Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Review - Exercises - Page 360: 45

Answer

The equation $~~~3x+2~cos~x+5 = 0~~~$ has exactly one real root.

Work Step by Step

$3x+2~cos~x+5 = 0$ Let $f(x) = 3x+2~cos~x+5$ The function $f(x) = 3x+2~cos~x+5$ is continuous and differentiable for all $x$. When $x=-3$, then $(3x+2~cos~x+5) \lt 0$ When $x=1$, then $(3x+2~cos~x+5) \gt 0$ Therefore, by the Intermediate Value Theorem, there is a number $c$ in the interval $(-3,1)$ such that $(3c+2~cos~c+5) = 0$. Thus the equation has at least one real root. Let's assume that the equation has at least two real roots $a$ and $b$. Then $f(a) = f(b) = 0$. According to Rolle's Theorem, there is a number $k$ such that $f'(k) = 0$. We can try to find $k$: $f'(x) = 3-2~sin~x = 0$ $sin~x = \frac{3}{2}$ However, there is no number $x$ such that $sin~x = \frac{3}{2}$. Therefore, the assumption that the equation has at least two real roots is incorrect. Therefore, the equation has exactly one real root.
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