University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.5 - Continuity - Exercises - Page 95: 54


Turn the equation into the function $f(x)=\cos x-x$ and apply the Intermediate Value Theorem for Root Finding.

Work Step by Step

$$\cos x=x$$ $$\cos x-x=0$$ To prove the equation has at least one solution, we can apply the Intermediate Value Theorem for Root Finding, which requires to us to do the followings: - Take the function $f(x)=\cos x-x$ - Find 2 values $x_1$ and $x_2$ such that $f(x_1)\lt0$ while $f(x_2)\gt0$ or vice versa. - Show that $f(x)$ is continuous on $[x_1, x_2]$. Considering the function $f(x)=\cos x-x$: - Domain: $(-\infty,\infty)$ 1) Continuity: On $(-\infty,\infty)$, we notice that $\lim_{x\to c}(\cos x-x)=\cos c-c=f(c)$ That means $f$ is continuous on $(-\infty,\infty)$. 2) Find $x_1$ and $x_2$: Here I graph $f(x)=\cos x-x$ for better findings, which I have also enclosed below. - At $x_1=0$: $f(0)=\cos0-0=1-0=1\gt0$ - At $x_2=\pi/2$: $f(\pi/2)=\cos(\pi/2)-(\pi/2)=0-\pi/2=-\pi/2\lt0$ And since $f$ is continuous on $R$, it is continuous on $[0,\pi/2]$ Therefore, according to the Intermediate Value Theorem, there must exist at least a value of $x=c\in[0,\pi/2]$ such that $f(c)=0$. In other words, the equation $\cos x-x=0$ or $\cos x=x$ has at least one solution in $[0,\pi/2]$.
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