The increasing interval of the function f x 丨 x 2丨 丨 x 1丨 is

Each absolute value of the shilling is equal to 0, and the zero point is found.

Let 丨x+2丨=0 get x= -2

Let 丨x-1丨=0 get x= 1

The two zeros are -2 and 1 respectively, and the following is discussed with the zero point as the boundary.

1. When x -2, both absolute values are negative, so.

f(x)=-(x+2)-(x-1)= -2x-1 This function is decreasing over the entire x-2.

2. When -2 x 1, the two absolute values are one positive and one negative, i.e., x+2 0x-1 0 so f(x)=x+2-(x-1)= 3 The function is parallel to the axis on the whole -2 x 1, neither increasing nor decreasing.

3. When x>1, both absolute values are positive, so.

f(x)=(x+2)+(x-1)= 2x+1 This function is incrementing over the entire x>1.

In summary, the increasing interval of the original function is [1, ).

x<-2

then f(x)=-2-x-x+1=-2x-1

is a decreasing function.

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

is a constant function.

x≥1f(x)=2+x+x-1=2x+1

is an incrementing function.

So the increment interval is [1,+

In summary, the increment interval is [1,+

0<1/2<1

So (1 2) x decreases.

So f(x) is added |x-1|Degression.

x-1 increments, -x+1 decrement.

So |x-1|=-x+1

So yes (- 1).

i.e. (x+1)(x-m)<0

m>0 then m>-1

So -11400 (1 2+3 4-1)=350

A: The actual production is 350 kg more than planned.

If you do this, you are asking questions, and it will be the same as asking questions in the future.

In general, let the function f(x) be defined in the domain d, and if for the values x1 and x2 of any two independent variables in an interval in the defined domain d, then the monotonic increase interval of the function f(x)=1 x 2 is (- 0) when x1

f(x)=1/x^2=x^(-2)

f'(x)=-2x^(-3)

f(x) is the additive function, then f'(x)>0

2x^(-3)>0

x^(-3)<0

x<0, so the increment interval is x<0

Derivative, so that the derivative is greater than zero, -2x (-3) >0 and x<0 so the increase interval is (- 0).

This is evident from the image method.

Categorical discussions. 1）(x>=2)

f(x)=2x-1 increment function.

2）（-1<=x<2)

f(x)=3 constant function.

3）(x<-1)

1-2x subtraction.

f(x)=|x+1|+|x-2|

2x-1,(x>=2)↑，3，（-1<=x<2);

1-2x,(x<-1)↓。

The increase interval for f(x) is [2, + the decrease interval is (- 1).

This is a composite function, let t= |x-1|,y=(1/2)^ t,∵|x-1|With respect to the straight line x=1 symmetry, t=|x-1|The monotonic increase interval is [1, + the monotonic decrease interval is (- 1].

y=(1 2) t is the subtraction function, and according to the principle of "same increase and different subtraction" of the composite function, it can be seen that

The monotonic decreasing interval of the original function is [1, + the monotonic increasing interval is (- 1).

x>=0, f(x)=|x|=x, monotonically incrementing.

x《Concealment=0, f(x)=-x, monotonous decreasing.

g(x)=2x-x 2=-(x-1) The 2+1 increment interval is to the left of the axis of symmetry, i.e., x<=1

1.Since the axis of symmetry is -b 2a = -1 2 and a>0, the function increments in the interval (-1 2,+

2.Let x<0, then -x>0, so f(-x)=e (-x) is also because of the odd function -f(-x)=f(x).

Therefore, f(x)=-f(-x)=-e (-x) and f(x) is an odd function jujube base, so f(0)=0

So f(x)={e x,x> filial piety 0

0 ,x=0

e^(-x), x<0

x<-2

2x-1,-2<=x<1

3, 1<=x

This function defines the domain as (-

The value range can be seen from the careful diagram of the stool [-3,3].

It is known from the image that it is a non-odd and non-even function.

x [-2,1).

lg^5+2*lg5*lg2+lg^2

LG5+LG2) 2=(LG10) 2=1 2=1 I wish you a high level of learning

1 (, infinity).

2 f(x)=e x x when x is greater than 0.

0 x isocal at 0.

E -xx Xiaokong Xiao Xiao at 0 o'clock fight defense draft.

f(x) is increasing at (0, positive infinity).

g(x)=2x-x 2=-(x-1) 2+1, increasing at (-infinity, 1).