Given that the function f (x) = 4x2-mx + 5 is an increasing function in the interval [- 2, + ∞), then the value range of M is () A. （-∞，-16）B. （-∞，16]C. （-∞，-16]D. （4，16）

Given that the function f (x) = 4x2-mx + 5 is an increasing function in the interval [- 2, + ∞), then the value range of M is () A. （-∞，-16）B. （-∞，16]C. （-∞，-16]D. （4，16）

The axis of symmetry of the function f (x) = 4x2 MX + 5 is x = M8, and the function f (x) = 4x2 MX + 5 is an increasing function in the interval [- 2, + ∞), ∵ M8 ≤ - 2, ∵ m ≤ - 16

If the function f (x) = 4x & # 178; - MX + 5 is an increasing function in the interval [- 2, + ∞), then the value range of F (1) is?

8x-m = 0, x = - 2, M is greater than or equal to - 16
F (1) is greater than or equal to 25

It is known that the function f (x) is an odd function defined on R, and when x > 0, f (x) = (&# 189;) &# 178;
Finding the analytic expression of function f (x)
Draw the function image and write the monotone interval of function f (x) according to the image
Wrong, right
It is known that the function f (x) is an odd function defined on R, and when x > 0, f (x) = (# 189;) to the x power

Because f (x) is an odd function, when X & lt; 0, f (x) = - f (- x) = - (#189;) - x power
Monotone interval (- ∞, 0) monotone decreasing, (0, + ∞) & nbsp; monotone decreasing
The shape is roughly like this

Let f (x) = x & # 178; - AlN (2x + 1) (x ∈ (-, # 189;, 1], a > 0)
(1) If the function f (x) is a decreasing function in its domain, the value range of a is obtained;
(2) Does the function f (x) have a minimum? If there is a minimum, point out the value of X when it reaches the minimum, and prove your conclusion

(1) Let f '(x) x (2x + 1) let t = x (2x + 1), X ∈ (- 1 / 2,1], t = 2 (x + 1 / 4) ^ 2-1 / 8. When x = 1, t has the maximum value of 3, so a > 3
(2) Let f '(x) = 2x-2a / (2x + 1), Let f' (x) = 0, we get 2x ^ 2 + x-a = 0, a > 0, so △ = 1 + 8A > 0, the equation has solution. The extreme point X1 = - 1 / 4 - √ (1 + 8a) / 4, X2 = - 1 / 4 + √ (1 + 8a) / 4, x1x2 = - A / 20, so x1

The function f (x) = x & #178; - 2|x| - 3 is known
① To prove that f (x) is an even function
② Draw an image of this function
③ Write the monotone interval of this function

(1)
f(x)={-x²-2x+3 (x≥0)
{-x²+2x+3 (x

If f (x + 2) = x & # 178; - 4, then f (3)=

x+2=t
x=t-2
f（t)=t^2-4t
t=3
f（3）=-3

Given the function f (x + 2) = x & # 178; - 4, then f (3) =

F (3) = the square of F (1 + 2) = 1 - 4 = - 3 &;

If the function F X defined on R satisfies FX = x & # 178; + 1, X ≤ 0; f (x-1) - f (X-2), x > 0, then the value of F (3) is?

f(0)=1,f(-1)=2
When x = 1, f (1) = f (0) - f (- 1) = 1-2 = - 1
When x = 2, f (2) = f (1) - f (0) = - 1-1 = - 2
When x = 3, f (3) = f (2) - f (1) = - 2 - (- 1) = - 1

If the function y = f (x) satisfies f (x + 2) = f (x), and X is [- 1,1], f (x) = x & # function
If the function y = f (x) satisfies f (x + 2) = f (x), and X is [- 1,1], f (x) = x & # 178; function g (x) = ㏒ 10x, then the number of zeros of function H (x) = f (x) - G (x) is

The specific explanation is as follows: because f (x + 2) = f (x), f (x) is a function with the minimum period of 2, and because f (x) = x ^ 2 when x is [- 1,1], the image of F (x) can be repeatedly shifted to the right in the x-axis direction with the period of 2, and the maximum value is 1, 3, 5, 7, 9, 11 Wait, max

If the function FX satisfies f (x + 2) = f (x), if x belongs to [- 1,1], FX = x & # 178; then if x belongs to [1,3], find the analytic expression of FX

f(x+2)=f(x)
It is known that f (x) is a periodic function, and the period is 2
When x belongs to [- 1,1], f (x) = x & # 178;
Move the function image two units to the right
So when x belongs to [1,3], f (x) = x & # 178;