Zero point when $f'(x)\gt c$ Suppose that the function $f:\mathbb R\to\mathbb R$ is continuously differentiable and that there is a positive number $c$ such that $f'(x)\ge c$ for all points $x$ in $\mathbb R$. Prove that there is exactly one number $x_0$ at which $f(x_0)=0$.
This question border me since it seems easy and obvious. I can prove the uniqueness of $x_0$ (since by mean value thm, if there are two zero points, there must be a point such that $f'(x)=0$). However, I cannot prove the existance of such $x_0$.
 A: We show first that if $x$ is positive, then $f(x)\ge f(0)+cx$. Let $g(x)=f(x)-f(0)-cx$. Then $g(0)=0$, and $g'(x)=f'(x)-c\ge 0$. So $g(x)$ is non-decreasing (Mean Value Theorem), which shows that $g(x)\ge 0$ for all positive $x$. 
It follows that there is a positive $x$ such that $f(x)\gt 1$. Indeed for any $M$ there is a positive $x$ such that $f(x)\gt M$. 
A similar argument shows that $f(x)\le f(0)+cx$ if $x$ is negative. It follows that there is an $x$ such that $f(x)\lt -1$. 
Thus by the Intermediate Value Theorem there is an $x_0$ such that $f(x_0)=0$. A small modification of the argument shows that for any $b$, there is an $x_b$ such that $f(x_b)=b$. 
A: If $x_1<x_2$, then $f(x_2) = f(x_1) + \int_{x_1}^{x_2} f'(t) dt \ge f(x_1) + c(x_2-x_1)$, hence $f$ is strictly increasing.
If $x>0$, $f(x)  \ge f(0) + cx$.
If $x<0$, $f(0) \ge f(x) + cx$.
Hence $f$ is strictly increasing (hence injective), $\lim_{x \to -\infty} f(x) = -\infty$, $\lim_{x \to +\infty} f(x) = +\infty$ (hence surjective, since continuous).
Since $f$ is a bijection, for any $y$ there exists a unique $x$ such that $f(x) = y$.
A: Alternative answer, using a contradiction argument rather than André's direct proof:
By the hypothesis, the function is increasing on $\mathbb R$. Suppose that $f(x)=0$ has no solution, hence $f$ is either positive either negative on $\mathbb R$ by continuity.
If it is negative, then we have that $\lim_{x\rightarrow+\infty} f(x) = l$, with $l\leq 0$. But then $\lim_{x\rightarrow +\infty} f'(x)=0$, which means by unfolding the definition of a limit that for each $\epsilon>0$, there exists some $x\in\mathbb R$ so that $f'(x)<\epsilon$, contradicting the hypothesis.
If $f$ is positive, a similar argument shows that $\lim_{x\rightarrow-\infty} f'(x)=0$, again contradicting the hypothesis on $f$.
