# How do you find the critical points of #y=x-2sinx# on the interval #[0, pi/2]#?

There is one critical point of

The critical points of a function occur where the derivative of the function is zero on the specified interval.

Find the derivative of

Set the derivative,

Within the interval from

Therefore, there is only one critical point of

The

We can check our answer graphically. The slope changes from negative to positive at approximately 1.05.

graph{x-2sinx [-10, 10, -5, 5]}

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To find the critical points of ( y = x - 2 \sin(x) ) on the interval ( [0, \frac{\pi}{2}] ), we need to first find the derivative of the function and then set it equal to zero to solve for critical points.

The derivative of the function is ( y' = 1 - 2 \cos(x) ).

Setting the derivative equal to zero and solving for ( x ), we get:

( 1 - 2 \cos(x) = 0 )

( 2 \cos(x) = 1 )

( \cos(x) = \frac{1}{2} )

This equation is satisfied when ( x = \frac{\pi}{3} ) and ( x = \frac{5\pi}{3} ) within the interval ( [0, \frac{\pi}{2}] ).

So, the critical points of ( y = x - 2 \sin(x) ) on the interval ( [0, \frac{\pi}{2}] ) are ( x = \frac{\pi}{3} ) and ( x = \frac{5\pi}{3} ).

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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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