High School Function Problems, Rush Rush !!

f(x-4)=f(x-2-2)=-f(x-2)=f(x), so the f(x) period is 4

f(40)=f(0)

f(25)=f(1)

f(7)=f(-1)

Odd functions, which are subtractive functions on [, are subtractive functions on [-2,2].

i.e. f(-1)> f(0)> f(1).

So there are: f(7)> f(40)>f(25).

You can use a small segment of y=-x or y=-x 3 to make a graph where f (x-2) = -f(x) is satisfied on r, so that x=x+2 gives f(x-2)=-f(x)=f(x+2), so the period is 4Draw the subtraction function of y=-x 3 over the interval [0,2], and use the odd function f(x) to get the function of y=-x 3 at [-2,0]. In this way, you can draw an extra positive direction for the period of the function, because 40 is ten times that of x=4, so f(4)=f(40)=0, f(25)=f(4x6+1) is less than 0; f(7)=f(4x1+3)=f(3) is greater than 0, so f(7)>f(40)>f(25).

It is simply not possible to deduce a symmetry about x=1!! Only when f(2-x)=f(x+2) can x=(2-x+x+2) 2=2But now it's f(x-2)=-f(x)=f(x+2), i.e. f(x-4)=f(x-2-2)=-f(x-2)=f(x), x can't be eliminated, there is no symmetry!!

f(x)=f(-x), and if x-2 is treated as x, then f(x-4) = f(x).

So f40=f(0)f(25)=f(1)f(7)=f3,f3=f-1,f-1=f1

Because 0-2 is a subtraction function, f40 = f7 f(25).

The period between the adjacent highest and lowest points is t 2 so t 2 = 7 12- 12

t= because 2 w=t = , so w=2 when f(x)=asin(wx+), the maximum and minimum values of the function are related to the value of a. Because a is greater than 0, the maximum value is a and the minimum value is -a

So a+b=3, -a+b=-1

So a=2, b=1

So f(x)=2sin(2x+)1

Substituting the point (12,3) as a function yields: sin 6+ = 1 so 6+ = 2+2k ,k z

And because |ф|/2

So = 3

So f(x)=2sin(2x+3)+1 because x[ 2, so 2x+ 3 [ 4 3,7 3 ].

Since the monotonically increasing interval of the sinusoidal function is [ - 2+2k , 2+2k ], the unitonic increasing interval of f(x) is [ 3 2, 7 3 ] and because f(x) = 2sin(2x+ 3)+1=0, so sin(2x+ 3)=-1 2

So x=5 3

From the question to know:

When x 3, f(x) = (1 2) x

When x<3, f(x) = f(x+1).

and 3=log2 2 =log2 8 The first 2 is the base, and 8 is the true number] so f(log2 3)=f(1+log2 3)=f(log2 2*3).

f(log2 6) log2 6 log2 8 =3】=(1/2)^(log2 12)

2^(-log2 12)

2^(log2 1/12)

i.e.: f(log2 3) = 1 12

Because log(2) 3<2 , we take the following formula known to become x=log(2)3+2=log(2)12, and bring it into the above formula. The answer is -12.

The derivative is obtained to the negative x power of f(x) derivative = 2ln2*2.

Since the negative x power of 2 is greater than 0 constant, f(x)>0 constant.

That is, the function is an increment function.

I'm sorry I haven't studied math for a few years, but I can give you some advice.

x∈r)..So the easiest way to do this is to bring in a value. Then you can solve it by drawing through the roots.

2 x is an increment function, so 1 2 x is a subtraction function, so -1 2 x is an increment function.

Definitional solution: f(x)=a-2 (2 x)+1,(x r)

Take x10 at will, so it's a single increment. Conversely, a decreases monotonically when a is less than or equal to 2.

Circle: x 2 + y 2-4 x = 0, i.e. (x-2) 2 + y 2 = 4 So, the center of the circle is (2,0) and the radius is 2

Let the center of the moving circle (x,y) and the radius be r, and the circle is tangent, then (x-2) 2+(y-0) 2=(r+2) 2 and tangent to the y axis, then |x|=r

The two-formula simultaneous equation yields y 2=4|x|+4x

When x>0, the trajectory equation of the dynamic circle equation is y 2=8x, (which is a parabolic equation)When x<0, the trajectory equation of the dynamic circle equation is y=0, (which is a ray).

