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February 18th, 2019, 10:19 PM   #1
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Exclamation 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.
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February 19th, 2019, 07:47 AM   #2
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Quote:
Originally Posted by MikeAndIke View Post
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.
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