# Proving a function $f : \mathbb{R} \rightarrow \mathbb{R}$ for which $f'(x) > 0$ has exactly one real root

Show a differentiable function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ for which there is $$c > 0$$ such that $$f'(x) \geq c$$ has exactly one real root

Here's my try -

First suppose there were two or more roots; choose any two of them and denote them by $$r_1$$, $$r_2$$. Then we would have $$f(r_1) = f(r_2) = 0$$. But by Rolle's Theorem, this would mean there exists a point $$p$$ in the interval $$(\min(r_1, r_2), \max(r_1,r_2))$$ such that $$f'(p) = 0$$, which contradicts our hypothesis. Hence, there are either no roots or one root.

Now suppose there are no roots. ... I don't know how to finish this part ...

• You say "hence there are either no roots or zero roots" but I don't understand a difference between these two things. Isn't a function with no roots the same as a function that has zero roots? – Clayton Apr 27 '19 at 17:44
• Yes I meant to say one root or no roots – user666614 Apr 27 '19 at 18:22

Unfortunately, this isn't true unless you have some more assumptions on $$f(x)$$. As a counterexample, $$f(x) = e^{x}$$ satisfies $$f'(x) = e^{x} > 0$$ for every $$x$$, but $$f(x)$$ as no real roots.
• As an addendum to this answer, the statement becomes true if $|f'(x)|\geq\varepsilon>0$ for some $\varepsilon>0$. – Clayton Apr 27 '19 at 17:45
For $$x\ge 0$$ we have $$\displaystyle f(x)=f(0)+\int_0^x f'(t)\mathop{dt}\ge f(0)+cx\to+\infty$$
For $$x<0$$ we have $$\displaystyle f(x)=f(0)-\int_x^0 f'(t)\mathop{dt}\le f(0)-c|x|\to-\infty$$