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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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,838 Thanks: 1481  Quote:
$y\left(\dfrac{\pi}{6}\right) = 2\sin^2\left(\dfrac{\pi}{6}\right) = 2\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{2}$ $y\left(\dfrac{\pi}{3}\right) = 2\sin^2\left(\dfrac{\pi}{3}\right) = 2\left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{2}$ 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.  

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absolute, absolute extrema, extrema, finding, mikeiketube, problem 
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