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October 15th, 2017, 12:59 AM   #1
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If y= arc cos(-(x^2)/2), then state the interval on which y is defined?
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October 15th, 2017, 01:30 AM   #2
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October 15th, 2017, 05:52 AM   #3
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$-1 \le -\dfrac{x^2}{2} \le 1$

$-\dfrac{x^2}{2} \le 1 \implies x^2 \ge -2$, 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 $|x| \le \sqrt{2}$, $\dfrac{\pi}{2} \le y \le \pi$
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Last edited by skeeter; October 15th, 2017 at 06:17 AM.
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