The function f (x) = 2-x ^ 2, G (x) = X. if the function f (x) = min {f (x), G (x)} is defined, the maximum value of F (x) is?

The function f (x) = 2-x ^ 2, G (x) = X. if the function f (x) = min {f (x), G (x)} is defined, the maximum value of F (x) is?

1、 Draw f (x) and G (x) in a rectangular coordinate system, a straight line and a parabola (opening downward)
2、 Find the intersection of the two curves (the curve also includes a straight line. I don't know what level the landlord is. Forgive me), that is, solve the solutions of the equations y = 2-x ^ 2 and y = X. the intersection of the two curves is a (- 2, - 2) and B (1,1)
3、 Look at the two curves. Between the intervals (- 2,1), f (x) > G (x), so f (x) is the smaller one, that is, f (x) = g (x) = x; Similarly, on (- ∞, - 2] and [1, + ∞), f (x)

It is known that the image of function f (x) is continuous on [a, b], which is defined as: F1 (x) = min {f (T) | a ≤ t ≤ x} (x ∈ [a, b]), F2 (x) = max {F（ I just want to ask, when x = a, how does F2 (x) - F1 (x) ≤ K (x-a) hold

When x = a, the right side of the inequality sign is 0, the left side F2 (x) represents the maximum value, and F1 (x) represents the minimum value. They only have a difference greater than or equal to 0, and the right side of the inequality sign is 0, and the inequality sign is less than or equal to, which can only be equal. At this time, the value of F (x) should be a constant, and K can be any positive integer. But K has the minimum value, so it is 1

It is known that the image of function f (x) is continuous on [a, b]. It is defined as: F1 (x) = min {f (T) | a ≤ t ≤ x} (x ∈ [a, b]), F2 (x) = max {f (t) | a ≤ t ≤ x} (x ∈ [a, b]), where min {f (x) | x ∈ D} represents the minimum value of function f (x) on D, and Max {f (x) | x ∈ D} represents the maximum value of function f (x) on D. if there is a minimum positive integer k, make F2 (x) - F1 (x) ≤ K (x-a) holds for any x ∈ [a, b], then the function f (x) is called the "k-order contraction function" on [a, b]. Given that the function f (x) = X2, (x ∈ [- 1, 4]) is the "k-order contraction function" on [- 1, 4], then the value of K is _

F1 (x) = X2, X ∈ [- 1, 0) 0, X ∈ [0, 4], F2 (x) = 1, X ∈ [- 1, 1) X2, X ∈ [1, 4] F2 (x) - F1 (x) = 1-x2, X ∈ [- 1, 0) 1, X ∈ [0, 1) X2, X ∈ [1, 4] when x ∈ [- 1, 0], 1-x2 ≤ K (x + 1), (K ≥ 1-x, K ≥ 2; when x ∈ (0, 1), 1 ≤ K (x + 1), (k

The functions f (x) = 1 - | 1-2x| and G (x) = (x-1) ^ 2 on the definition field [0,1] are known, and min {x1, x2. Xn} is the minimum value in x1, x2.xn The functions f (x) = 1 - | 1-2x| and G (x) = (x-1) ^ 2 on [0,1] are known, and min {x1, x2.xn} is the minimum value in x1, x2.xn, (1) find the function analytical formula of F (x) = min {f (x), G (x)}. (2) find the value range of F (x)

After mapping,
analysis
When x > 1 / 2, f (x) = 1-2x + 1 = 2-2x
When x

Let the functions f (x) and G (x) be continuous at point C, and prove that the function K (x) = max {f (x), G (x)} is also continuous at point C Please kindly do me a favor

x->c+,lim K(x)=lim max(F(x),G(x))=max(F(c),G(c));
x->c-,lim K(x)=lim max(F(x),G(x))=max(F(c),G(c));
Left limit = right limit

Given the function f (x) = x ^ 2-1, G (x) = - x, let H (x) = max [f (x), G (x)] (i.e. the greater of F (x) and G (x)), then the minimum value of H (x) is What is your answer?

It's easy to see from the drawing that what you want is the intersection to the right of F (x) and G (x)
Let f (x) = g (x), then x ^ 2-1 = - x, get x = (- 1 + root 5) / 2 or (- 1 - root 5) / 2, and find the previous one, that is, x = (- 1 + root 5) / 2. At this time, H (x) is the smallest, which is - (- 1 + root 5) / 2 = (1 - root 5) / 2