Prove or disprove Average rate of change Prove or disprove
For any Real valued function $f$ whose domain is all real numbers and continuous on $\mathbb{R}$
there is no function $f$ exist satisfy the following condition:
If $[a,b]$ and $[c,d]$ are two distinict intervals then the average rate of change on $[a,b]$ which is defined as $\frac{f(b)-f(a)}{b-a} $ doesn't equal average rate of change on $[c,d]$. i.e. a function $f$ is continuous on $\mathbb{R}$ whose average rates of change are all distinct
 A: Intuition. If you draw the line connecting $(a,f(a))$ and $(b,f(b))$, you must be able to freely slide this line around without hitting the graph of $f$ at two other points, else you would have another interval with the same average rate of change. It turns out that because $f$ is continuous, it is impossible for this to occur.

Let $f$ be continuous with distinct rates of change on every interval. We will try to show this leads to a contradiction.
Let $a<b$.
Let $g(x) := f(x) - (\frac{f(b)-f(a)}{b-a}(x-a) + f(a))$ and note that $g(a)=g(b)=0$ and $g$ is also continuous.
Outline:

*

*Show that $g(x)\ne 0$ for any $x \in (a,b)$ (i.e. $g(x) \ne 0$ if $a<x<b$).

*Use the intermediate value theorem to deduce that either

*

*$g(x) > 0$ for all $x \in (a,b)$, or

*$g(x) < 0$ for all $x \in (a,b)$.



*Use the intermediate value theorem to show that there exist distinct points $c<d$ in $(a,b)$ such that $g(c)=g(d)$.

*Show that $\frac{f(d)-f(c)}{d-c} = \frac{f(b)-f(a)}{b-a}$, a contradiction.


Commentary: the bulk of the proof is essentially working with a special case where $f(a)=f(b)$ and the average rate of change is $0$ on this interval (steps 1-4 above). The transformation from $f$ to $g$ is used to reduce the general case to this special case; you may recognize this transformation as how one proves the mean value theorem using Rolle's theorem.
A: Take a special case where $f(a)=f(b)=0, a\lt b$. Now, if $f(x)=0$ for $x\in [a,b]$, the proof is trivial. Otherwise, there is $x\in(a,b)$ such that $f(x)\ne 0$, and WLOG suppose that $y=f(x)\gt 0$. Now, pick any $z\in(0,y)$. Per intermediate value theorem, there is $c\in(a,x)$ such that $f(c)=z$, and also there is $d\in(x,b)$ such that $f(d)=z$. Thus, $\frac{f(b)-f(a)}{b-a}=0=\frac{f(d)-f(c)}{d-c}$.
For the general case, for a given function $f$ and given two points $a\lt b$, take the function $g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$. As $g(a)=g(b)=0$, as above we can find $c,d,a\lt c\lt d\lt b$ such that $g(c)=g(d)$. However, this means that $f(c)-f(a)-\frac{f(b)-f(a)}{b-a}(c-a)=f(d)-f(a)-\frac{f(b)-f(a)}{b-a}(d-a)$, which is easily seen to be equivalent to $\frac{f(b)-f(a)}{b-a}=\frac{f(d)-f(c)}{d-c}$.
