A real valued function $f$ is defined on the interval $(-1,2)$. A point $x_0$ is said to be a fixed point of $f$ if $f(x_0)=x_0$. Suppose that $f$ is a differentiable function such that $f(0)>0$ and $f(1)=1$. Show that if $f'(1)>1$, then $f$ has a fixed point in the interval $(0,1)$.
My approach:
Let $h(x)=f(x)-x$. Now since $f$ is continuous on $[0,1]\implies h$ is continuous on $[0,1]$ and since $f$ is differentiable on $(0,1)\implies h$ is differentiable on $(0,1)$.
Now $h(0)=f(0)-0=f(0)>0$ and $h(1)=f(1)-1=0$.
Also $h'(x)=f'(x)-1$ $\forall x\in (0,1)$.
This implies that $h'(1)=f'(1)-1>0.$
Therefore by applying MVT on the function $h$ on the interval $[0,1]$ we can conclude that $\exists c\in(0,1)$ such that $$h'(c)=\frac{h(1)-h(0)}{1-0}=-h(0)<0.$$
Therefore we have $h'(c)<0$ and $h'(1)>0$. Therefore by applying IVT on the interval $[c,1]$, we can conclude that $\exists c_1\in (0,1)$ such that $h'(c_1)=0$.
How to proceed after this?
Since $f'(1) = L > 1$, we get that $\lim_{x \to 1^-} \frac{1-f(x)}{1-x} = L> 1$, because this is the left-sided derivative.
Thank you @Franklin for asking for a more detailed answer. What follows is precisely that.
Let us use the $\epsilon-\delta$ definition of limits to prove the following statement :
To prove this, we will use the definition of the left sided limit $\lim_{x \to 1^-}$ in the expression $\lim_{x \to 1^-} \frac{1-f(x)}{1-x} = L > 1$. Let us pick $\epsilon = \frac{L-1}{2}$.
By definition (of the left hand limit), there exists a $\delta>0$ such that for all $\delta > (1-x) > 0$, we have $$\left|\frac{1-f(x)}{1-x} - L\right|< \epsilon = \frac{L-1}{2}$$
Using the triangle inequality, for every $x$ such that $\delta > (1-x) > 0$, $$ \frac{1-f(x)}{1-x} > L - \frac{L-1}{2} = \frac{L+1}{2} > 1 $$ as $L>1$. Now, if $x_0$ is any point such that $\delta > 1-x_0>0$, then from above, it follows that $$\frac{1-f(x_0)}{1-x_0} > 1 \implies f(x_0)< x_0$$
where the implication follow by simple algebraic rearrangement.
We are in position to apply the intermediate value theorem. A common trick in proving fixed-point theorems (at the rudimentary level) is to use the intermediate value theorem with the function $g(x) = f(x)- x$, since $g(x)=0$ if and only if $f(x)=x$.
Keeping this heuristic in mind, we let $g(x) = f(x)-x$. Then, by choice, $f(x_0)<x_0$ implies that $g(x_0)<0$, and $f(0)>0$ implies that $g(0)>0$. In particular, we have found an interval $[0,x_0]$ such that :
$g(0)>0$.
$g(x_0)<0$.
$g$ is continuous on $[0,x_0]$, since $f$ is differentiable , hence continuous on $[0,x_0]$, the function $h(x)=x$ is continuous on $[0,x_0]$, and $g = f-h$ is the difference of two continuous functions on $[0,x_0]$.
Thus, we can apply the intermediate value theorem to the function $g$ on $[0,x_0]$. This gives us a number $0<y<x_0$ such that $g(y)=0$. By definition of $g$, $f(y)=y$. Finally, since $y \neq 0$ (because $f(0) > 0$) and $x_0<1$, it follows that $y \in (0,x_0) \subset (0,1)$.
That is, we have proved that $y \in (0,1)$ is a fixed point of $f$, as desired.