Does the derivative of a differentiable function attain its maximum and minimum?

Solution 1:

Edit: there are indeed differentiable functions with bounded derivative that still fail to attain extrema.

For example

$$ f(x) := \begin{cases} x^2(x-1) \sin (1 /x) & \text{ if } 0<x\leq 1\\ 0 & \text{ if } x= 0 \end{cases} $$ has derivative

$$ f'(x) := \begin{cases} (3x^2-2x) \sin (1 /x) - (x-1)\cos(1/x)& \text{ if } 0<x\leq 1\\ 0 & \text{ if } x= 0 \end{cases}. $$

As $x\searrow 0$ the maxima and minima tend to $\pm 1$ but can never attain it. See graph below ($f$ on the left; $f'$ on the right). And you can play around with the exponent and the other term to create different envelopes, of course.

Original: adapted from Pugh's Real Mathematical Analysis, p.157: Consider $$ f(x) := \begin{cases} x^{3 /2} \sin (1 /x) & \text{ if } 0<x\leq 1\\ 0 & \text{ if } x= 0 \end{cases} $$ whose derivative is $0$ at $x=0$ and $$ f'(x) = \frac{3}{2}\sqrt{x} \sin (1 /x) - \frac{1}{\sqrt{x}}\cos (1 /x) \qquad \text{for } x \neq 0. $$ That $1 /\sqrt{x}$ blows up.

Solution 2:

No, it's not true. For instance, define $f:[-1,1]\to\Bbb R$ by $$f(x)=\begin{cases}x^2\sin\frac{1}{x^2} & \text{ if }x\ne 0\\ 0 & \text{ if }x=0\end{cases}$$ This function is differentiable on the whole of $[-1,1]$, but its derivative away from $0$ is $$2x\sin\frac{1}{x^2}-\frac{2}{x}\cos\frac{1}{x^2}$$ which is unbounded above and below.

A related question (and perhaps the one you wanted to ask) is: if the derivative is bounded, does it necessarily attain those bounds? It seems that Skorpion's answer says no.

Solution 3:

"I think this is intuitively true since we can always find a ..."

More generally: there exists a bounded derivative on $[0,1]$ that not only has no extrema, it has no relative extrema.

Dangerous words "intuitively true" but, alas, we ---as either mathematics students or mathematicians--- are frequently led astray by our intuition. Derivatives are especially tricky in that regard. For example does your intuition tell you that a differentiable function could be nowhere monotone? (i.e., not monotone in any interval).

So to elaborate on the excellent constructions you have seen in the other answers let me push your intuition to its limits:

(i) If $f:[a,b]\to \mathbb R$ is differentiable, then it is possible that $f'$ is unbounded in most subintervals. There will have to be a dense set of intervals on which it is bounded. [In one example given in another answer there was only one point of unboundedness.]

(ii) If $f:[a,b]\to \mathbb R$ is differentiable with $f'$ bounded then there may not be a maximum or a minimum [as you see in another answer].

(iii) ...are you ready? There exist differentiable functions $f:[a,b]\to \mathbb R$ with $f'$ bounded that do not have any relative extrema whatsoever! [The examples you see here have lots of relative extrema.]

I learned (iii) in a paper by my friend Richard O'Malley [Approximate maxima. Fund. Math. 94 (1977), no. 1, 75–81.]

"A.M. Bruckner has communicated to this author an example of a bounded approximately continuous function which has no relative extrema."

Naturally I expected Andy would include that construction in his well-known treatise on derivatives [Bruckner, Andrew, Differentiation of real functions. Second edition. CRM Monograph Series, 5. American Mathematical Society, Providence, RI, 1994.]

Can't find it there. Neither can Andy. He did send me a sketch of the proof. If you are really interested in derivatives then this is likely something you want to know. In the meantime think about a bounded derivative that has no max, no min, no relative max, no relative min. Intuitive?