Let f (x) = ax ^ 2 + BX (a is not equal to 0) satisfy the following conditions: F (x) = f (- 2-x), the equation f (x) = x has two equal real roots (1) find the analytic expression of F (x) (2) if the inequality π ^ f (x) > (1 / π) ^ (2-tx) is in | t|

Let f (x) = ax ^ 2 + BX (a is not equal to 0) satisfy the following conditions: F (x) = f (- 2-x), the equation f (x) = x has two equal real roots (1) find the analytic expression of F (x) (2) if the inequality π ^ f (x) > (1 / π) ^ (2-tx) is in | t|

（1）
∵ f (x) = f (- 2-x), f (x) is a quadratic function
∴f(0)=f(-2)
The axis of symmetry of F (x) is x = - 1
That is - B / (2a) = - 1, B = 2A
f(x)=ax²+2ax
The equation f (x) = x has two equal real roots
That is ax & # 178; + 2aX = x,
That is, ax & # 178; + (2a-1) x = 0 has two equal real roots
∴2a-1=0
∴a=1/2
∴f(x)=1/2x²+x
(2)
Inequality π ^ f (x) > (1 / π) ^ (2-tx)
That is, π ^ f (x) > π ^ (TX-2)
That is f (x) > TX-2
That is 1 / 2x & # 178; + x > TX-2
That is, xt-1 / 2x & # 178; - X-2 X & # 178; - 2x + 4 > 0 = = > x ∈ R
The value range of real number x is x-3 + √ 5
The difference between the definition of junior high school function and senior high school function
I know the difference between definition and concept
There is a difference, junior high school definition only emphasizes that x is an independent variable, y is a dependent variable, and the learned functions such as first-order, quadratic, proportional, inverse function and other basic functions
In senior high school, the concept of mapping is used in the definition of functions, and the functions learned are also upgraded, including exponential, logarithmic, trigonometric and other functions that are required in the college entrance examination
difficulty
It is known that the vertex coordinates of the quadratic function y = f (x) are (- 32,49), and the difference between the two real roots of the equation f (x) = 0 is equal to 7
Let y = a (x + 32) 2 + 49, let y = 0, we can get AX2 + 3ax + 94A + 49 = 0, ∧ X1 + x2 = - 3, x1 · x2 = a + 4 × 494A, ∧ x1-x2 | = (x1 + x2) 2 − 4x1x2 = 9 − 9A − 49 × 4 = 7, the solution is a = - 2369
What are the essential differences between the concept of function in junior high school, the concept of function in senior high school and the concept of function in university?
The concept of function is essentially the relationship between an "object" and another "object". An object can have one or two "variables". It depends on how much you can understand and whether you can use them. This is the difference and connection
The essence is that junior high school students are different from senior high school students and college students. Different people give you different courses_ (laughter)~
Junior high school definition: function is a corresponding relationship, the main research analytic
High school definition: function is a kind of corresponding relationship based on two nonempty number sets, studying the three elements of function
The definition of function in university is the same as that in high school, but the research scope is wider, including multivariate function, complex variable, real variable function and so on
Junior high school, high school function is fur! It's the most basic part. University function has engineering significance and practical use! You want to be an engineer? University function must be! University of all kinds of functions to learn ~ ~ ~ after the course to use! It's all connected! Work hard, young man!
You'll get it when you get into the math department of the University. You'll know better when you graduate from college.
Given that the function f (x) = x / ax + B (a is not equal to 0), f (2) = 1, and the equation AX & # 178; + (B-1) x = 0 has two equal real roots, find f (x)
Don't copy and paste on Baidu, OK? Some questions are wrong
From the function f (x) = x / ax + B (a is not equal to 0), f (2) = 1, we know that f (2) = 2 / (2a + b) = 1, i.e. 2A + B = 2. ① from the function f (x) = x / ax + B (a is not equal to 0), f (2) = 1, we know that f (2) = 2 / (2a + b) = 1, i.e. 2A + B = 2
Basic concepts of function in junior high school
If the quadratic function y = f (x) satisfies f (3 + x) = f (3-x), and the equation f (x) = 0 has two real roots x1, X2, then X1 + X2 is equal to?
