Questions about function solving! A: A lot of points!

This is the law summarized by drawing: k>0, the image of the primary function y=kx+b passes through one or three quadrants, k<0, the image of the primary function y=kx+b passes through two quadrants, and b is the ordinate of the intersection of the image of the primary function y=kx+b and the y-axis, when b is a positive number.

When b is negative, the image of a function intersects with the y-axis and the negative half-axis, so y=kx+b (k is a positive number and b is also a positive number), and its coordinates are connected to a straight line in 1 2 3 quadrants. y=kx-b (k is a negative number b is also a negative number), and its coordinates are in a straight line within 2 3 4 quadrants.

Key 1: The intersection of the line and the y-axis, that is, the longitudinal intercept, i.e. b, b is positive, then the intersection is above the x-axis, and vice versa.

2: The angle between the straight line and the x-axis (on the right), that is, the slope k, if k is positive, then the angle is acute, and k is negative, then the angle is obtuse.

Let K go plus or minus and B take plus or minus, and draw four straight lines.

You'll find out if you draw the picture yourself.

The first question: because when x is positive, k and b are both positive, and y cannot be negative, it can only be in the 1 2 3 quadrant.

The second question: because x is negative, k and b are negative, and y cannot be negative, so it can only be in 2 3 4 quadrants.

b represents the intersection of the straight line and the y-axis, the regular is on the y-axis positive semi-axis, k represents the slope of the straight line, and the regular means that the straight line is at an acute angle to the x-axis positive semi-axis.

In summary, the straight line is no more than 4 quadrants.

In the same way, when k and b are negative, the straight line intersects the negative half axis of the y axis and is obtuse at the positive half axis of the x axis.

Therefore, it is only 1 quadrant.

f(x)=-f(x-2)

So f(2)=-f(2-2)=-f(0).

For odd functions, f(0)=0

So f(2)=0

Yes, f(x) is an odd function defined on r, f(0) = 0, f(-2) = -f(2).

and f(x-2) = -f(x), so that x=0, f(-2) = -f(0)=0, f(2)= -f(-2)=0

f(x)=lnx/x

f ′(x) = [1/x*x-lnx]/x² = (1-lnx)/x²

At x [1 2,1], f (x) = (1-lnx) x 0 is constant and f(x) is continuously and monotonically increasing.

f(1/2)=ln(1/2)/(1/2)=2ln(1/2)f(1)=0

f(x) is bounded at [1 2,1], and 2ln(1 2) f(x) 0

option a is correct.

x<-1

then 1-x <0

2x<0

So f(x)=1

i.e. 1>1, not true.

2x<0

So (1-x) 1>1

x²-1)²>0

x≠ 1 so -10

2x>=0

So (1-x) 1>(2x) +1

x²-1)²-2x)²>0

x²+2x-1)(x²-2x-1)>0

The zero point is -1- 2, 1- 2, -1 + 2, 1 + 2, so x<-1- 2, 1- 21 + 2

So 0<=x<-1+ 2

x>1 is 1-x <0

2x>=0

So 1>(2x) +1

2x)²<0

It is not established to sum up.

1

When x>=0, f(x)=x 2+1, and when x<0, f(x)=1, then (0,+ is monotonically increasing.

So the inequality f(1-x 2)>f(2x), i.e., 1-x 2>2x, and x>=0

So x 2 + 2 x - 1 < 0 and x > = 0

So 0<=x<-1+ 2

Solution: (1) Looking at the figure, we can see that the symmetry line is x=-1d(-2,3)(2) is obtained by two points (-2,3)(1,0).

So the expression for the function is x+y-1=0

3) Because the parabola crosses 3 points (1,0),(0,1),(0,3) there are 9*A - 3 b + c = 0, a + b + c = 0, c = 3

The solution yields a=-1 and b=-2

x^2-2x+3=0

Deriving it yields f(x).'=-2x-2 when x=1 f(x).'=-4 (slope).

So the straight line is y=-4x+4 and the image is.

When the minimum value of 1 2, 2 is obtainedRange [, obtained in x or 1 x... case the function values are close.

(1) When x is a positive integer, 1When x=1, [x]=1, [1x]=1, f(x)=1 2, i.e., the minimum value of f(x) is 1 2.

2。When x>=2, [x]=x, [1 x]=0, f(x)=(x+1 x) (x+1), let y=f(x)<1, then y=(x 2+1) (x 2+x), i.e., (y-1)x 2+yx-1=0 has a solution on x>=2, let g(x)=(y-1)x 2+yx-1

g(0)=-1, g(2)=6y-5, when the axis of symmetry x=-y 2(y-1) is in (0,2], that is, y<=4 5, only g(2)>=0 can be satisfied, and y>=5 6Therefore, this is not true.

When the axis of symmetry x=-y 2(y-1) is in [2, positive infinity), i.e., y>=4 5, only g(2)>=0 is satisfied, and y>=5 6 is solved, so the minimum value of y is 4 5

2) From (1), the range of the function f (x) when x is a positive integer is [4 5,1) when x is a fraction, 1When x is in the interval (0,1), [x]=0,(1 x-1)<=1 x]<1 x,,,.

The range of the function f (x) is (0,2).

2.When x is in the interval (1, positive infinity), [1 x]=0, x-1<=[x]<=x, the range of the function f (x) is .

In summary, the range of the function f (x) is [4 5,2].