# Change in the number of positive zeros of a continuous function

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.

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 $$0$$.
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.