Prove if $f'(x)\geq 1$ then $\exists c$ such that $f(c)=0$. Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable on $\mathbb{R}$ If $f'(x)\geq 1$ for all $x\in \mathbb{R}$, then there exists a $c\in \mathbb{R}$ such that $f(c)=0$. 
I realised that since $f'(x)\geq 1$, $f$ is monotonously increasing, ie: $x\leq y\Rightarrow f(x)\leq f(y)$, and I also realised if $f(x)>0$ for all $x$ or $f(x)<0$ for all $x$ then by IVT there wouldn't be $c$ such that $f(c)=0$. But I dunno where to go from there, nor do i know how to put it into a reasonable proof. Help?
 A: Let $y=f(x) > 0$. By mean value theorem, we then have that there exists $c \in (x-y,x)$ such that
$$f'(c) = \dfrac{f(x)-f(x-y)}{x-(x-y)} \geq 1 \implies f(x) - f(x-y) \geq y \implies f(x-y) \leq f(x)-y = 0$$
Hence, by intermediate value theorem, there exists $z \in [x-y,x]$ such that $f(z) = 0$.
Argue similarly for $y = f(x) < 0$.
A: If there is no point where $f(x)$ vanishes then because of intermediate value property $f(x)$ maintains a constant sign for all $x$.
We will start with the case when $f(x) > 0$ for all $x$. We are given that $f'(x) \geq 1 > 0$ so that $f(x)$ is increasing and hence it follows that $\lim_{x \to -\infty}f(x) = L$ exists and $L \geq 0$. If $x < 0$ then $$f(x) - f(2x) = -xf'(\xi)\tag{1}$$ for some $\xi \in (2x, x)$. If we take limit as $x \to -\infty$ of above equation we see a simple contradiction. The LHS tends to $L - L = 0$, but since $f'(x) \geq 1$ for all $x$ the RHS tends to $\infty$. It follows that we can't have $f(x) > 0$ for all $x$. Thus there will be some point when $f(x) \leq 0$ and we will thus have a change of sign and $f(x)$ would vanish.
Same way we can argue if we assume $f(x) < 0$ for all $x$. In this case however we need to consider limits when $x \to \infty$.
From the above line of argument it is also clear that the result holds even if we replace the condition $f'(x) \geq 1$ with $|f'(x)| \geq k > 0$ for some fixed $k$.
In other words if an everywhere differentiable function maintains a constant sign then its derivative must take values arbitrarily close to $0$.
