My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum

Thanks Tree2Thanks
  • 2 Post By skeeter
LinkBack Thread Tools Display Modes
October 15th, 2017, 01:59 AM   #1
Joined: Apr 2017
From: India

Posts: 34
Thanks: 0


If y= arc cos(-(x^2)/2), then state the interval on which y is defined?
shashank dwivedi is offline  
October 15th, 2017, 02:30 AM   #2
Senior Member
Joined: Feb 2016
From: Australia

Posts: 1,734
Thanks: 605

Math Focus: Yet to find out.
Wolfram|Alpha: Computational Knowledge Engine)
Joppy is offline  
October 15th, 2017, 06:52 AM   #3
Math Team
Joined: Jul 2011
From: Texas

Posts: 2,773
Thanks: 1427

$-1 \le -\dfrac{x^2}{2} \le 1$

$-\dfrac{x^2}{2} \le 1 \implies x^2 \ge -2$, which is true for all $x \in \mathbb{R}$

$-1 \le -\dfrac{x^2}{2} \implies x^2 \le 2 \implies -\sqrt{2} \le x \le \sqrt{2}$

intersection of the two solution sets is the latter ... $|x| \le \sqrt{2}$

Since $-\dfrac{x^2}{2} \le 0$ for $|x| \le \sqrt{2}$, $\dfrac{\pi}{2} \le y \le \pi$
Thanks from topsquark and shashank dwivedi

Last edited by skeeter; October 15th, 2017 at 07:17 AM.
skeeter is offline  

  My Math Forum > College Math Forum > Differential Equations

differential, functions

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
functions paul19 Calculus 5 November 21st, 2016 02:15 PM
Derivatives, trignometric functions and exponential functions Nij Calculus 2 November 25th, 2015 07:20 AM
Functions help lamhmh Applied Math 2 June 19th, 2011 10:59 PM
Functions hoyy1kolko Algebra 1 January 7th, 2011 06:37 AM
help / functions sel Calculus 3 October 21st, 2008 04:48 AM

Copyright © 2018 My Math Forum. All rights reserved.