# What are the extrema of #f(x)=-sinx-cosx# on the interval #[0,2pi]#?

Since

Solve:

Now, either use the unit circle or sketch a graph of both functions to determine where they are equal:

On the interval

hope that helps

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To find the extrema of ( f(x) = -\sin(x) - \cos(x) ) on the interval ([0, 2\pi]), we first find the critical points by setting the derivative of ( f(x) ) equal to zero:

( f'(x) = -\cos(x) + \sin(x) )

Setting ( f'(x) = 0 ):

( -\cos(x) + \sin(x) = 0 )

( \sin(x) = \cos(x) )

Solving for ( x ):

( x = \frac{\pi}{4}, \frac{5\pi}{4} )

Now, we evaluate the function at the critical points and at the endpoints of the interval ([0, 2\pi]):

( f(0) = -\sin(0) - \cos(0) = -1 )

( f\left(\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = -\sqrt{2} )

( f\left(\frac{5\pi}{4}\right) = -\sin\left(\frac{5\pi}{4}\right) - \cos\left(\frac{5\pi}{4}\right) = -\sqrt{2} )

( f(2\pi) = -\sin(2\pi) - \cos(2\pi) = -1 )

The extrema occur at the critical points ( \frac{\pi}{4} ) and ( \frac{5\pi}{4} ), where the function takes on the value ( -\sqrt{2} ). The minimum value of the function is ( -\sqrt{2} ), and the maximum value is ( -1 ).

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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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