The center of the circle x 2+y 2-4x=0 is (2,0), the radius is 2, and the center of the moving circle is (x,y) and the radius is r.

x-2)^2+y^2=r^2

x|=r, Polaroid(x-2) 2+y 2=x 2

Simplification yields y 2=4(x-1).

Let the coordinates of the center of the circle be (x,y).

Since tangent to the y-axis, the radius is the absolute value of x.

Then the centroid distance of the 2 circles is the square of (x-2) + the square of y is equal to the absolute value of x + 2, and then the classification and discussion can be solved by combining the image.

This curve is a parabola and the negative semiaxis of x.

The center of the circle of garden x square + y square - 4x = 0 is (

1.When the garden is on the right side of the y-axis, the center of the garden is equal to the distance between the center of the known garden and the y-axis According to the definition of parabola (standard parabolic equation: if the distance from each point on the curve to a certain point and a straight line is equal, then the curve is parabolic), then the equation is x-squared = 8y (x= =0).

2 The geometry is derived from y=0 (x is less than 0), which also meets the conditions, so the solution is obtained.

f'(x)=-2(2 xln2) 2 x+1) 2<0 function f(x) is a monotonic subtraction function in the deterministic deficiency of the spike domain.

Let g(x)=f(x)-1=2 (2 x+1)-1=(1-2 x) (1+2 x).

g(-x)= 1-2^(-x)]/1+2^(-x)] 2^x-1]/[1+2^x]=-1-2^x)/(1+2^x)

The function g(x) is the Lubqi function.

The existence of a -1 makes the function f(x)+a odd.

Solution: y=3+2·3 (x+1)-9 x=3+2*3 x*3 1-(3 2) x=3+6*3 x-(3 x) 2

Let t=3 x,1 3<=t<=9

y=-t 2+6t+3,1 3<=t<=9y, the opening is downward, and the axis of symmetry is t=3

So the maximum value is y=12, when t=3 is taken.

When t=1 3, y=-44 9; When t=9, y=-24, the minimum value is y=-24, and when t=9, it is taken.

So the maximum value is y=12, which is taken when x=1;

The minimum value is y=-24 and x=2.

y=3+2 3 (x+1)-9 x=3+6 3 x-3 (2x), let 3 x=t [1 3,9],y=-t +6t+3=-(t-3) 12t=3, the function takes the maximum value of 12, and when t=9, the function takes the minimum value of -24

Let t=x 3, t=[1 3,8], so that y=3+6t-t 3, draw a sketch of the 6t and t 3 figures, and the maximum and minimum values will come out.

x 3 is a single increment function.

I don't know if it's beyond the content of the higher one function,

Finding the derivative directly will make the derivative zero, and then find the boundary value x=-1 y=44 9 x=2 y=-24, and compare it to get x=1, the maximum y=12, x=2, and the minimum y=-24 as the answer.

I think this topic is what I did in junior high school.

ax1^2+bx1+c=0

ax2 2+bx2+c=0, so -ax1 2=bx1+c, the same way, ax2 2=bx2+c

Let f(x) = (a 2) x 2 + bx + c

then f(x1)=ax1 2 2+bx1+c

f(x2)=ax2^2/2+bx2+c

Put -ax1 2=bx1+c

ax2^2=bx2+c

Substitution gets. f(x1)=-a*x1^2/2

f(x2)=3ax2^2/2

Because x1, x2 is not equal to 0

So x1 20, x2 20

Unary quadratic equations.

So a 20

So f(x1)*f(x2).

3a^2*x1^2*x2^2/40

That is, f(x1) and f(x2) are positive and negative.

So there must be a point between x1 and x2 that intersects with the x-axis.

So there must be one between x1 and x2.

The first problem can use the inequality a+b>=2 root sign (ab) (a>0, b>0, and the equal sign when a=b). Since x<-2, x+2<0. The original formula becomes y=(x+2)+1 (x+2)-2<2-2=0 (note that the domain of x is <-2, so there is no =) The value range is;

In the second problem, a variable t=root(1-2x) and t>=0 can be used to make x=(1-t 2) 2, and the original formula can be reduced to y=(1-t 2+2t) 2, which can be solved according to the range of the quadratic function of t>=0.

The first one is split into y=1-1 x+2 because x -2 so x+2 0 so -1 x+2 0 so y 1