0
$\begingroup$

Context

In the previous question, we were asked to prove Rolle's Theorem: If f is continuous on (a,b) and f (a) = f (b) = 0 then f (c) = 0 for a certain c in (a,b).

Question

In the case of I = [0, 1], give an example of a function fitting the description of part (a), non-constant on any sub-interval of [0,1], such that f '(c) = 0 for infinitely many c in (0,1)

My first guess of a function that fit the description was f(x) = sin (1/x), but that doesn't work because it isn't defined on [0,1], not continuous, f(0) =/=0 and f(1) =/= 0

Any help would be appreciated. Thank you!

$\endgroup$

2 Answers 2

1
$\begingroup$

At least for $x \not = 0$, you know $-1 \le \sin\left(\frac1x\right) \le 1$, and that it is differentiable, and that it changes sign infinitely often in $(0,1)$

So find some non-constant differentiable function which is always non-negative on $[0,1]$ with $g(0)=g(1)=0$, such as $g(x)=x(1-x)$

Now multiply these together

So $$f(x)=x(1-x)\sin\left(\frac1x\right)$$ for $x \not = 0$. To deal with continuity, define $f(0)=\lim_{x\to 0} f(x) = 0$. This will now meet the conditions as it too is differentiable in $(0,1)$ and changes sign infitiely often.

It looks like this red curve, bounded by $\pm g(x)$ in grey enter image description here

$\endgroup$
1
  • $\begingroup$ Thank you! That makes sense and helps a lot $\endgroup$
    – Kelly
    May 6, 2018 at 23:07
0
$\begingroup$

Take $I$ and divide it in two subintervals of length $1/2$, take the fist of those subintervals and fit a parabola passing through the points $(0,0)$ and $(1/2,0)$. Now take the other subinterval of lenght $1/2$, divide it into two subintervals of lenght $1/4$, take the first of those and fit another parabola passing trough the pints $(1/2,0)$ and $(3/4,0)$. Iterate this procedure infinitely many times and you will get the desired function.

If you need an explicit formulae you can write it like this: $$f(x)=(x- \frac{1}{4})^2- \frac{1}{4} \quad \mbox{if} \ x\in [0,\frac{1}{2}]$$ $$f(x)=\Big[x-\Big(\sum_{i=1}^n \frac{1}{2^i}+\frac{1}{2^{n+2}}\Big)\Big]^2-\Big(\frac{1}{2^{n+2}}\Big)^2 \quad \mbox{if} \ x \in \Big[\sum_{i=1}^n \frac{1}{2^i},\sum_{i=1}^{n+1}\frac{1}{2^i}\Big] \quad \forall n \geq 1$$ $$f(x)=0 \quad \mbox{if} \ x=1$$

$\endgroup$

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.