Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Review - True-False Quiz: 24


This is true.

Work Step by Step

Consider the function $f(x)=x^{10}-10x^2+5$. We see that $f(0)=0^{10}-10\cdot0^2+5=5$ and $f(1)=1^{10}-10\cdot 1^2+5=1-10+5=-4.$ So we have that $f$ is continuous everywhere and changes the sign when $x$ goes from $0$ to $1$. This means that there is at least one zero of $f$ i.e. a root of the given equation in the interval $(0,1)$, by the intermediate value theorem. Since $(0,1)$ is the subset of $(0,2)$ this means that there is also a root in $(0,2)$.
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