The decreasing interval of the function y = 1 / √ 12-x-x & # 178; is

The decreasing interval of the function y = 1 / √ 12-x-x & # 178; is

The decreasing interval is that x increases and Y decreases, that is, only the denominator part of the function increases
Let f (x) = 12-x-x & # 178; then it is transformed into finding the increasing interval of this function and f (x) > 0
F (x) = - (x + 4) (x-3) the opening of the image is downward, and the axis of symmetry x = - 1 / 2
According to the f (x) image, the interval is (- 4, - 1 / 2]
Suggest that the owner draw a picture, it is easy to find out the answer

Find the monotone interval of function y = (1 / 2)

Y = (1 / 2) ^ t, is decreasing function on t ∈ R, and decreasing function on T > = 0
Increasing when t = x ^ 2, X0
So y = (1 / 2) ^ (x ^ 2), in (- ∞, 0] is an increasing interval, and (0, ∞) is a decreasing interval

If y = (a squared-1) x is a decreasing function on R, then the value range of a------

It is impossible for a parabola to be monotonic in the range of real numbers unless it is a piecewise function
If it's a piecewise function, then
When x is less than zero, y monotonically decreases if and only if A-1 is greater than zero. The solution is a ∈ {a │ A is less than - 1 or a is greater than 1}
When x is greater than or equal to zero, the necessary and sufficient condition for monotone decreasing of Y is that A-1 is less than zero, and the solution is a ∈ {a │ - 1 is less than a greater than 1}

If the function f (x) = 4x ^ 2-mx + 5-m is an increasing function in the interval [- 2, + infinity] and a decreasing function in the interval (- infinity, - 2}), then the value of real number m is?

According to the image and meaning of the function, we know that x = - 2 is the symmetry axis of the function f (x) = 4x ^ 2-mx + 5-m, - - M) / (2 * 4) = - 2, M = - 16