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

Answer

Turn the equation into the function $f(x)=\cos x-x$ and apply the Intermediate Value Theorem for Root Finding.
1536148654

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 $[0,\pi/2]$.
Small 1536148654
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.