Find the monotonic interval y sin 2x 3 y sin 2x 2 3 .

First of all, it is clear that the increase interval of y=sinx is: [- 2+2k , 2+2k ]; The subtraction interval is: [ 2+2k ,3 2+2k ].

Therefore, (1) from -2+2k <=2x- 3<= 2+2k:- 12+k <=x<=5 12+k, that is, y sin(2x 3) increases in the interval [- 12+k ,5 12+k]; From 2+2k <=2x 3<=3 2+2k:

5 12+k <=x<=11 12+k, that is, y sin(2x 3) minus interval is [5 12+k, 11 12+k].

2) Because x is negative in 2x 2 3, the increase interval is: 2+2k <= 2x 2 3<=3 2+2k:

5 12-k <=x<= 12-k, that is, y sin( 2x 2 3) increases in the interval is: [-5 12-k, 12-k]; In the same way, the subtraction interval can be derived from - 2+2k <= 2x 2 3<= 2+2k.

In short, as long as we remember the increase and decrease interval of y=sinx, and treat the u in SINU (where x is a positive value in u) as x and substitute it into the increase and decrease interval in turn, we can get the desired increase and decrease interval;

If not, if x in u is negative, then substituting the increase and decrease interval in turn, then the decrease interval and the increase interval are obtained one by one.

1) 2x 3 [- 2+2k, 2+2k] on monotonic increase.

2x 3 [ 2+2k, 3 2+2k ] on monotonic minus.

2) 2x 2 3 [- 2+2k, 2+2k] on monotonous increase.

2x 2 3 [ 2+2k , 3 2+2k ] on monotonic subtraction.

The rest is sorted out by yourself.

Note: The questioner has poor basic knowledge of functions, so I hope to redouble my efforts!

1) 2x 3 [ 2+2k ,3 2+2k ] monotonically reduce and then simplify to find x.

2) This function needs to use the induction formula to change the coefficient of x -2 to 2, and then examine the monotonic interval before the solver has a problem.

Find the monotonic interval: y e (2x 2 3) y lgsin(2x 3).

1) Analysis: The function y e (2x-2 3), which is defined in the domain of r and lets u=2x-2 3, is a one-time function, monotonically increased;

y=e u, monotonically increased.

2) Analysis: The function f(x) lgsin(2x- 3) is fixed and slowed sedan sin(2x- 3)>0==>00;When x (5 12, 2 3), f'(x).

y=2sin( 6-4x)=-2sin(4x- 6), from 2k - 2 4x- 6 2k + 2,k hole z is z : k 2- 12 x k 2 + 6, k z y=2sin (sell and merge 6-4x) monotonically decreasing interval is: medium flutter [k 2- 12, k 2 + 6](k z) by 2k + 2 4x- 6 2k + 3 2....

Find the axis of symmetry -b 2a = 1 2

Then look at the function image opening direction downwards.

Therefore, the negative infinity to the symmetry of the orange finch axis Qing fiber for the monotonous delivery of the honor of Wu imitation increase.

So the increasing interval is (-infinity, 1 2].

The axis of symmetry to positive infinity is monotonically decreasing.

So the decreasing interval is [1 2, +infinity).

f(x)= 2x^2-3x+1

f'(x) =4x-3

f'(x)=0

4x-3=0

x=3/4f''(x) =4 > potato danhe0 (min)f(x)= 2x 2-3x+1 monotonous delay.

Increase =[3 4, +

Meiosis small = (-3 4].

Quadratic functions can be solved quickly with the axis of symmetry.

The axis of symmetry is x=-(3) 4=3 4

The opening refers upwards to the socks and.

Therefore, the monotonic increase interval is [3 4, + the monotonically decreasing good opening interval is (- Weiji, 3 4).

Solution: y 2x 3-5x 2+3, then it defines the domain x r;

y 6x 2-10x 6x (x-5 3) makes y 0, then x1 0, x2 5 3

When x 5 3, y 0, the original function y 2x 3-5x 2+3 monotonically increases;

When 0 x 5 3, y 0, the original function y 2x 3-5x 2+3 monotonically decreases;

When x 0 and y 0, the original function is monotonically incremented.

The monotonically increasing interval of the function y 2x 3-5x 2+3 is (- 0] [5 3, and the monotonically decreasing interval is (0,5 3).

y'=6x 2-10x=3(x-5)x=0, x=0 or x=5 3

When x<0, y'>0, the function increases monotonically.

When 05 3, y'>0, the function increases monotonically.

The derivative of the equation y'=1+2x, when y'=1+2x=0→x=－1/2

And y'=1+2x is monotonically increasing in any interval, and we can get: y when x 1 2'0, so the function y x-2 x decreases monotonically on (- 1 2); When x 1 2 y'0, so the function y x-2 x increases monotonically on (-1 2,+.