On quadratic function
Let X1 be the root of the equation f (x) = 0, then f (x1) = 0
Because f (3 + x) = f (3-x)
So f (x1) = f [3 + (x1-3)] = f [3 - (x1-3)] = f (6-x1) = 0
So x = 6-x1 is also the root of the equation f (x) = 0, that is, X2 = 6-x1
So X1 + x2 = 6
How to define the concept of function in junior high school
Elementary function is a function which is generated by power function, exponential function, logarithmic function, trigonometric function, anti trigonometric function and constant through finite rational operation and finite function combination, and can be expressed by an analytical formula
Function is not defined in junior high school
It seems that there is no high school definition: each number in a has a unique corresponding quantity in B
In a change process, there are two variables X and Y. for each value of X, the variable y has a unique value corresponding to it. We call x an independent variable, y a dependent variable, and y a function of X
That's what we teach
Given the function f (x) = x power of a + B (a > 0, and a is not equal to 1, it is a real number), what is the analytic expression of F (x) when the image passes through points (1,7) and (0,4)
F(x)=a^x+b
F(1)=a+b=7
F(0)=a^0+b=1+b=4
b=3 a=4
F(x)=4^x+3
Given the function a (x) = f (x) + G (x), where f (x) is the positive proportional function of X and G (x) is the inverse proportional function of x)
And a (1 / 3) = 16, a (1) = 8
(1) Find the analytic expression of a (x) and point out the domain of definition
(2) Finding the range of a (x)
Let f (x) = K1 * x; G (x) = K2 / X
a(1/3)=1/3*k1+3*k2=16
a(1)=k1+k2=8
The solution is K1 = 3
k2=5
a（x）=3x+5/x
Domain (x is not equal to 0)
Range (negative infinity, positive infinity)
Let f (x) = ax, G (x) = B / X
A (1 / 3) = A / 3 + 3B = 16, a (1) = a + B = 8, a = 3, B = 5
A (x) = 3x + 5 / x, the definition field is x not equal to 0
(2) The range is (- ∞, - 2 √ 15] ∪ [2 √ 15, + ∞)
a（x）=3x+5/x
We know that a (1 / 3) = f (1 / 3) + G (1 / 3) = 16,
a（1）=f（1）+g（1）=8 ，
Let f (x) = KX g (x) = H / X
So (1 / 3) K + H / (1 / 3) = 16
k+h=8
So k = 3, H = 5
You should be able to do it yourself
F (x) is a positive proportional function of X
Let f (x) = KX
G (x) is the inverse proportional function of X
Let g (x) = m / X
Then a (x) = f (x) + G (x) = KX + m / X
And a (1 / 3) = k * 1 / 3 + m / (1 / 3) = K / 3 + 3M = 16
a(1)=k*1+m/1=k+m=8
It is easy to get k = 3
M=5
∴a（x）=3x+5/x
∵ g (X... expansion)
F (x) is a positive proportional function of X
Let f (x) = KX
G (x) is the inverse proportional function of X
Let g (x) = m / X
Then a (x) = f (x) + G (x) = KX + m / X
And a (1 / 3) = k * 1 / 3 + m / (1 / 3) = K / 3 + 3M = 16
a(1)=k*1+m/1=k+m=8
It is easy to get k = 3
M=5
∴a（x）=3x+5/x
∵ g (x) = 5 / X is an inverse scale function, and the definition field of ∵ g (x) is x ≠ 0
For f (x) = 3x, R is defined
To sum up. The definition field of a (x) = f (x) + G (x) is x ≠ 0
a（x）=3x+5/x
PS: for ax + B / X. it's the easiest and easiest to get the range by using the mean inequality (or you can think of it as a hook function)
When x > 0. a(x)=3x+5/x≥2√(3x*5/x)=2√15
(if and only if 3x = 5 / x, the equal sign holds. That is, when x = (√ 15) / 3)
When x < 0. -x＞0.a(x)=-(-3x-5/x)≤-2√（（-3x）*5/(-x)）= -2√15
If and only if - 3x = - 5 / x, the equal sign holds. When x = - (√ 15) / 3)
∪ a (x) ∈ (negative infinity, - 2 √ 15] ∪ [2 √ 15, positive infinity)