Can you help me solve the math function problem in the first year of high school? Thank you.

From the second condition, we can know that his axis of symmetry is the line x=-1.

So we will primitiveize f(x)=a(x+b 2a) 2-1-b 2 4a

From condition 1, we can see that the image opening is upward, so a>0, then -b 2a=-1 and -1-b 2, 4a=-2

Solve a=1 b=2

f（x）=x^2+2x-1

f(x)=x^2-2x-1-kx^2-2kx+k=(1-k)x^2-2(1+k)x+k-1

To be -2 to 2 is a subtraction function.

1) Let's first discuss that when 1-k=0, i.e., k=1 f(x)=-4x, the subtraction function holds within the condition.

2) When 1-k>0 i.e. k<1 the image opening is upward, it is necessary to make it a subtraction function in the condition, i.e. [-2,2] is to the left of the axis of symmetry.

Therefore, the axis of symmetry x=(1+k) (1-k) is greater than or equal to 2, and the range of k is [1, 3,1).

3) When 1-k<0 i.e. k>1 image opens the countryside.

To make it a subtractive function within the condition, i.e. [-2,2] is to the right of the axis of symmetry.

So the axis of symmetry x=(1+k) (1-k) is less than or equal to -2 and the range of k is (1,3).

Therefore, the value range of k is [1, 3, 3].

Untie; (1) From the condition It is easy to know that the axis of symmetry of f(x) is the straight line x=-1, i.e., -b (2a)=-1, and b=2a......(1) From the condition f(x) [2, that is, the function has a minimum value of -2, i.e., (4ac-b) 4a=-2......（2）

The solution of Simultaneous (1) and (2) gives a=1 and b=2

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

2）f（x）=f（-x）-kf（x）=x²-2x-1-kx²-2kx+k=（1-k）x²-2x（1+k）-（1-k）

When k=1, f(x)=-4x, which decreases monotonically on r, which is in line with the topic;

When k≠1, the axis of symmetry of f(x) is the line x=(1+k) (1-k), δ4(1+k) +4(1-k) 0

If k 1, then the image of f(x) is open upward, and there is at this time.

1+k) (1-k) 2, solution k 1 3, k [1 3,1) If k 1, then the image opening of f(x) is downward, and there is .

1+k) (1-k) -2, k -3, k (1, in summary, k [1 3,

The axis of symmetry is the line x=-1.

Primitiveize as f(x)=a(x+b 2a) 2-1-b 2 4a

From condition 1, we can see that the image opening is upward, so a>0, then -b 2a=-1 and -1-b 2, 4a=-2

So a=1 b=2

f（x）=x^2+2x-1

f(x)=x^2-2x-1-kx^2-2kx+k=(1-k)x^2-2(1+k)x+k-1

To be -2 to 2 is a subtraction function.

1) Let's first discuss that when 1-k=0, i.e., k=1 f(x)=-4x, the subtraction function holds within the condition.

2) When 1-k>0 i.e. k<1 the image opening is upward, it is necessary to make it a subtraction function in the condition, i.e. [-2,2] is to the left of the axis of symmetry.

Therefore, the axis of symmetry x=(1+k) (1-k) is greater than or equal to 2, and the range of k is [1, 3,1).

3) When 1-k<0 i.e. k>1 image opens the countryside.

To make it a subtractive function within the condition, i.e. [-2,2] is to the right of the axis of symmetry.

The axis of symmetry x=(1+k) (1-k) is less than or equal to -2 and the range of k is (1,3).

The value range of k is [1, 3, 3).

1) Use the assignment method, so that m=n=0, substitute f(0+0)=f(0)f(0), f(0)=f(0), f(0)=f(0) or 1, when f(0) royal servant Xun = 0, any x r, there are f(x)=f(x)f(0)f(0)=0, at this time, f(x) is always 0, contradicting the question conditions, so f(0)=1

2) Let x<0, then -x>0, f(x+(-x))=f(x)talkingf(-x)=f(0)=1, f(x)=1 f(-x), because -x>0, so f(-x)>0, so 1 f(-x)>0, that is, when x<0, f(x)>0. When combined with the known x>0, f(x) >0, and f(0)=1>0, so x r, there is always f(x)>0

3) Because f(x) constant "0, you can use the business law, for any x1, x2=x1+a,a>0,f(x2) f(town this x1)=f(x1+a) f(x1)=f(x1)f(a) f(x1)=f(a)=f(a), because a>0, so f(a)<1, so f(x2).< f(x1) so f(x) is a subtraction function on r.

It is known in the problem that for any real number m, n exists f(m+n)=f(m)*f(n), then m,n can take any real number, you can set m=0, n=2 (you can set a real number at will, but one of m,n must be set to 0, because this problem makes the rock chaotic f(0)=1), then f(0+2)=f(2)*f(0), that is, f(2)=f(2)*f(0), and it can be seen from the question that f(bend 2) 0 is divided by f(2) at the same time. f(0)=1

The answer to the subject's screenshot is the same, but m is not set to a specific number, and both solutions are OK.

