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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 $f$ ?

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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.

On the other hand, as you observed in the question there is at least one fixed point by Brouwer's fixed point theorem, so the correct answer is (B). Strictly speaking, (A) is correct as well, but I suppose (B) is the better answer.

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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.

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  • $\begingroup$ 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.) $\qquad$ $\endgroup$ – Michael Hardy Apr 3 '16 at 19:27
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    $\begingroup$ @MichaelHardy That is a better approach than mine indeed. Still, mine works. $\endgroup$ – Ian Apr 3 '16 at 19:29
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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 $1$.

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