The quadratic function f (x) satisfies the following conditions: (1) f (1 + x) = f (1-x); (2) the maximum value of F (x) is 15; (3) the sum of two cubes of F (x) = 0 is 32. Find f (x); if x belongs to [- 1,4], find the maximum value of F (x); find the minimum value of F (x) in the interval [M, M + 2]

The quadratic function f (x) satisfies the following conditions: (1) f (1 + x) = f (1-x); (2) the maximum value of F (x) is 15; (3) the sum of two cubes of F (x) = 0 is 32. Find f (x); if x belongs to [- 1,4], find the maximum value of F (x); find the minimum value of F (x) in the interval [M, M + 2]

(1) f (1 + x) = f (1-x); (2) the maximum value of F (x) is 15; (3) the sum of two cubes of F (x) = 0 is 32;
Because f (1 + x) = f (1-x), the axis of symmetry is x = 1
Since the maximum value of F (x) is 15, f (1) = 15
Let y = ax & sup2; + BX + C
-b/2a=1 b=-2a
y=ax²-2ax+c=0
15=a-2a+c
a+15=c
The original formula is y = ax & sup2; - 2aX + 15 + a
According to Weida's theorem, x1x2 = 1 + 15 / A, X1 + x2 = 2
x1^3+x2^3=x1^3-(-x2^3)=(x1+x2)(x1²-x1x2+x2²）
=2（4-3-45/a)=32
1-45/a=16
a=-3
The quadratic function is y = - 3x & sup2; + 6x + 12
If x belongs to [- 1,4], find the maximum value of F (x)
The maximum value is f (1) = 15
The minimum value of F (x) in the interval [M, M + 2]
Classified discussion
① M < 1 < m + 2 minimum f (m) or F (M + 2)
② 1 < m minimum f (M + 2)
③ 1 ＞ m + 2 minimum f (m)