Find y 2sin 3 2x monotonically increasing interval

The monotonic increase interval of the function y=sinx is (2k -1 2 , 2k +1 2 ), k is an integer, and the monotonic decrease interval is (2k +1 2 , 2k +3 2 ), and k is an integer.

y=2sin(3-2x) where the x coefficient is negative, so there are two solutions:

1）y=2sin(π/3-2x)=y=-2sin(2x-π/3)

So 2x-3 is in (2k +1 2 , 2k +3 2 ), note that it is a monotonic reduction interval of y=sinx. The solution is available (k +5 12 , k +11 12).

2) y=2cos[ 2-( 3-2x)], and then use the property of y=cosx to solve the problem, and do not say more.

Pay attention to the period, now the solution is k, to indicate that k is an integer, in addition, the interval of the problem solution can be an open interval, can also be a closed interval, half open and half closed is also correct, depending on the requirements of the question to do.

First find the derivative of the function, then let the derivative be greater than 0, and find its interval.

The answer is (k +5 12 , k +11 12).

Note: A derivative is a formula, not a number.

y=2sin( 3-2x)=-2sin(2x- 3)At 2x- 3 [2k + loss 2,2k +3 2], i.e., x [k +5 12,k +11 6], the function increases.

So the monotonic increase interval of y=2sin( 3-2x): [k +5 12,k +11 pin Lu Chong 6],k z

y=2sin(2x+π/6)

Monotonic increase interval:

Let 2k - 2 2x + 6 2k + 2, k z get k - 3 x k + 6, k z

So the monotonic increase interval is (k - 3, k + 6), k z

Find y=2sin(3x-4) and x[-2,2] find the monotonically increasing interval. x belongs.

A good Hazeng interval is 2 k- 2 3x- 4 2 k+ 2 <>

There's more than one monotonically increasing interval here

Why k=1 or 2?

It's counting! Because here k can take different values

k is a constant!

Here, the value of k only needs to be satisfied, so that x is only in the interval [-2 ,2], but here is the sin increasing interval.

Yes! Find the increasing interval of sinx

For example. The monotonically increasing interval of sinx is.

Got it, so let's find y=2sin(-3x- 4) to find the monotonically increasing interval.

Hello! This is the last question that was not finished

The final increment interval is the union of these six intervals!

The second problem calculates the range of values of x.

Then take k is equal to , as long as the value of k is taken so that x satisfies the interval between [-2 ,2] and meets the requirements

Finally, there is the union of these intervals.

In the previous question, remember to take k is equal to 0, and then the calculated interval should also be incorporated!

Okay, it's a bit slow today

To solve this kind of problem, you have to be patient, and you can't miss the k value!

y=2sin(-2x-π/6)

2sin(2x+π/6)

When 2 +2k 2x+ 6 3 2 +2k, that is, 6 +k x 2 3 +k, the monotonic increase interval is ( 6 + k, Xiao Chong accompanies pure 2 3 + k )

Solution x belongs to [0, ].

i.e. 0 x

i.e. 0 2x 2

i.e. 3 2x+ 3 2 + 3=7 3, i.e. when 3 2x+ 3 2.

y=sin(2x+3) is the increment function.

0 x 12 at this point

Or when 3 2 2x + 3 7 3.

y=sin(2x+3) is the increment function.

At this time 7 12 x

That is, the monotonic increase interval is and.

The hardest part is when you certify what you are.

If you have any questions, look for Shenma.

The increment range is: 2k 2 2x 3 2k 2

i.e.: k 5 12 x k 12

The increase interval is: [k 5 12, k 12], k z