Given the function f (x) = x ^ 3 + X, the solution set of inequality f (2-x ^ 2) + F (2x + 1) > 0 is

Given the function f (x) = x ^ 3 + X, the solution set of inequality f (2-x ^ 2) + F (2x + 1) > 0 is

F '(x) = 3x ^ 2 + 1 > 0, so f (x) is an increasing function on R
Because f (- x) = - f (x), f (x) is an odd function
Therefore, the inequality is: F (2x + 1) > F (x ^ 2-2)
Get: 2x + 1 > x ^ 2-2
That is, x ^ 2-2x-3

Known function f (x) = x2-2x-8, G (x) = 2x2-4x-16, (1) Find the solution set of inequality g (x) ＜ 0; (2) If f (x) ≥ (M + 2) x-m-15 holds for all x > 2, find the value range of real number M

From G (x) = 2x2-4x-16 < 0, x2-2x-8 < 0, that is, (x + 2) (x-4) < 0, and the solution is - 2 < x < 4. Therefore, the solution set of inequality g (x) < 0 is {x| - 2 < x < 4}; (2) Because f (x) = x2-2x-8, when x > 2, f (x) ≥ (M + 2) x-m-15 is established, then x2-2x-8 ≥ (M + 2) x-m-15

If the function f (x) = {2x ^ 2 + 1 (x0) is known, then the inequality f (x) - X

x> 0, f (x) - x = - 2x-x = - 3x

It is known that the X-Power of the exponential function f (x) = (A-1) is a subtractive function on R, and the X-Power of inequality (2a-1) is greater than the 1 + 2x power of (2a-1)

The subtraction function is 0

Subtractive interval of the square power of 2x-x under the root sign of function y = 2

Let y = f (z) = 2 √ Z, where z = g (x) = 2x-x ² （2x-x ² ≥0）,
Then y = f (g (x)) = 2 √ (2x-x) ²) , The definition field is 0 ≤ x ≤ 2;
Because f (z) = 2 √ Z is a monotonically increasing function,
Therefore, in order to make f (g (x)) monotonically decrease in a certain interval, G (x) = 2x-x ² (0 ≤ x ≤ 2) monotonically decreasing in this interval;
Because, G (x) = 2x-x ² (0 ≤ x ≤ 2) is a parabola, the opening of the parabola is downward, and the axis of symmetry is x = 1,
It can be obtained that when 1 ≤ x ≤ 2, G (x) decreases monotonically,
So, the function y = 2 √ (2x-x) ²) The monotonic decreasing interval of is [1,2]

It is known that the X-Power of function f (x) = 2 minus 1 is greater than the X-Power of upper 2 plus 1 (1) Judge the monotonicity of F (x) and prove it (2) Find the inverse function of F (x)

(1) Y = (2 ^ x-1) / (2 ^ x + 1) = 1-2 / (2 ^ x + 1), which is easy to find as an increasing function. The proof is as follows:
Y '= 2 ^ (x + 1) LN2 / (2 ^ x + 1) ^ 2 > 0, so the conclusion is valid
Proof can also be completed by definition. (omitted)
(2) First solve 2 ^ x = (1 + y) / (1-y), so there is x = log2 (1 + y) / (1-y),
The inverse function is y = log2 (1 + x) / (1-x)