Prove without using mean value theorem Let $a$ and $b$ be the fixed points of a differentiable function $f$, meaning that $f(a)=a$ and $f(b)=b$. If it is given that $f'(x)$ is never equal to 1, then prove that such a function does not exist.
Using the mean value theorem, on $a$ and $b$, we have for some $c$:
$$f'(c)=\frac{f(a)-f(b)}{a-b}=\frac{a-b}{a-b}=1$$
Which is clearly a contradiction. It is a simple application of the mean value theorem. However, I want to see where its provable without using the MVT (or the equivalent Rolle's Theorem), more from the definition of derivative. But I am not able to see any way out to.
If you ask for the motivation, I think it is interesting to prove simple applications of the MVT from something more elementary.
 A: Assuming you mean that $f'(x)$ is never $1$, the only thing I can think of is essentially to reprove Rolle and the MVT (that is, the MVT is pretty elementary):
Let $g(x)=f(x)-x$. Then $g(a)=0$ and $g(b)=0$. If $g(x)$ is not identically $0$, then it must attain a maximum or a minimum at some point $c$ between $a$ and $b$. Then $f'(c)-1=g'(c)=0$, but we said that $f'(x)$ is never $1$.

Here is another approach using integrals, but I don't think it is more elementary than the MVT (and it requires $f$ to be continuously differentiable):
If $g'(x)\gt0$ for all $x$ between $a$ and $b$, then $g(b)-g(a)=\int_a^bg'(x)\,\mathrm{d}x\gt0$. Likewise for $g'(x)\lt0$ for all $x$ between $a$ and $b$. If $g'(d)\gt0$ and $g'(e)\lt0$ for some $d$ and $e$ between $a$ and $b$, then the Intermediate Value Theorem says that $g'(c)=0$ for some $c$ between $d$ and $e$.
We might be able to lift the requirement that $f$ be continuously differentiable, but I think we need to use the MVT to show what kinds of discontinuities a derivative can have.
