Exercise 5.3.7

Question

A fixed point of a function $f$ is a value $x$ where $f(x)=x$. Show that if $f$ is differentiable on an interval with $f^{\prime}(x) 
eq 1$, then $f$ can have at most one fixed point.

Ulisse Mini · Accepted Answer

Suppose for contradiction that $x,y$ are fixed points of $f$ with $x < y$, then apply MVT on $[x,y]$ to get
  $$
  f'(c) = \frac{f(x) - f(y)}{x - y} = \frac{x - y}{x - y} = 1
  $$
  But we know $f'(c) 
e 1$, therefore finding more than one fixed point is impossible.

Exercise 5.3.7

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Exercise 5.3.7

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