(1) According to the information given in the question: a, quadratic function; B, the minimum value is-1/4.
According to the above two conditions, the standard analytical formula of quadratic function can be listed, namely
f(x)=A(x+B)^2- 1/4
Then pass through two points according to the function and substitute into the above analytical formula, two unknowns and two equations. This should be no problem for you.
(2) The information given by the topic is limited, and the quadratic function does not give the maximum value, so the first step to solve this problem should be to set the standard analytical formula of the function, namely:
F (x) = a (x+b) 2+c ———— Since no extreme value is given, the constant term can only be set as the unknown c.
Therefore, to solve this problem, three unknowns must have three equations, that is, three points must be obtained, and the topic has given an important information, that is, a point, that is, f(0)=2, that is, the function must pass through this point (0,2). At this time, you need to get the other two according to this feature and f(x-2)=f(x)+4 of the function.
A, let x=2 and substitute f(x-2)=f(x)+4 to get f (0) = f (2)+4 = = > f (2) =-2.
B, let x=0 and substitute f(x-2)=f(x)+4 to get f (-2) = f (0)+4 = = > f (-2) = 6.
Therefore, the function must pass through three points (-2,6), (0,2) and (2,2) and be substituted into f (x) = a (x+b) 2+c, and three unknowns of three equations will be solved.
(3)a, let x=2, and substitute it into solvable f(0)
B, let x=4 and substitute it into solvable f(2)
C, you may be misled by x-2 and X, but they are just variables and have no real relationship. To change our thinking, we just need to regard X in f(x) as T. Is this problem easy to solve? So as long as shilling t=x-2, substitute f(x-2) = X 2-4x+8, and finally get f(t). Since they are all variables, it doesn't matter what letters you use to replace them, then you can restore T to X, but when you restore T to X, don't change it to t+2. After all, they are all variables. You just need to simply change t to X.
Methods: let t=x-2, then x=t+2. (Note: at this time, it is not the final variable replacement, but the operation between variables, so the transformation in the process should be carried out through four operations, and only the final variable replacement can be directly replaced. One is the operation in the process, and the other is the replacement of the result, which is different. )
Substituting f (x-2) = x 2-4x+8, we can get
f(t)=(t+2)^2-4(t+2)+8
=t^2+4t+4-4t-8+8
=t^2+4
Results: F (x) = x 2+4 substitution with variables.
(4) According to the characteristics of parity function, we can find the solution:
Odd number function: f(x)=-f(-x)
Even function: g(x)=g(-x)
Because f(x)+g(x)= 1/x- 1, we can get it by substituting x=-x into the above formula.
F (-x)+g (-x) =-1/x-1formula a.
According to the characteristics of parity function, Formula A can be transformed into:-f (x)+g (x) =-1/x-1.
f(x)+g(x)= 1/x- 1
-f(x)+g(x)=- 1/x- 1
Both equations need f(x). Just cancel g(x). You must know how to cancel it yourself.
(5) The answer to this question should be a piecewise function if I get it right. Let me talk about my own solution first:
It is mentioned in the stem that f(x) is a odd function over real number field R, then f(x)=-f(-x) is satisfied.
When the analytical formula of x>0 is f(x) = x 2+x+ 1. Assuming that X satisfies this analytical formula in the whole real number field (both bold assumption and careful verification are possible), we can use this analytical formula to verify the values of two mutually opposite X's and see if their F (x) values are opposite numbers, and then we can know whether our hypothesis is correct. Take simple x= 1, x=- 1 to calculate.
When x= 1, f (1) =1+1= 3;
When x=- 1, f (-1) =1+1=1; Obviously, f (1) =-f (-1) = = >; The assumption is not true, so we can know that when X.
Because the topic only gives x>0, it is obvious that the topic wants us to give the overall analytical formula on the real number field R, so the variable X is required.
Make t<0, then-t > 0, due to-t > 0, obviously can satisfy the definition domain of f (x) = x 2+x+1,so we can substitute -t=x into the above formula, and after substitution, we get: f (-t) = (-t) 2-t+ 1.
=t^2-t+ 1——————————& gt； Formula b
Because f(t) also satisfies the odd function in the real number field R, then substituting f(-t)=-f(t) into formula B, we can get:
F (t) =-t 2+t- 1, variable substitution can be obtained:
F (x) =-x 2+x-1,and the domain at this time is: x.
Play x>0 and x
f(x)=x^2+x+ 1 x & gt； 0
f(x)=-x^2+x- 1 x & lt； 0
This problem seems to ignore a point, that is, x=0, which also belongs to the real number field R. In fact, there is a jump at the point where x=0, that is, f(0)= 1, f(-0)=- 1, because f(0) and f(-0) are actually the same point, but
f(x)=x^2+x+ 1 x & gt； 0 = = = = = = = => The image is a semi-parabola in the first quadrant.
f(x)=-x^2+x- 1 x & lt； 0 = = = = = = = => The image is a semi-parabola below the fourth quadrant.
This answer is for personal opinion and reference only. Please don't take it personally. If you have any other questions, you can also ask me, thank you!