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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 ...

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    $\begingroup$ 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? $\endgroup$ – Clayton Apr 27 '19 at 17:44
  • $\begingroup$ Yes I meant to say one root or no roots $\endgroup$ – user666614 Apr 27 '19 at 18:22
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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.

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  • $\begingroup$ As an addendum to this answer, the statement becomes true if $|f'(x)|\geq\varepsilon>0$ for some $\varepsilon>0$. $\endgroup$ – Clayton Apr 27 '19 at 17:45
  • $\begingroup$ Hi, I updated my original post with more assumptions. Sorry about that. $\endgroup$ – user666614 Apr 27 '19 at 18:22
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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 $\displaystyle f(x)=f(0)-\int_x^0 f'(t)\mathop{dt}\le f(0)-c|x|\to-\infty$

And you can conclude.

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