Property of Antiderivatives

Question

Consider two differentiable functions $f:\mathbb R \rightarrow \mathbb R$ and $g:\mathbb R \rightarrow \mathbb R$. Suppose that there exists a $\delta$ such that for $x>\delta$, $f'(x)>g'(x)$. I am trying to show that this implies there exists a $\delta_2$ such that for $x>\delta_2$, $f(x)>g(x)$. So far, I have used $f(x)=\int_{\delta}^x f'(t)dt + f(\delta)$ and similarly for $g(x)$ to show that this obviously holds if $f(\delta)>g(\delta)$. I am now trying to construct the case where $f(\delta)<g(\delta)$.

Answer 1

Consider $f(x) = -2e^{-x},$ and $g(x) = 1-e^{-x}$

Then $f(x) < g(x)$ for all $x$ and $f'(x) > g'(x)$ for all $x.$

Answer 2

This is more of a long comment than an answer, but I wanted to make two points

1. Expanding on @RyszardSzwarc comment if it wasn't clear: let $f(x)=\arctan(x)-\pi/2$ and $g(x)=0$. Then $f'(x)=\frac{1}{1+x^2}$ is strictly greater than $g'(x)=0$ for all $x$, but $f(x)<g(x)$ for all $x$.
2. Suppose now that $f'(x)>g'(x)$ for all $x$ and $f(0)=g(0)$. Then $f(x)>g(x)$ for all $x>0$ since \begin{align} f(x)&=f(x)-0\\ &=f(x)-g(x)\\ &=f(0)+\int_0^x f'(t)dt-\left(g(0)+\int_0^x g'(t) dt\right) \\ &=f(0)-g(0)+\int_0^x\left(f'(t)-g'(t)\right)dt\\ &\int_0^x\left(f'(t)-g'(t)\right)dt. \end{align} Then you need the surprisingly non-trivial fact that the integral of a positive function is positive. It is true that integrable functions are continuous almost everywhere, and once you know $f'(x)-g'(x)$ is continuous at some point, it's relatively easy to show $\int_0^x\left(f'(t)-g'(t)\right)>0$ for all $x>0$.