In this paper, we have a problem about the function of higher one. We know that the quadratic function f (x) satisfies f (0) = 1 and f (x + 1) - f (x) = 2x 1. Find the analytic expression of F (x)

In this paper, we have a problem about the function of higher one. We know that the quadratic function f (x) satisfies f (0) = 1 and f (x + 1) - f (x) = 2x 1. Find the analytic expression of F (x)

If f (x + 1) - f (x) = 2x, then f (x) - f (x-1) = 2 (x-1), f (x-1) - f (X-2) = 2 (X-2) F (1 〉 - f (0 〉 = 2x0 all add up to f 〈 x) - f 〈 0) = 2 〔 0 + 1 + +X-1)=X（X-1）
It is known that f (x) is a function defined on R. for any x, y belonging to R, f (x + y) + (X-Y) = 2F (x) f (y), and f (o) is not equal to 0
(1) Verification: F (0) = 1
(2) Judging the parity of function f (x)
1. Let x = y = 0, so the meaning is: F (0) + F (0) = 2 (f (o)) ^ 2 --- > 2F (0) = 2 (f (o)) ^ 2, because f (0) ≠ 0 --- > F (0) = 12.2f (x) f (y) = f (x + y) + F (X-Y) 2F (x) f (- y) = f (X-Y) + F (x + y) -- > 2F (x) f (y) = 2F (x) f (- y) -- > F (0) f (- y) -- > F (y) = f (- y) by any of Y
It is known that the image of quadratic function passing through points a (3,3), B (4,0) and the origin O.P. is a moving point on the image of quadratic function
As shown in the figure, it is known that the image of the quadratic function passes through points a (3,3), B (4,0) and the origin O. P. it is a moving point on the image of the quadratic function. The vertical line passing through point P is the vertical line of X axis, and the vertical foot is d (m, 0), and it intersects with the straight line OA at point C
(1) The analytic expression of quadratic function is obtained;
(2) When the point P is above the line OA, the maximum value of the line PC is obtained;
(3) When m > 0, explore whether there is a point P, so that △ PCO is an isosceles triangle. If there is, find out the coordinates of p; if not, explain the reason
(1) Let y = ax (x-4), substituting point a coordinates (3,3) into a = - 1, the analytic expression of the function is y = - X & # 178; + 4x, (2) 0 ＜ m ＜ 3, PC = Pd-Cd, = - M & # 178; + 3M, = - (M-3 / 2) &# 178; + 9 / 4, ∫ - 1 ＜ 0, the opening is downward, the maximum value of ∫ is, when d (3 / 2, 0), pcmax = 9 / 4, (3) when 0 ＜ m ＜ 178
What about the picture
The function defined on R is y = f (x), f (0) is not equal to 0. When x > 0, f (x) > 1, and for any a, B belonging to R, f (a + b) = f (a) times
f（b）
Question: (1) proof: for any x belonging to R, f (x) > 0
Problem (2) f (x) is an increasing function on R (3) if f (x) is multiplied by F (2x-x ^ 2) > 1, find the value range of X
(1) It is proved that: for any a, B, f (a + b) = f (a) f (b) since a, B can be any real number, let a = b = 0 have: F (0 + 0) = f (0) f (0), that is: F (0) = f (0) & sup2;, ∧ f (0) = 0 or 1, and f (0) ≠ 0, ∧ f (0) = 1, let x < 0, then: let a = x, B = - X have: F (x
It is proved that let a = 0, B = 1, so f (0 + 1) = f (0) * f (1) get f (0) = 1
Because when x > 0, f (x) > 1, so when - x0, f (x) > 1, so 0
What does the mathematical function f (x) mean?
F (x) is a function with X as the independent variable. For example, y = x can also be written as f (x) = x, which means the same. F (a) = 0, which means that in this function f (x), when x = a, the function value is 0. The factorial theorem is to find a satisfying the condition of F (a) = 0, and the process of finding can be calculated orally
It's a function expression
It can be understood as a substitute symbol
For example, if f (x) = 2x + 3, then any X in the domain of definition is applicable to 2x + 3, and Y is equal to 2x + 3; for example, if x = 3, then y = f (3) = 9
Do you understand?
Ha ha, my own understanding
f(x)=。。。
Equivalent to junior high school
y=...
