It is proved by definition that the function f (x) = x + 1 / X is an increasing function on [1, + infinite sign]

It is proved by definition that the function f (x) = x + 1 / X is an increasing function on [1, + infinite sign]

Suppose 1=

If the functions f (x) and G (x) are odd and f (x) = AF (x) + BG (x) + 2 has a maximum value of 5 on (0, + ∞), then f (x) is ()
Options: A. minimum - 5 B. maximum - 5 C. minimum - 1 D. maximum - 3

If f (x) and G (x) are both odd functions, then AF (x) + BG (x) is also an odd function. Let H (x) = AF (x) + BG (x), H (x) be an odd function
Then f (x) = H (x) + 2
The image of H (x) is symmetrical about the origin, and the image of F (x) is obtained by translating the image of H (x) upward one unit
So, the symmetry of F (2) is about X
Suppose that f (x) has a maximum value on (0, + ∞), and the point is (m, 5), f (x) is centrosymmetric with respect to point (0,2)
Then according to the symmetry, the minimum value of F (x) on (- ∞, 0) is (- m, - 1)
So, choose C
If you don't understand, please hi me,

Given that f (x) and G (x) are odd functions, H (x) = AF (x) + BG (x) + 2 has a maximum value of 5 in the interval (0, + ∞), then the minimum value of H (x) in (- ∞, 0) is ()
A. -5B. -1C. -3D. 5

Let f (x) = H (x) - 2 = AF (x) + BG (x), then f (x) is an odd function. When ∵ x ∈ (0, + ∞), H (x) ≤ 5, ∵ x ∈ (0, + ∞), f (x) = H (x) - 2 ≤ 3. When x ∈ (- ∞, 0), - x ∈ (0, + ∞), ∵ f (- x) ≤ 3 ⇔ - f (x) ≤ 3 ⇔ f (x) ≥ - 3. ∵ H (x) ≥ - 3 + 2 = - 1, so B is selected

F (x) and G (x) are all odd functions. H (x) = AF (x) + BG (x) + 2 has a maximum value of 5 in (0 ~ positive infinity) and a minimum value of H (x) in (negative infinity ~ 0)

F (x) = AF (x) + BG (x) is an odd function
There is a maximum of 3 on (0, + ∞)
F (x) has a minimum value of - 3 on (- ∞, 0)
So h (x) = AF (x) + BG (x) + 2 has a minimum value of - 3 + 2 = - 1 on (- ∞, 0)

Both f (x) and G (x) are odd functions defined on R, if f (x) = AF (x) + BG (x) + 2, the maximum value on (0, + infinity) is 5
minimum value

Let u (x) = f (x) - 2 = AF (x) + BG (x), then the maximum value of U (x) on (0, + ∞) is 3,
Both f (x) and G (x) are odd functions defined on R
U (x) is also an odd function
The minimum value of U (x) on (- ∞, 0) is - 3
That is to say, the minimum value of AF (x) + BG (x) on (- ∞, 0) is - 3, ■
The minimum value of F (x) = AF (x) + BG (x) + 2 is - 1 in (- ∞, 0)

It is known that f (x) and G (x) are odd functions defined on R, if the maximum value of F (x) - AF (x) + BG (x) + 2 on (0, + infinity) is 5,
Find the minimum value of F (x) on (negative infinity, 0)
I saw some answers. Where did - 2 come from,
F(-2)=af(-2)+bg(-2)+2=5
Both f (x) and G (x) are odd functions

Both f (x) and G (x) are odd functions defined on R. f (- x) = - f (x) g (- x) = - G (x)
F (x) = AF (x) + BG (x) + 2 = 5 A (x) + BG (x) = 5-2 = 3 F (- x) = AF (- x) + BG (- x) + 2 = - [AF (x) + BG (x)] + 2 = - 3 + 2 = - 1, the minimum value is - 1

If the functions f (x) and G (x) are odd functions on R, H (x) = AF (x) + BG (x) + 2 has the maximum value 5 on the interval (0, positive infinity), then H (x)
What is the minimum value on the interval (negative infinity, 0)

Let X1 in H (x1) be (0, + ∞), and H (x1) = 5
So - X1 in H (- x1) is in (- ∞, 0)
Because f (x) and G (x) are odd functions, f (- x) = - f (x), G (- x) = - G (x)
H（-X1）
=af(-X1)+bg(-X1)+2
=-af(X1)-bg(X1)+2
=-［af(X1)+bg(X1)+2］+4
=-H（X1）+4
Because H (- x1) - 5 + 4 = - 1
The minimum value is - 1

It is known that the function f (x) defined in the interval (- 1,1) is both an odd function and a decreasing function. Try to find f (x ^ 2-2) + F (3-2x)

First of all - 1

It is known that the domain of definition of odd function f (x) is (- 2,2), and f (x) is a decreasing function on its domain, if f (A-2) + F (3a-2)

Solution: - 2

(1) Find the maximum or minimum of the following functions. ① y = 2x2-3x-5 ② y = - x2-3x
(1) Find the maximum or minimum of the following functions
①y=2x2-3x-5 ②y=-x2-3x+1
The process should be detailed

(1) F (x) '= 4 * x-3 = 0, we get x = 3 / 4, because f (x) is a parabola with an opening upward, so f (3 / 4) is the minimum, f (3 / 4) = - 98 / 16 (2) f (x)' = - 2 * x-3 = 0, we get x = - 3 / 2. Because f (x) is a parabola with an opening downward, so f (- 3 / 2) is the maximum, f (3 / 2) = 13 / 4