What are the local extrema and inflection points for #y = 2sin (x)  cos^2 (x)# for #[0, 2π]#?
There are inflection points because the concavity changes:
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To find the local extrema and inflection points of ( y = 2\sin(x)  \cos^2(x) ) on the interval ([0, 2\pi]), we first find the critical points by setting the derivative equal to zero and solving for (x). Then we test these critical points to determine whether they are local extrema or inflection points by analyzing the second derivative.

Find the first derivative: [ \frac{dy}{dx} = 2\cos(x) + 2\sin(x)\cos(x) ]

Set the first derivative equal to zero and solve for (x): [ 2\cos(x) + 2\sin(x)\cos(x) = 0 ] [ \cos(x)(2 + 2\sin(x)) = 0 ] [ \cos(x) = 0 ] or ( 2 + 2\sin(x) = 0 )
For ( \cos(x) = 0 ), solutions are (x = \frac{\pi}{2}) and (x = \frac{3\pi}{2}).
For (2 + 2\sin(x) = 0), we get ( \sin(x) = 1 ), which gives (x = \frac{3\pi}{2}).
So, the critical points are (x = \frac{\pi}{2}) and (x = \frac{3\pi}{2}).

Find the second derivative: [ \frac{d^2y}{dx^2} = 2\sin(x) + 2\cos^2(x)  2\sin^2(x) ]

Evaluate the second derivative at the critical points: [ \frac{d^2y}{dx^2} \left(\frac{\pi}{2}\right) = 2 ] [ \frac{d^2y}{dx^2} \left(\frac{3\pi}{2}\right) = 2 ]
Since the second derivative is negative at both critical points, they correspond to local maxima.
There are no inflection points on the interval ([0, 2\pi]).
Therefore, the local extrema for (y = 2\sin(x)  \cos^2(x)) on ([0, 2\pi]) are:
 Local maximum at (x = \frac{\pi}{2}) with (y = 2)
 Local maximum at (x = \frac{3\pi}{2}) with (y = 2)
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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