Quadratic function f (x) = ax ²+ The coefficients a, B and C of BX + C are not equal to each other. If 1 / A, 1 / B and 1 / C form an equal difference sequence, a, C and B form an equal ratio sequence, and the maximum value of F (x) on [- 1,0] is - 6, then a= Hope to have a detailed answer. Thank you

Quadratic function f (x) = ax ²+ The coefficients a, B and C of BX + C are not equal to each other. If 1 / A, 1 / B and 1 / C form an equal difference sequence, a, C and B form an equal ratio sequence, and the maximum value of F (x) on [- 1,0] is - 6, then a= Hope to have a detailed answer. Thank you

1. It is known from the equal difference that 2 / b = 1 / C + 1 / B, that is, 2Ac = AB + BC
2. From the equal ratio: C ²= ab
3. Substitute 2 into 1: a = (c + b) / 2
4. Substituting 3 into 2: 2C ²- cb-b ²= 0 is (2C + b) * (C-B) = 0
Because ABC is not equal to each other, B = - 2C = 4A, C = - 2A
5. The function expression is formulated as f (x) = a (x + B / 2a) ²+ [(4ac-b ²）/ 4a]
According to the simplification of B = 4A and C = - 2A
f（x）=a（x+2) ²- 6a
From this, we can find the symmetry axis of F (x)
hypothesis
1) . a > 0, then f (x) takes the maximum value when x = 0
f（x）=a（0+2） ²- 6a=-6
a=3
b=12
c=-6
2）a

Quadratic function f (x) = ax ²+ The image opening of BX + C is downward, and (- a) is satisfied. B.C is an isometric sequence, a, B, (A-C) Quadratic function f (x) = ax ²+ The image opening of BX + C is downward, and (- a) is satisfied. B.C is an equal difference sequence, and a, B, (A-C) is an equal ratio sequence. Try to find the solution set of inequality f (x) ≥ 0

-a+c=2b
a-c=-2b
a(a-c)=b ²
a(-2b)=b ²
-2a=b
-a+c=-4a c=-3a
f(x)=ax ²+ BX + C opening downward, so a = 0
ax ²- 2ax-3a>=0
x ²- 2x-3

Known function f (x) = loga (a ^ x-1) (a > 0 and a ≠ 1) 1) verify that the image of function f (x) is on one side of Y axis 2) find the value range of X that makes f (x) > 0

Tell the landlord about the steps
① F (x) is on one side of the y-axis, indicating that the definition domain is on the positive or negative half axis of X. calculate the definition domain
② There is the definition field and judgment function calculated above. This function is a composite function of elementary functions, which can be seen
The detailed process is as follows:
① 00 or x0,
When 0

Given the function f (x) = loga (a ^ x-1) (a > 0 and a ≠ 1) 1), it is proved that the image of function f (x) is on one side of Y axis (2) any two points on the image of function f (x)

The first a in the question should be based on a
According to the definition field of logarithmic function, so a (a ^ x-1) > 0,
Because a > 0
So a ^ x < 1
So when 00
When 1, the image of the function f (x) is on one side of the Y axis
Suppose X1 if 0A ^ X2, a ^ x1-1 > A ^ x2-1
So loga (a ^ x1-1) so [loga (a ^ x1-1) - loga (a ^ x2-1)] / (x1-x2) > 0
If [loga (a ^ x1-1) - loga (a ^ x2-1)] / (x1-x2) > 0
Therefore, the slope of any two points on the function f (x) image is greater than 0

1 function y = loga (2x-1) + 2 a > 0 and a ≠ 1 what is the constant crossing point?

When x = 1, y = loga1 + 2 = 2
Crossing point (1,2)

Find the solution of a mathematical problem: the known function f (x) satisfies f (x + 1) = (1 + F (x)) / (1-f (x)). If f (1) = 2010, find f (2011)

This is obviously a recursive formula. You can first know that f (x + 2) = (1 + F (x + 1)) / (1-f (x + 1)), and then substitute f (x + 1) = (1 + F (x)) / (1-f (x)) into the above formula. You can know that f (x + 2) = - 1 / F (x). If you continue to recurse, you can know that f (x + 2) = f (X-2), that is, f (x + 4) = f (x), so f (2011) = f (3 + 4x502) = f (3) Don't let me talk about it next