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> 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 $(0,1)$. 
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**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?