How many fixed points are there for $f:[0,4]\to [1,3]$ Let , $f:[0,4]\to [1,3]$ be a differentiable function such that $f'(x)\not=1$ for all $x\in [0,4]$. Then which is correct ?
(A) $f$ has at most one fixed point.
(B) $f$ has unique fixed point.
(C) $f$ has more than one fixed point.
Here, $f:[0,4]\to [1,3]\subset [0,4]$ is continuous and $[0,4]$ is compact convex. So by Brouwer's fixed point theorem $f$ has a fixed point. But I am unable to use the condition $f'(x)\not=1$ and how I can conclude that how many fixed points are there for $f$ ?

You don't need anything nearly as strong as Brouwer's fixed-point theorem in order to show existence of at least one fixed point.  Notice that $f(0) - 0 \ge 1 > 0$ and $f(4) - 4 \le -1 < 0$, so $f(x) - x=0$ for some $x\in(0,4)$ by the intermediate value theorem.
Now the question is whether there could be more than one fixed point, say $f(x_1)=x_1$ and $f(x_2)=x_2$.  The mean value theorem then tells us that
$$ 1 = \frac{f(x_1) - f(x_2)}{x_1-x_2} = f'(c) $$
for some $c$ between $x_1$ and $x_2$.  But it was given that $f'(c)$ cannot be equal to $1$.
