2

Say that $f$ is differentiable and that the derivative of $f$ does not equal $1$ on $(-\infty, \infty)$. Show that there is at most one real number a such that $f(a)=a$.

In order to solve this I am required to use the mean value theorem.

I understand that this will be true when $f(x)=x$, or when $f(x)-x = 0$. Thus if I could show that $f(x) -x !=0$ at any point this would be proven.

Let $g(x) = f(x)-x$

$\dfrac{\mathrm dg}{\mathrm dx} = \dfrac{\mathrm df(x)}{\mathrm dx}$, which does not equal one

$\dfrac{\mathrm df(x)}{\mathrm dx} \neq \dfrac{f(b)-f(a)}{b-a}$, by the mean value theorem

thus:

$b-a\neq f(b)-f(a)$

This is as far as I can get. I know I'm really close, and any help would be greatly appreciated.

grg
  • 1,017
slehmann
  • 25

1 Answers1

6

Suppose to the contrary that there exist numbers $a$ and $b$, with $a\lt b$, such that $f(a)=a$ and $f(b)=b$.

By the Mean Value Theorem, there is a $c$ between $a$ and $b$ such that $$\frac{f(b)-f(a)}{b-a}=f'(c).$$ But $f(a)=a$ and $f(b)=b$. It follows that $$\frac{f(b)-f(a)}{b-a}=\frac{b-a}{b-a}=1.$$ So $f'(c)=1$, contradicting the fact that $f'(x)$ is nowhere equal to $1$.

André Nicolas
  • 507,029
  • It seems like you don't need to include $g$ here. – Cameron Williams Nov 24 '15 at 04:24
  • @Cameron Williams: It may be better to use $g$. We get $g'(c)=0$ for some $c$. Since $g'(x)=f(x)-1$, this shows that $f'(c)=1$, contradicting the fact that $f'(x)$ is never $1$. The $g(x)$ was a residue, I changed my mind in the middle and forgot to delete the let $g(x)=$. Thanks! – André Nicolas Nov 24 '15 at 04:27
  • Yeah. I think it's six one way, half a dozen of the other. I would have gone the $g$ route personally, but this is cleaner in some sense. – Cameron Williams Nov 24 '15 at 04:28
  • Thank you so much, that makes complete sense. – slehmann Nov 24 '15 at 04:29
  • @slehmann: While in this case we didn't need $g$, it is often an excellent approach for many other problems to isolate 'discrepancies' in this manner. We know the 'boundary' case is $f(x) = x$ so we see how $f$ differs from this. It is especially useful when the 'boundary' case is not so simple as here. – user21820 Nov 24 '15 at 09:58