# 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 $f$ ?

Suppose that there are two fixed points $a<b$. By the mean value theorem, there is some $c\in (a,b)$ such that $$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}=\frac{b-a}{b-a}=1$$ contrary to the hypothesis on $f^{\prime}$. So there is at most one fixed point.
Note that a derivative has the intermediate value property. Therefore if $f'(x)$ is never $1$, then $f'(x)>1$ for all $x$ or $f'(x)<1$ for all $x$. In the first case, the function $g(x)=f(x)-x$ is increasing, so it can have at most one root; in the second case it is decreasing and the same applies.
• The intermediate value property of derivatives (Darboux's theorem) is not actually needed here. The IVT and the MVT together are enough. (See my posted answer for the reason.) $\qquad$ – Michael Hardy Apr 3 '16 at 19:27
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 $1$.