Let f be twice differentiable on all of R. Is it true that for every ξ there exist x1 and x2 with f'(ξ) = (f(x2) − f(x1))/(x2 − x1)? Prove it or give a counterexample.

Step-by-step solution

1. 1

   Notice this is the converse of the mean value theorem, not the MVT itself. The MVT says every chord slope equals some derivative value. The statement here asks the reverse: that every derivative value is realised as a chord slope. Converses of true theorems need separate proof — or a counterexample.

2. 2

   Find the obstruction. By the fundamental theorem of calculus, every chord slope is an average of the derivative:

   $\frac{f(x_2)-f(x_1)}{x_2-x_1}=\frac{1}{x_2-x_1}\int_{x_1}^{x_2}f'(t)\,dt$

   An average of $f'$ over an interval of positive length can never exceed the supremum of $f'$, and it can equal that supremum only if $f'$ is constant on the interval. So if $f'$ attains a strict maximum at a single point $\xi$, that value $f'(\xi)$ can never be a chord slope.

3. 3

   Exhibit a counterexample. Take

   $f(x)=\arctan x,\qquad f'(x)=\frac{1}{1+x^{2}},\qquad f''(x)=\frac{-2x}{\left(1+x^{2}\right)^{2}}$

   This $f$ is twice (indeed infinitely) differentiable on all of $\mathbb{R}$, as required.

4. 4

   Show the value f′(0) = 1 is unattainable. For any $x_1<x_2$,

   $\frac{\arctan x_2-\arctan x_1}{x_2-x_1}=\frac{1}{x_2-x_1}\int_{x_1}^{x_2}\frac{dt}{1+t^{2}}<\frac{1}{x_2-x_1}\int_{x_1}^{x_2}1\,dt=1$

   The inequality is strict because $\tfrac{1}{1+t^{2}}<1$ for all $t\neq 0$, and a single point cannot make the integrals equal. So no chord slope ever reaches $1=f'(0)$.

   $\boxed{\text{The statement is FALSE; }f(x)=\arctan x\text{ with }\xi=0\text{ is a counterexample}}$

5. 5

   Check the numbers. The largest chord slopes for $\arctan$ come from short intervals straddling $0$: for $[-h,h]$ the slope is $\tfrac{2\arctan h}{2h}=\tfrac{\arctan h}{h}$, which equals $0.9967$ at $h=0.1$ and $0.99997$ at $h=0.01$ — approaching $1$ but never reaching it.

6. 6

   Note when the statement is true. If $f'$ attains its value at $\xi$ somewhere other than an isolated extremum — for instance if $f'$ is monotone, or if $f'(\xi)$ lies strictly between $\inf f'$ and $\sup f'$ — then the intermediate value theorem applied to the chord-slope function does produce suitable $x_1,x_2$. The failure is confined to values of $f'$ at strict global extrema.