University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 96: 67

Answer

Take a function $g(x)=f(x)-x$ and apply the Intermediate Value Theorem to show that there exists a value $c\in[0,1]$ that $g(c)=0$.

Work Step by Step

We examine a function : $$g(x)=f(x)-x$$ From the exercise, we know that $f(x)$ is continuous on $[0,1]$, and function $y=x$ is also continuous on $[0,1]$. According to the properties of continuous functions, $g(x)=f(x)-x$ is also continuous on $[0,1]$. According to the exercise, on $[0,1]$, $0\le f(x)\le1$ Therefore, $f(0)-0\ge0$, while $f(1)-1\le0$. That means, for $x=0$, $g(0)\ge0$, and for $x=1$, $g(1)\lt0$ According to the Intermediate Value Theorem, since $g(x)$ is continuous on $[0,1]$, there exists a value $x=c\in[0,1]$ such that $g(c)=0$, or in other words, $$f(c)-c=0$$ $$f(c)=c$$
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