# What are the critical values, if any, of # f(x)= x cos x −3 sin x +2x in [0,2pi]#?

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To find the critical values of ( f(x) = x \cos x - 3 \sin x + 2x ) in the interval ([0, 2\pi]), we first find its derivative and then solve for where the derivative equals zero or is undefined.

Taking the derivative of ( f(x) ) with respect to ( x ):

( f'(x) = \frac{d}{dx}[x \cos x] - \frac{d}{dx}[3 \sin x] + \frac{d}{dx}[2x] )

( f'(x) = \cos x - x \sin x - 3 \cos x + 2 )

Setting ( f'(x) ) equal to zero and solving for ( x ):

( \cos x - x \sin x - 3 \cos x + 2 = 0 )

Combining like terms:

( -2\cos x - x\sin x + 2 = 0 )

Now, there's no algebraic way to solve this equation for ( x ), so numerical methods or graphical methods are typically used to find approximate solutions within the interval ([0, 2\pi]).

Once you've obtained these approximate solutions, you can check each one to see if it corresponds to a maximum, minimum, or inflection point by using the second derivative test or by examining the behavior of ( f'(x) ) around those points.

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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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