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 February 18th, 2019, 10:19 PM #1 Newbie   Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus Finding the absolute extrema problem So, this one is more challenging for me to do, since I normally work with equations/functions without sin or pi involved. The problem:y =2sin^(2)x over [-pi/6,pi/3]. How am I suppose to find the absolute extrema with sin and pi involved? Please show steps on how to do so. Last edited by skipjack; February 19th, 2019 at 07:49 AM. February 19th, 2019, 07:47 AM   #2
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Quote:
 Originally Posted by MikeAndIke So, this one is more challenging for me to do, since I normally work with equations/functions without sin or pi involved. The problem:y =2sin^(2)x over [-pi/6,pi/3]. How am I suppose to find the absolve extrema with sin and pi involved? Please show steps on how to do so.
note the function values at the endpoints of the given interval ...

$y\left(-\dfrac{\pi}{6}\right) = 2\sin^2\left(-\dfrac{\pi}{6}\right) = 2\left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{2}$

$y\left(\dfrac{\pi}{3}\right) = 2\sin^2\left(\dfrac{\pi}{3}\right) = 2\left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{2}$

now determine critical values of $x$ on the open interval ...

$y' = 4\sin{x}\cos{x} = 0 \implies x = 0$

$y(0) = 2\sin^2(0) = 0$

therefore, on the given interval

$y\left(\dfrac{\pi}{3}\right) = \dfrac{3}{2}$ is the absolute maximum

$y(0) = 0$ is the absolute minimum. Tags absolute, absolute extrema, extrema, finding, mikeiketube, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post suhdude Calculus 1 February 21st, 2016 05:22 PM factory1o1 Calculus 0 July 5th, 2012 05:55 AM ProJO Calculus 2 March 28th, 2011 12:00 PM sansar Calculus 1 April 12th, 2009 07:05 AM shepherdm1270 Calculus 0 March 24th, 2008 09:12 AM

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