There are a few common themes and concepts to consider when it comes to the problem of higher level functions:

1.Definition and properties of functions: Understand the basic definition of functions, the characteristics of function images, and the various properties of functions, such as definition domains, value ranges, parity, monotonicity, etc.

2.Images and Transformations of Functions: Learn how to depict the image of a function through the formulas or characteristics of a function, and understand the basic transformations of the function image such as moving, stretching, and flipping.

3.Function Operations: Master the four rules of function operations, the concepts of composite functions and inverse functions, and how to perform operations and combinations between functions.

5.Exponential and logarithmic functions: Learn about the definitions, properties, and image characteristics of exponential and logarithmic functions, as well as how to solve exponential and logarithmic equations.

6.Trigonometric functions: familiar with the definition, properties, and image characteristics of trigonometric functions, and master the relationship between trigonometric functions and the basic methods of solving trigonometric equations.

7.Application of Functions: Understand the application of functions in practical problems, such as function modeling, the application of functions in physics, economics, etc.

These are just the basic directions of some of the higher level function problems, and the specific questions depend on the textbook and curriculum requirements. Si Wu Block.

All painted on the surface of the next trip, look! (all marked).

<>5, y=x -1 (the third function in the second row on your dust paper) <>

6. (first in the third row) <>

7. (The third faction Zhendan ranks second) <>

8. (third in the third row) <>

In this problem, it means that no matter how a changes, the image of this logarithmic function will always pass through this point, so if the value of the function is not affected by a, it is enough to make loga(x+2)=0, in this case, x=-1, y=3, that is, the image of the function is constant through the point (-1, 3).

（-1,3）

The fixed point of the logarithmic function is that the true number is 1, and no matter how many bases it is, the logarithm must be zero

The logarithm of 1 is equal to 0, and the fixed point has nothing to do with a, so the fixed point is (-1,3).

Let x+2=1 that is, x=-1, and y=3+0=3, so that the fixed point (-1,3) is constant

(-1,3) When x is equal to a negative time, the logarithm of the log base a is 0, so y is equal to 3. This is the case for this kind of problem, first let the logarithm equal to 0 and then find y, remembering that log with a base and 1 logarithm is 0.

(-1,3) is the standard function over (1,0), but this function moves 2 left and 3 up!

The one above me can basically be regarded as a qualified proof process, but the latter conclusion is a bit careless, and it is indeed monotonous and decreasing. Hope you understand

The question is Does b belong to (0,1)?

The element in b corresponding to the 60 degrees of the element in a is the root number of 2 thirds

The element in a, which corresponds to the root number two of the element in b, is 45°

Define the domain as r, which means that ax 2+ax+1 is everstable at zero, and let the function g(x)=ax 2+ax+1, so the function has no real root in r.

That is, the discriminant formula of the root is less than 0, a 2-4a<0

0. Hope it helps

The mapping from a to b is "find sinusoid", i.e. sina = b

1) sin60°=b, then b = third of the root number of the half.

2) sina = the root number of the second part, then a = 45°.

Because f(-1)=0, and f(x) 0 holds for any real number x.

So f(x) takes the minimum value at x=-1, then the parabolic opening is upward and the axis of symmetry is x=-1

That is, a>0 and x=-b 2a =-1 are solved: b=2a>0 and f(-1)=a-b+1=0

Solution: a=1, b=2

So f(x)=x +2x+1=(x+1) when x>0, f(x)=f(x)=(x+1) When x<0, f(x)=-f(x)=-x+1) In summary: The expression for f(x) is:

x+1)²,x＞0

f（x)=（x+1)²,x＜0

When 4x + 1 x + 2 and 4x + 1 2x + 4 i.e. x 1 3, f(x) =4x + 1 , the maximum value is f(1 3) =7 3 when x + 2 4x + 1 and x + 2 2x + 4 i.e. 1 3 x 2 3, f(x) =x + 2 , and the maximum value is f(2 3) =8 3

When -2x + 4 4x + 1 and -2x + 4 x + 2

That is, at x 2 3, f(x) = 2x + 4, and the maximum value is f(2 3) =8 3

In summary: The maximum value of f(x) is 8 3

Proof: Set x1>x2>0then x1 x2>1, so f(x1 x2)<0f(x1)-f(x2).

f(x1/x2*x2)-f(x2)

f(x1/x2)+f(x2)-f(x2)=f(x1/x2)

0 is f(x1) so f(x) is a subtractive function on r+.

This kind of problem can be done by building a functional model.

Proof: Let this function be f(x)=log(a)x

Because f(3)= -1

So a=1 3

So f(x)=log(1 3)x

Because 0 a 1

So this logarithmic function is a subtraction function on the fact that x is a positive real number (i.e. r+, as you are talking about).