Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises - Page 292: 21


The equation $~~x^3-15x+c = 0~~$ has at most one root in the interval $[-2,2]$

Work Step by Step

$x^3-15x+c = 0$ Let $f(x) = x^3-15x+c$ $f'(x) = 3x^2-15$ Then $f'(x) \lt 0$ for all $x$ in the interval $[-2,2]$ The function $f(x)$ is continuous and differentiable for all $x$. Let's assume that the equation has at least two roots $a$ and $b$ in the interval $[-2,2]$. Then $f(a) = f(b) = 0$. According to Rolle's Theorem, there is a number $k$ in the interval $(a,b)$ such that $f'(k) = 0$. However, this contradicts the fact that $f'(x) \lt 0$ for all $x$ in the interval $[-2,2]$ Therefore, the equation has at most one root in the interval $[-2,2]$
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