Given info regarding $f$ show it is "$1-1$" and more 
If $f:\mathbb R\to\mathbb R$ one function with $f(\mathbb R)=\mathbb R$ which is differentiable and for it applies $f'(x)\neq0, \forall x\in \mathbb R$. Also, $C_f$ (the graph of $f$) passes through the points $A(1,2)$and $B(0,1).$
I) Show that $f$ is a "$1-1$" function.
II) Solve the equation: $f^{-1}(f(1+\ln x)-1)=0$.
III) Show that there is at least one point $M$ of $C_f$ such as that the tangent is perpendicular to $ε:x+y-1=0$.
IV) If $f'$ is continuous then find $\lim\limits_{x\to -\infty}{e^x \over f(x)}$.

Personal work:
I) Because $f$ is differentiable then it will be continuous.
$\forall x \in \mathbb R$ it applies $f'(x)\neq0 \Rightarrow f(x)\neq c$ because if $f'(x)=0$, then $f(x)=c.$
ΙΙ) Because $C_f$ passes through the points $A(1,2)$ and $B(0,1)$ then it will apply $$f(0)=1, f(1)=2$$
 A: Hints:


*

*Assume that there exist $x,y\in\mathbb{R}$ such that $f(x)=f(y)$ and use MVT.

*Can you use the two points given and the fact that $f$ is $1-1$ to make $f,f^{-1}$ "disappear" from the equation?

*Translate the phrase "perpendicular to $(\varepsilon):x+y-1=0$"; this means that we need a tangent line that has a slope $\lambda$ such that:
$$\lambda\cdot\lambda_{\varepsilon}=-1\Leftrightarrow\lambda(-1)=-1\Leftrightarrow\lambda=1.$$
Can you continue from here?

*This is a little bit tricky, but I'd prefer you to have solved the other three questions first to give you some hint, so when you're done with the rest, please inform - if, of course, you haven't solved it yourself, already.


As for your work:


*

*It isn't sufficient to show that $f$ is not constant to show that is is $1-1$. There are many (infinitely many) non-constant functions that are not $1-1$ e.g.
$$f(x)=\sin x, g(x)=x^2,h(x)=e^{-x^2},\dots$$

*You also need the fact that, since $f$ is $1-1$ it is invertible with its inverse, $f^{-1}$ satisfying the following equation:
$$f^{-1}(f(x))=x\ \forall x\in\mathbb{R}.$$

A: I) This is subtle.
"$f'(x) = 0; \forall x \in \mathbb R$" does mean that for any $x$ whatsoever then $f'(x) = 0$.  And as you know $f'(x) = 0$ means that "$f(x) = c; \forall x\in \mathbb R$ for some constant $c$".
!!!BUT!!   $f'(x) \ne 0; \forall x \in \mathbb R$" does **!!!NOT!!!* mean "It is not true that $f'(x) = 0$ for all $x \in \mathbb R$".  It means, "for any $x \in \mathbb R$ it will never be the case that $f'(x) = 0$.
So if, for instance $f(x) = 2x^2 + 3x$ then $f'(x) = 4x + 3$.  Then "it is not true that $f(x) = 0$ for all $x$" is a true statement.  $f'(-\frac 34) = 0$ and "for any $x$ it will never be the case that $f'(x) = 0$" is a false statement.
It is the later interpretation, that is meant.
So to prove I).
You need to prove that if $f(a) = f(b)$ then $a = b$.
Suppose not.  Suppose that $a < b$ and $f(a) = f(b)$.  THen by the Mean Value Theorem there exists a $c$ so that $a < c < b$ and $f'(c) = \frac {f(b) - f(a)}{b-a} = \frac {0}{b-a} = 0$.  But there is no $c$ so that $f'(c)=0$ so this is impossible.
So if $f$ is differentiable and $f'(x)$ is never equal to $0$ for any $x$ then $f$ is 1-1.
Now the point is that if $f$ is 1-1, then that means there exists a function $f^{-1}$ so that $f^{-1}(f(x)) = x$ for any value of $x$.
So
II)  $f^{-1}(whatever) = 0$ means that $f(0) = whatever$.  And $f(0)= 1$.
So $f^{-1} (f(1 + \ln x) -1 ) = 0$ means that 
$f(1+ \ln x) - 1 = f(0) = 1$ 
$f(1 + \ln x) = 2$
Now $f(1) =2$ and $f$ is 1-1.  So that means if $f(1+ \ln x) = 2 = f(1)$ then .... what ....
III) Okay, the slope of $x + y - 1 = 0$ is the slope of 
$y = -x + 1$ is $-1$.
So a line perpendicular to $y=-x + 1$ will have a slope of $-\frac 1{-1} = 1$.
So you must prove that there is some $x$ where $f'(x) = 1$.
Hint: $f(0)= 1$ and $f(1) = 2$.  If we connected those in a line the slope would be $1$.  Do we have a theorem  that says there is an $x$ so that $f'(x)$ will have the same slope as $(0, f(0))$ and $(1, f(1))$?
IV)  What do we know about the behavior of $e^x$ as $x \to -\infty$.  What does that say about $\lim_{x\to -\infty} \frac {e^x}{h(x)}$?  Will we ever have a problem?  If so, what about L'hopital?
