What are the points of inflection of #f(x)=xcosx # on the interval #x in [0,2pi]#?
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To find the points of inflection of ( f(x) = x \cos(x) ) on the interval ( x ) in ([0, 2\pi]), we need to find where the second derivative changes sign or equals zero. The second derivative is ( f''(x) = 2 \cos(x) - x \sin(x) ). We then solve ( f''(x) = 0 ) for ( x ) in the interval ([0, 2\pi]) and determine the intervals where the sign changes. Solving ( f''(x) = 0 ) yields solutions at ( x = \pi/2 ) and ( x = 3\pi/2 ). Then, we evaluate the sign of ( f''(x) ) in each interval between these points. The intervals where ( f''(x) ) changes sign indicate points of inflection. Therefore, there is one point of inflection at ( x = \pi/2 ) and another at ( x = 3\pi/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.
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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