Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Review - Exercises: 33

Answer

1) Prove that the function $f(x)=x^5-x^3+3x-5$ is continuous on its domain. 2) Calculate $f(1)$ and $f(2)$ 3) Show that $0\in(f(1),f(2))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(1,2)$ so that $f(a)=0$. That means there is a root of the equation in interval $(1,2)$.

Work Step by Step

$$x^5-x^3+3x-5=0$$ interval $(1,2)$ *STRATEGY: 1) Prove that the function $f(x)=x^5-x^3+3x-5$ is continuous on its domain. 2) Calculate $f(1)$ and $f(2)$ 3) Show that $0\in(f(1),f(2))$. From that, according to the Intermediate Value Theorem, we can show that there must be a number $a$ in the interval $(1,2)$ so that $f(a)=0$. That means there is a root of the equation in interval $(1,2)$. 1) Since $f(x)$ is a polynomial function, it is continuous at any number in its domain, which is $R$. Therefore, the interval $(1,2)$ lies in the domain of $f(x)$. 2) Calculate: $f(1)=1^5-1^3+3\times1-5=1-1+3-5=-2$ $f(2)=2^5-2^3+3\times2-5=25$ 3) Since $f(1)\lt0$ and $f(2)\gt0$, $0$ lies in the interval $(f(1), f(2))$. According to the Intermediate Value Theorem, since $f(x)$ is continuous on $(1,2)$, there must a number $a$ in $(1,2)$ that $f(a)=0$. That means there must be a root of the equation in the interval $(1,2)$.
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