Freshman math help me I don t know how to do these questions

1 f(2)=8a+2b=6;f(-2)=-8a-2b=-6 This problem can also be proved to be an odd function by f(-x)=-f(x).

2 This is a piecewise function, which can be divided into four parts: (provided that x>=0) when x<=1, y=x

When 1<=x<=2, y= (1+x2)When 2<=x<=3, y= [1+(3-x) 2]When 3<=x<=4, y=4-x

3 From the conditions given in the question:

f(2)=f(1)+f(1)=2f(1)=1 f(1)=1/2 f(1)=2f(1/2) f(1/2)=1/4

f(x+1)=f(x)+f(1)=f(x)+1 2f(x+1)+f(x)=2f(x)+1 2>=1 yields: f(x)>=1 4

Since f(x) is an increasing function over the interval (0,+.

So, x>=1 2

Because it is an odd function, f(-2)=-f(2)=-6

There are four cases, in the AB segment: y=x, in the BC segment: y=root number (x 2-2x+2), in the CD segment, y = root number (x 2-6x+10), and in the AD segment: y=4-x

f(2x+1)>=f(2), i.e., there are 2x+1>=0 and 2x+1>=2, resulting in x>=1 2

Let f(x)=ax 3+bx,a, b r, and f(2)=6, find the value of f(-2).

f(-2)=a(-2)^3-2b=-(a2^3+2b)=-f(2)=-6

The edge of the square ABCD with a side length of 1 starts from the vertex A and passes through B, C, D and then back to A, let X represent the distance traveled by the point P, and Y represents the length of the line segment Pa, and find the analytical expression of the function of Y about X.

y=x 0≤x≤1

y=√[1+(x-1)^2] 1≤x≤2

y=√[1+(x-2)^2] 2≤x≤3

y=4-x 4≤x≤4

The function defined on r y=f(x) satisfies: f(x)+f(y)=f(x+y), f(2)=1

f(x) is an increment function over the interval (0,+, if f(x+1)+f(x) 1, find the range of x values.

Solution: f(x)+f(y)=f(x+y).

Let x=0, y=0

f(0)=0

Then let x=0, y=-x

f(x)+f(-x)=f(0)=0

So f(x) is an odd function.

And because f(x) is an increasing function over the interval (0,+.

So f(x) is an increasing function over the interval (-.

f(x+1)+f(x)=f（2x+1）≥1=f(2)

And f(x) is an increasing function over the interval (-.

2x+1≥2

x≥1/2

This is the odd function f(-x)=-f(x)=-6

Piecewise function: f(x) = 0 when x<0

x when 0<=x<1

Root number (1+(x-1) 2) When 1<=x<2 root number (1+(x-3) 2) When 2<=x<34-x when 3<=x<4

0 when x>=4

f(x+1)+f(x)>=1

f(x+1+x) >= f(2)

2x+1 >= 2

x >= 1/2

f(-2)=-8a-2b=-6

2.Represented by a piecewise function.

y=x,[0,1]

y=sqrt[1+(x-1)^2],[1,2]y=sqrt[1+(3-x)^2],[2,3]y=4-x,[3,4]

1f(x) is in [0, + increments, f(2)=1

2x+1>=1, i.e. x>=

(1) When x<0 is set, then -x>o

So f(-x) = -x(1-x).

And because f(x) is an odd function on r.

So f(x) = f(-x) = x(1-x).

So f(x)=x(1+x) x 0

f(x)=x(1-x) x<0(2) quadratic function plotting. Monotonically increasing on r.

x<0, -x>0

f(x)=-f(-x)=-(-x)(1-x)=x(1-x) is parsely f(x)=x(absolute value of 1+x).

At x>0, it increases monotonically, and when x<0, it decreases monotonically.

（1） x(1+x) x>=0f(x)= {

x(1-x) x<=0

Reason: When x<=0, because of the odd function, f(x)=-f(-x)Because -x>=0, f(-x)=-x(1-x)So f(-x) = -x(1-x).

2) Increment on r.

The monotonic interval is r

(1) f(x)=-x(1+x) (x<0), f(x)=x(1+x) (x=>0), 2) monotonically increasing in the interval of the function (negative infinity, positive infinity).

Note: a b denotes a to the b power.

a@b denotes the b-ths of a.

2. It is easy to know the coordinates of the three intersection points.

a0,bb-1+(1-b)@2,0

c-1-(1-b)@2,0

The perpendicular bisector equations for the chord AB and BC are obtained respectively.

Its intersection point is the center of the circle, and the solution is the center of the circle o

1,b-1/2

Find the radius from the points o and a.

The equation is (x+1) 2+(y-(b+1) 2) 2=1+((b-1) 2) 2

Problem 3: From the equation, we can know that the points 0,1 are on the circle and have nothing to do with b.

Then the point 0,1 is the request.

Solution: Let the quadratic function f(x)=x 2+2x+b and the three intersection points of the two coordinate axes be a, b, and m., respectively

then =4-4b>0 and b≠0, i.e., b<1 and b≠0

Let x=0, get y=b

Let y=0, to get x=-1+ 1-b or x=-1-1-b

i.e. a(-1+ 1-b,0), b(-1-1-b,0), m(0,b).

Since the two points a and b are symmetrical with respect to the straight line x=-1, the center c must be on the straight line x=-1.

Let c(-1,m).

By |ca|=|cm|=r to get 1-b+m 2=1+(m-b) 2=r 2 to get m=(b+1) 2, r 2=b 2-2b+5

Therefore, the equation for the circle c is (x+1) 2+[y-(b+1) 2] 2=b 2-2b+5(b<1 and b≠0)