Fixed point of $f:[0, 4] \to [1,3]$ such that $f'(x) \neq 1$

Question

Let $ f:[0, 4]\to [1,3]$ be a differentiable function such that $f'(x)\neq 1$ for all $ x \in [0,4]$ Then the function $f$ has

1. Atmost one fixed point
2. Unique fixed point
3. No fixed point
4. More than one fixed point

As far I guess the function is a contraction mapping cause the range is smaller than the domain. Hence has unique fixed point. But I am unable to prove and unable to give more logical description. Is my thinking totally wrong? Thanks in advance.

Answer 1

First note that $f(x)-x$ is positive at $x=0$ and negative at $x=3$ and hence, by the intermediate value property it has at least a fixed point.

Next, by Darboux Theorem, $f'$ satisfies the Intermediate Value Property. Since $f'(x) \neq 1$ for all $x \in [0,4]$ you have either

• $f'(x) <1$ for all $x \in [0,4]$ or
• $f'(x) >1$ for all $x \in [0,4]$

Show that the second is not possible, and hence $f'(x) <1$ for all $x \in [0,4]$.

Now, you can prove that the fixed point is unique. Indeed, if $a,b$ are fixed points, apply the Mean Value Theorem on $[a,b]$.