# Derivative of positive, continuously differentiable function is positive close to zero

Consider a continuously differentiable function $$f: [0, 1] \to \mathbf{R}_{\geq 0}$$, such that $$f(0)=0, f'(0)=0$$ and $$f(x)>0$$ for $$x>0$$.

I want to prove that there exists $$\bar{x}>0$$ such that $$f(x)$$ is increasing on $$[0, \bar{x}]$$. (or, even better, strictly increasing)

This question is nearly identical to this Derivative of positive function is positive close to zero However, apart from continuity, I have additionally continuous derivatives and, because of that, cannot make the counterexample proposed there to work. The answer to that question is (paraphrasing): 'for a function oscillating near 0 between $$x^2$$ and $$x^4$$ we have $$f(0)=0$$, $$f(x)>0$$ for $$x>0$$ and the function is continuous'. However, I cannot construct such a function that would not violate continuity of the derivative. For example, consider: $$f(x)= (\sin(1/x)+1)x^2+(\cos(1/x)+1)x^4.$$ The derivative of this function oscillates between $$0$$ and $$1$$ and thus is not continuous (whilst all the other conditions are met).

In short: is the continuity of the derivative sufficient to make the original statement true, or does there exist a counterexample with continuous derivatives?

for $$f(x) = x^4 \left(\sin(\frac{1}{x})\right)^2$$ you get $$f^\prime(x)= 4x^3 \left(\sin(\frac{1}{x})\right)^2 - 2x^2 \sin(\frac{1}{x})\cos(\frac{1}{x}) = 2x^2\sin(\frac{1}{x})\left(2x\sin(\frac{1}{x})- \cos\frac{1}{x} \right)$$
This can be clearly continuously extended to $$x=0$$ , and if $$x$$ becomes small behaves like $$-2x^3\sin(\frac{1}{x}) \cos(\frac{1}{x})$$ from which it is not difficult to see that $$f^\prime$$ will have arbitrary many points where it is $$<0$$.
Now this function is not positive but only $$\ge0$$, but by adding, say $$x^{10}$$, you get a positive function (for $$x>0$$) with similar properties for $$f^\prime$$.