The known function f (x) = | log3 ^ x |, 0

The known function f (x) = | log3 ^ x |, 0

Set 0

If the function f (x) = absolute value lgx, if 0 < a < B and f (a) = f (b), then the value range of a + 2B
I calculate 2 √ 2 to positive infinity, but the answer is 3 to positive infinity. I use the mean inequality to calculate that 2B + 1 / B is greater than or equal to 2 √ 2. What's wrong?

2 √ 2 is the value when a = √ 2, but actually the range of a is 0

Given function f (x) = LG (x ^ 2 + ax-a-1)
The following propositions are given
The function f (x) has a minimum
2 when a = 0, the range of function f (x) is r
3 if the function f (x) increases monotonically in the interval [2, positive infinity], then the value range of real number a is a > = - 4
The correct proposition is
Say everything right and wrong

y0=x^2+ax-a-1 A
y=lgy0 B
Formula A, axis of symmetry x = - A / 2:
1. From the image of a function, we know that: when x ≤ - A / 2, X ↑ Y0 ↓. B is a monotone increasing function, Y0 ↓ y ↓, that is, X ↑ y ↓; when x ≥ - A / 2, X ↑ Y0 ↑, combined with B monotone increasing, we can get x ↑ y ↑. It seems that there is a minimum value, Y0 = x ^ 2 + ax-a-1 = (x + A / 2) ^ 2 - (5 / 4A ^ 2 + 1). Obviously, the minimum value is just below the x-axis. So, sorry, it's wrong to have a minimum value for function f (x);
2,1 has mentioned that the function f (x) = LG (x ^ 2 + ax-a-1) is a function that decreases first and then increases if Y0 > 0 is not considered. How the range of value can be r has nothing to do with a = 0. No matter how much a is equal to, the range of value can not be r, let alone the restriction of Yo > 0;
3, monotone increasing interval [2, + infinity). We mentioned that x ≥ - A / 2 is monotone increasing, but considering the condition of Y0 > 0, we can only take the right intersection of Y0 = x ^ 2 + ax-a-1 and X axis as the starting point of monotone increasing, because the middle section of two intersections of function Y0 = x ^ 2 + ax-a-1 is below X axis
When Y0 = 0, x = omitted, and the root will not be marked, which is recorded as x1, x2;
+If the root is less than 2, note that it can't be + followed by = 2, because half of the edge is [, so a = - 4 of 3 may be wrong;
The actual range can be calculated

It is known that the maximum value of the quadratic function is 2, the vertex of the image is on the line y = x + 1, and the image passes through the point (3, - 1)
No, thank you

The maximum value of quadratic function is 2, that is, y = 2, and the vertex of image is on the line y = x + 1
By substituting y = 2 into y = x + 1, the vertex coordinates of the x = 1 parabolic curve are (1,2)
Let the analytic expression of quadratic function be y = a (x-1) & sup2; + 2, and substitute x = 3, y = - 1
The analytic expression of a = - 3 / 4  quadratic function is y = - 3 / 4 (x-1) & sup2; + 2