# What are the points of inflection of #f(x)=xsinx # on the interval #x in [0,2pi]#?

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To find the points of inflection of ( f(x) = x \sin(x) ) on the interval ( x ) in ([0,2\pi]), we first need to find the second derivative of ( f(x) ), then find the values of ( x ) where the second derivative changes sign from positive to negative or vice versa.

First derivative: ( f'(x) = x \cos(x) + \sin(x) )

Second derivative: ( f''(x) = -x \sin(x) + 2 \cos(x) )

Now, we need to find where ( f''(x) = 0 ) or does not exist, and then determine the sign of ( f''(x) ) around those points.

Setting ( f''(x) = 0 ), we solve ( -x \sin(x) + 2 \cos(x) = 0 ). This equation doesn't have solutions in the interval ( [0,2\pi] ).

Next, we determine the sign of ( f''(x) ) in the intervals determined by the critical points of ( f''(x) ).

We observe the sign changes of ( f''(x) ) around the critical points. These sign changes indicate points of inflection.

Analyzing the sign of ( f''(x) ) around the critical points in the interval ( [0,2\pi] ), we find that there are points of inflection at ( x = \frac{\pi}{2} ) and ( x = \frac{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.

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