For the function f (x), if there exists x0, such that f (x0) = x0, then x0 is called the fixed point of the function. The quadratic function f (x) = ax ^ 2 + (b-2) x + B + 1 is known

For the function f (x), if there exists x0, such that f (x0) = x0, then x0 is called the fixed point of the function. The quadratic function f (x) = ax ^ 2 + (b-2) x + B + 1 is known

This is the third time that I have answered this question, and you don't write it all down. Please see here. This is a new type of question that puts forward new concepts and investigates the ability of innovation and understanding. The college entrance examination must take. (1) - 2,3 is the fixed point, then f (- 2) = - 2, f (3) = 3, the solution is a = 1, B = 6F (x) = (x + 6) / X. the zero point of F (x) is

For function f (x), if there exists x0 ∈ r such that f (x0) = x0 holds, then x0 is called the Tiangong No.1 point of F (x). It is known that the two Tiangong No.1 points of function f (x) = AX2 + (B-7) x + 18 are - 3 and 2 respectively. (1) find the value of a, B and the expression of F (x); (2) when the domain of definition of function f (x) is [T, t + 1] (T > 0), the maximum value of F (x) is g (T), and the minimum value of F (x) is g (T) (t) And find the expression of H (T) = g (T) - G (T)

(1) According to the meaning of the problem, we get f (- 3) = - 3, f (2) = 2; that is, 9A + 21-3b + 18 = - 3, 4A + 2b-14 + 18 = 2, the solution is a = - 3, B = 5 ∫ f (x) = - 3x2-2x + 18 (2) ∫ f (x) axis of symmetry is & nbsp; X = - 13 < 0 ∫ f (x) is a monotone decreasing function in [T, t + 1] (T > 0), G (T) = f (x) max = f (T) =

Given x > 1, find the minimum value of function y = (2x ^ 2-x + 1) / X-1
Where x belongs to (1, positive infinity)

(2x+1)(x-1) +2 2 2
y= ----------------------- = 2x+1 + -------------- = 2(x-1) + -------------- +1
x-1 x-1 x-1
two
≥ 2*√(2(x-1)*--------) +1 =5
x-1
if and only if
two
2 (x-1) = --- the time equal sign holds
x-1
That is, when x = 2, y has a minimum value of 5
[middle school mathematics and chemistry answer group]