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 October 15th, 2017, 01:59 AM #1 Newbie   Joined: Apr 2017 From: India Posts: 23 Thanks: 0 Functions If y= arc cos(-(x^2)/2), then state the interval on which y is defined?
 October 15th, 2017, 02:30 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,506 Thanks: 502 Math Focus: Yet to find out.
 October 15th, 2017, 06:52 AM #3 Math Team   Joined: Jul 2011 From: Texas Posts: 2,699 Thanks: 1357 $-1 \le -\dfrac{x^2}{2} \le 1$ $-\dfrac{x^2}{2} \le 1 \implies x^2 \ge -2$, 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.

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