The image in the mapping is a kind of correspondence;
Set is easier to understand;
All values of X are a set;
All the values of F (x) are a set;
Each x gets a corresponding f (x) after a certain operation;
For example: F (x) = 3x + 2;
That is to say, X is multiplied by 3 and then added to 2 to get f (x)... Expansion
The image in the mapping is a kind of correspondence;
Set is easier to understand;
All values of X are a set;
All the values of F (x) are a set;
Each x gets a corresponding f (x) after a certain operation;
For example: F (x) = 3x + 2;
That is to say, X is multiplied by 3 and then added to 2 to get f (x) folded
F is a corresponding way. For example, f (x) is the value of the independent variable X after some operation. For example, if f (x) = 2x + 1, then the corresponding way represented by F is the double of the independent variable plus 1; if f (x) = x ^ 2, then f means that the operation is to square the independent variable. In the future, you will encounter many representations like g (x) or F (x) and H (x), which only represent a kind of correspondence. In addition, f (x) can not only express the function relation (such as f (x) = 2x + 1), but also express the value containing in general (that is, when... Is expanded)
F is a corresponding way. For example, f (x) is the value of the independent variable X after some operation. For example, if f (x) = 2x + 1, then the corresponding way represented by F is the double of the independent variable plus 1; if f (x) = x ^ 2, then f means that the operation is to square the independent variable. In the future, you will encounter many representations like g (x) or F (x) and H (x), which only represent a kind of correspondence. In addition, f (x) can not only express the functional relationship (for example, f (x) = 2x + 1), but also express the numerical value in general (that is, when x is a certain value, f (x) represents the result calculated in the corresponding way at that time). Put it away
The function defined on R is y = f (x), f (0) is not equal to 0. When x > 0, f (x) > 1. For a and B belong to R, f (a + b) = f (a) + F (b)
1 prove that f (0) = 1, 2 for any x belongs to R, f (x) > 0.3 prove that f (x) is an increasing function on R. 4 if f (x) multiplies f (2x-x cottage) > 1, find the value range of X
In my opinion, there should be some mistakes in the title, which should be followed by F (a + b) = f (a) * f (b), so that every question can be done. (1) let a = b = 0, then from F (a + b) = f (a) * f (b), f (0) = F & sup2; (0), f (0) = 1 or F (0) = 0 and f (0) ≠ 0, so f (0) = 1, the proposition proves (2) when x > 0, from the known f (x) > 1 > 0
1. F (1 + 0) = f (1) + F (0) f (1) > 0, so f (0) = 0
F (a + b) = f (a) + F (b)
It's multiplication
F
The title is right! We're doing the same thing
Let f (x) = {above is 1-x square, X ≤ 1; below is x square + X-2, x > 1, find f (1 / F (2)
When ∵ x > 1, f (x) = x ^ 2 + X-2,
∴f(2)=4+2-2=4
∴f(1/f(2))=f(1/4)
∵x
If it's 1-x
It is known that f (x) is a function defined on R. for any x, y belonging to R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0
Judging the parity of function
F (0 + 0) + F (0-0) = 2F (0) f (0), and f (0) is not equal to 0, so f (0) = 1 is not an odd function, and f (x + x) + F (x-x) = 2F (x) f (x), f (x-x) + F (x + x) = 2F (x) f (- x), then 2F (x) f (x) = 2F (x) f (- x), and f (x) is not always zero, so f (x) = f (- x), the function is even
Given the function f (x) = x + A / x + A, X belongs to [1, positive infinity), and A0 is constant, find the value range of real number a
① Let X1 > x2
f（x1）-f（x2）=x1+a／x1-x2-a／x2
=（x1-x2）（1-a／x1x2）
∵x1＞x2 ∴x1-x2＞0
∵ x1, X2 ≥ 1, a is less than 1, so a / x1x2 < 1
∴（1-a／x1x2）＞1
If f (x 1) > F (x 2), f (x) is an increasing function
② If M satisfies f (3m) > F (5-2m), f (x) is an increasing function
∴3m＞5-2m ∴m＞1
③g（x）=x²+ax+a
Let f (x) = g (x) + 2x + 1.5 = x & sup2; + (a + 2) x + A + 1.5
F (x) is always greater than 0 in [2,5]
∴F（2）＞0,F（5）＞0
∴a＞-19／6
(1) Simple increase because x is simple increase, a / X is simple increase, so f (x) is simple increase
(2) 3M > 5-2m, 3M > = 1 5-2m > = 1
SO 2 = > m > 1
(3) x^2 + (2+a)x + a+1.5 > 0
Discussion according to the situation, (1) 20
If you ask a question like this, it will be a meeting, but how about the next one
Given that the function f (x) satisfies f (x-1) = loga (x + 1) \ \ (3-x) (a > 0 and a is not equal to 1), (1) find the analytic expression of F (x)
Let X-1 = t
x=t+1
f(t)=loga(t+2)/(2-t)
f(x)=loga(x+2)/(2-x)
do you understand?
I don't know. I'm telling you