0

Suppose f is a differentiable function on an interval (a,b) and that f′ takes on both positive and negative values on (a,b). Prove that f′ must take on the value 0 as well.

According to a hint, the idea is to show that there exist a minimum or a maximum between the points f(x) and f(y) which are the points whose derivative takes on positive and negative values. Assume WLOG that $f'(x)>0$ and $f'(y)<0$ and $a<x<y<b$. Now the issue I am running into is that I can't say that there is a neighbour of f(x) that it is increasing since it is possible that f(x) can have a positive derivative at a point without it increasing in a neighbour of that point. How can I proceed?

david h
  • 201
  • 1
    https://math.stackexchange.com/q/54843. This is https://en.wikipedia.org/wiki/Darboux%27s_theorem_(analysis), and the Wikipedia page has also a proof. – Martin R Mar 24 '22 at 16:08
  • 1
    The result is true, but not trivial... There are differentiable functions whose derivative is not continuous, so the intermediate value theorem (for continuous functions) cannot be, in general, applied to this situation. – PierreCarre Mar 24 '22 at 16:12
  • @PierreCarre how is the statement true? Isn’t $f(x)=|x|$ a counter-example which proves the statement false? – Radial Arm Saw Mar 24 '22 at 16:15
  • @RadialArmSaw $|x|$ is not differentiable. – PierreCarre Mar 24 '22 at 16:17
  • In regards to your penultimate sentence. It is a highly unintuitive fact that there exist differentiable functions that are nowhere monotone (i.e., there is no subinterval whatsoever on which they are monotonic). – B. S. Thomson Mar 24 '22 at 17:17
  • @PierreCarre oh- I forgot. Sorry – Radial Arm Saw Mar 24 '22 at 17:56

1 Answers1

1

$f$ is continuous on the compact interval $[x,y]$, hence assumes maximum value in this interval. If the maximum is at $x$, then $$ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^+}\frac{\,\overbrace{f(x+h)-f(x)}^{\le 0}\,}{\underbrace{h}_{>0}}\le 0,$$ contradicting the assumption $f'(x)>0$. If the maximum is at $y$, then $$ f'(y)=\lim_{h\to0}\frac{f(y+h)-f(y)}{h}=\lim_{h\to 0^-}\frac{\,\overbrace{f(y+h)-f(y)}^{\le 0}\,}{\underbrace{h}_{<0}}\ge 0,$$ contradicting the assumption $f'(y)>0$. Hence the maximum is an an interior point $z$ of the interval $[x,y]$, and hence $f'(z)=0$.

Hagen von Eitzen
  • 382,479
  • The question is a duplicate, an identical proof has been given before: https://math.stackexchange.com/a/192440/42969. – Martin R Mar 24 '22 at 16:23
  • @Hagen von Eitzen Hi. the extreme value theorem tells us that f assumes a minimum or a maximum on [x,y] but we don't know whether it is a maximum or a minimum. Now from basic calculus, we intuitively know that the extreme value in this specific case should be a maxima, but are we able to prove that here? – david h Mar 24 '22 at 18:00