3
$\begingroup$

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 ...

$\endgroup$
  • 1
    $\begingroup$ 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 at 17:44
  • $\begingroup$ Yes I meant to say one root or no roots $\endgroup$ – effunna9 Apr 27 at 18:22
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ As an addendum to this answer, the statement becomes true if $|f'(x)|\geq\varepsilon>0$ for some $\varepsilon>0$. $\endgroup$ – Clayton Apr 27 at 17:45
  • $\begingroup$ Hi, I updated my original post with more assumptions. Sorry about that. $\endgroup$ – effunna9 Apr 27 at 18:22
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.