# What are the absolute extrema of #y=cos^2 x - sin^2 x# on the interval [-2,2]?

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To find the absolute extrema of ( y = \cos^2(x) - \sin^2(x) ) on the interval ([-2, 2]):

- Find the critical points by setting the derivative ( y' ) to zero and solving for ( x ).

[ y' = \frac{d}{dx}(\cos^2(x) - \sin^2(x)) ]

[ y' = -2\cos(x)\sin(x) ]

Setting ( y' ) to zero:

[ -2\cos(x)\sin(x) = 0 ]

This gives critical points at ( x = -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, ) etc., but we only need to consider the critical points within the interval ([-2, 2]).

- Evaluate ( y ) at the critical points and at the endpoints of the interval to find the absolute extrema.

[ y(-2) = \cos^2(-2) - \sin^2(-2) ] [ y(2) = \cos^2(2) - \sin^2(2) ] [ y(-\frac{\pi}{2}) = \cos^2(-\frac{\pi}{2}) - \sin^2(-\frac{\pi}{2}) ] [ y(0) = \cos^2(0) - \sin^2(0) ] [ y(\frac{\pi}{2}) = \cos^2(\frac{\pi}{2}) - \sin^2(\frac{\pi}{2}) ] [ y(\pi) = \cos^2(\pi) - \sin^2(\pi) ] [ y(\frac{3\pi}{2}) = \cos^2(\frac{3\pi}{2}) - \sin^2(\frac{3\pi}{2}) ]

Evaluate each of these expressions to determine the minimum and maximum values of ( y ) on the interval ([-2, 2]).

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When evaluating a one-sided 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 one-sided 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 one-sided 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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