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October 15th, 2017, 01:59 AM  #1 
Member Joined: Apr 2017 From: India Posts: 34 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: 1,734 Thanks: 605 Math Focus: Yet to find out.  
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 $x \le \sqrt{2}$, $\dfrac{\pi}{2} \le y \le \pi$ Last edited by skeeter; October 15th, 2017 at 07:17 AM. 

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