A fill in the blank question about a function, a fill in the blank question about a function

1) Method: This topic will be very simple to use as a diagram.

2) Steps: 1) Draw y=|x+1|with y=|x-2|From the auxiliary line perpendicular to the x-axis in the image, we can see that there is a figure of y=2-x on the left side of point a, and on the top of the figure of y=x+1, so when x<=, take the figure y=2-x as the expression of the function f(x).

3) In the same way, y=x+1 is obtained > x

4) From the monotonicity of the graphs, we can see that their intersection point a(, which is the point where the minimum value of the function f(x) is sought.

The answer is: 3) Supplement: From steps (2) and (3), we can know that the function f(x) is a piecewise function, and when x<=, f(x)=2-x; When x >, f(x)=x+1

First, divide the x range into 3 parts, x<=-1, -1=2, because -1 and 2 make x+1 and x-2 equal to 0

1. x<=-1, f(x)=max=2-x (two piecewise functions, you can see it when you draw an image), the minimum value is 3

2, x>=2, f(x)=max=x+1, the minimum value is 33, -1 two primary functions intersect at the point (, so.

1 In summary, the minimum value is:

At first glance, I thought it was a minimum of 0, but it was actually 3 2

If the absolute value of x+1 is greater than or equal to the absolute value of x-2, then the absolute value of x+1 is the result of max(); If it is less than the absolute value of x-2, it is the result.

The minimum absolute value is 0, but there is a limit to the value of x because of the explanation of the inequality mentioned above. In the end, it's 3 2.

In this kind of problem, you make two numbers equal, and the number you get matches the question.

x+1|=|x-2|

x+1）^2=(x-2)^2

x=so the minimum value is.

Draw the image y=x=1, y=x-2Flip the part below the x-axis, take the maximum value in the middle, and min, which is obtained when x=, so it is.

1. The two points are brought into the analytic formula.

y is the ordinate and x is the abscissa.

The solution is a=2 and b=9

a+b=11

2. a<0, the function is over.

Second, four quadrants, and vice versa.

1. Three quadrants.

b<0, the function is over.

Three, four quadrants, and vice versa.

1. 2 quadrants.

Therefore, it is only the first quadrant.

3. Proportional function, the formula is: y=kx

Then -2m-14=1

m = over two quadrants, k<0

The analytic formula is y=

4. See intersection, continuous system of equations.

y=kx+b

y=-3x-4

Because it intersects on the y-axis, x=0

b=4k cannot be evaluated.

5、q=30-5t(0<=x<=6)

y=1/2cos(πx+π/3)-sin(πx+5π/6)=1/2cos(πx+π/3)-sin(πx+π/3+π/2)=1/2cos(πx+π/3)- cos(πx+π/3)= -1/2cos(πx+π/3).

2kπ≤πx+π/3≤2kπ+π2k-1/3≤x≤2k+2/3(k∈z)

So the monotonically increasing interval of the function is [2k-1 3, 2k+2 3] (k z).

The intersection of the x-axis is (3 2,0) and the intersection of the y-axis (0,-6) passes through one or three or four quadrants, and y increases with the increase of x.

The secondary parabolic opening is upward, and the left side of the axis of symmetry decreases monotonically, where the axis of symmetry = -2(a-1) 2=1-a, to make f(x) monotonically decreasing in (-4), only 1-a 4, i.e., a -3

Solution: f(x) is an even function f(x)=f(-x) so b- (2-a)=0

So a=2-b2

The ordinate of the intersection of f(x) and y-axis is:

a+b=2-b2 + b=-(b- 1 2) 2 +9 4When b=1 2, the maximum value is 9 4

S1 plus shaded is equal to the product of A's abscissa and ordinate, and S2 plus shaded is equal to B's product of abscissa and ordinate. From the function, the product of the abscissa and the ordinate is constant 3So s1 + s2 = 4

That constant is not very clear, suppose it is three, y=3 x, then x*y=3, that is, the box of a, the box of b. And all such boxes are 3 in area, so s1 + shadow = 3, s2 + shadow = 3, so s1 plus s2 = 5

y=2/xy=kx+1

2/x=kx+1

kx^2+x-2=0

Because there is always a common point, that is, there is always a solution to the above formula, and the discriminant formula is used.

b^2-4ac>=0

1-4*k*(-2)>=0

8k>=-1

k>=-1/8

Very simple.

Matching method, match 2ax-x in the root number of the denominator to a -(x-a), replace x-a with t, it becomes the form of integral sign (1 sqrt(a -t)dt), and replace t with acosu, then the integral sign ((1 asinu)d(acosu)) gives the integral sign (-1*du), which is u+c. Therefore, the original function is.

arccos(t/a)+c=-arccos((x-a)/a)+c；If you replace t with asinu, you get arcsin((x-a) a)+c. Can be verified.

Check out the points table! See if this can be used, and change the corresponding constant.