Monotone increasing interval of function y = ax + B / X (a > 0, b > 0)

Monotone increasing interval of function y = ax + B / X (a > 0, b > 0)

The function y = ax + B / X (a > 0, b > 0) is an odd function. It is a "√" shape in the first quadrant, and the image in the third quadrant is symmetrical with the image about the origin in the first quadrant
Its monotone increasing interval is (- ∞, - √ (B / a)], [√ (B / a), + ∞)

The function y = 2 ^ [(- x ^ 2) + AX-1] is monotonically increasing on the interval (- ∞, 3), and a is obtained
Please write the detailed process

Let f (x) = 2 ^ [(- x ^ 2) + AX-1] obviously: F (x) > 0
Let: X1 > X2, x1, X2 belong to (- ∞, 3)
In order to make f (x) monotonically increasing, that is, f (x1) / F (x2) > 1
f（x1）/f(x2)=2^[(-x1^2)+ax1-1]/2^[(-x2^2)+ax2-1]>1
Just: [(- X1 ^ 2) + ax1-1] - [(- x2 ^ 2) + ax2-1] > 0 (because f (x) is an exponential function)
That is: (x2-x1) * (x1 + x2-a) > 0
And (x2-x1)

If the monotone increasing interval of function f (x) = AX2 − 1x is (0, + ∞), then the value range of real number a is______ ．

F ′ (x) = (ax − LX) ′ = a + 1x2, from the meaning of the question, a + 1x2 ≥ 0 is constant on X ∈ (0, + ∞), so a ≥ - 1x2 is constant on X ∈ (0, + ∞), so a ≥ 0. So the answer is: a ≥ 0

Monotone increasing interval of function y = cos (2x - π / 4)

If the simple increasing interval of cosx is [2K π - π / 2, 2K π], then 2K π - π / 2