=\frac{1 + (x - x^3/6)/10}{(x + 1)^2}$$
for real $x\ge0$. Then
$$a f(x) + (x + b) f'(x)=\frac{2-x^2}{20 (x+1)},$$
so that the lhs is $1$, whereas
$$f'(x)=-\frac{x^3+3 x^2+6 x+114}{60 (x+1)^3},$$
so that the rhs is $0$, less than the lhs.


