Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.8 Continuity - 1.8 Exercises - Page 93: 58

Answer

a) $f(-1)=-1 $ $f(0)=3$ There is a root of the equation $f(x) = x^5 − x^2 + 2x+ 3$, in the interval $(-1, 0)$. b) There is a root between $-0.88$ and $-0.87$, that is, in the interval $(0.88, 0.87).$

Work Step by Step

a) Given: $f(x) = x^5 − x^2 + 2x+ 3$, The function $f(x) = x^5 − x^2 + 2x+ 3$, which is continuous on the interval $[-1, 0]$, $f(-1) = (-1)^5 − (-1)^2 + 2(-1)+ 3$ $=-1−1-2+3 0$ $=-4+3 $ $f(-1)=-1 $ and $f(0) =(0)^5 − (0)^2 + 2(0)+ 3$ $=0-0+0+3$ $f(0)=3$ Since $3>0 >-1$ there is a number $c$ in $(-1, 0)$ such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a root of the equation $ x^5 − x^2 + 2x+ 3=0$, in the interval $(-1, 0)$. b) $f(-0.88) ≈ -0.062 <0$ and $f(-0.87) ≈ 0.0047> 0,$ so there is a root between $-0.88$ and $-0.87$, that is, in the interval $(0.88, 0.87).$
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