The monotonically decreasing interval of the function f (x) = |logax| (a > 0 and a ≠ 1) is __

The monotonically decreasing interval of the function f (x) = |logax| (a > 0 and a ≠ 1) is __

The image of function f (x) = |logax| is obtained by turning the image below the X axis symmetrically to the image above the X axis,
When 0 ＜ a ＜ 1, as shown in the figure:
．
At this time, f (x) = |logax| (0 < a < 1) is a subtractive function on the interval (0,1]
Similarly, when a > 1, the function image is also shown in the above figure:
At this time, f (x) = |logax| (a > 1) is a subtractive function on the interval (0,1]
To sum up, the monotonic decreasing interval of function f (x) = |logax| (a > 0 and a ≠ 1) is (0,1],
So the answer is: (0, 1]

The image of function f (x) decreases monotonically at (- OO, 0), (1 / 2, + OO) and increases monotonically at (0,1 / 2], then G (x) = the monotonic interval of F (logax)

When a > 1, let logax be (- OO, 0), x < 1, let logax be (1 / 2, + OO), x > A ^ (1 / 2)
Therefore, it decreases monotonically at (0,1), (a ^ (1 / 2), + OO)
Monotonically increasing in (1, a ^ (1 / 2))
When a < 1, the same is true, but the opposite is true, because logax decreases monotonically
Therefore, it increases monotonically at (0,1), (a ^ (1 / 2), + OO)
Monotonically decreasing in (1, a ^ (1 / 2))

If the function f (x) satisfies f (logax) = x + 1 / X (a > 0 and a ≠ 1), the analytical formula and monotone interval of the function f (x) are obtained

Let logax = t, then
x=a^t
f(t)=a^t+1/a^t
therefore
f(x)=a^x+1/a^x
The function is an even function. According to the image,
1. A (0,1)
X belongs to [0,1] increment
(1, + infinity) decrement
[- 1,0) decreasing, (- infinity, - 1) increasing;
2.a>1
X belongs to [0,1]
(1, + infinity) increasing
[- 1,0) increment,
(- infinity, - 1) decrement

1. Given that f (x) = logax (x ≥ 1) is a monotone decreasing function on (- infinite, + infinite), find the value range of A. 2. Given that a > 0, the x times of F (x) = 3 / A + the x times of a / 3 are even functions on R, find the value of A

1.0 I can't understand the second question. You can type it with the formula editor in the future, which will be much more eye-catching

Let f (x) = ex x， (1) Find the monotone interval of function f (x); (2) If k > 0, find the solution set of inequality f '(x) + K (1-x) f (x) > 0

（1）∵f（x）=ex
x
∴f′(x)＝−1
x2ex+1
xex＝x−1
x2ex
From F '(x) = 0, x = 1,
Because when x < 0, f '(x) < 0;
When 0 ＜ x ＜ 1, f '(x) ＜ 0; When x > 1, f '(x) > 0;
Therefore, the monotone increasing interval of F (x) is: [1, + ∝; the monotone decreasing interval is: (- ∞, 0), (0, 1]
(2) By F '(x) + K (1-x) f (x) = x − 1 + KX − kx2
x2ex=(x−1)(−kx+1)
x2ex＞0，
Get: (x-1) (kx-1) ＜ 0,
Therefore, when 0 < K < 1, the solution set is: {x|1 < x < 1
k}；
When k = 1, the solution set is: φ；
When k > 1, the solution set is: {x|1
k＜x＜1}．

Let the function f (x) = (e ^ x) / x 1 find the monotone interval of F (x). 2 if k > 0, find the solution set of inequality f '(x) + K (1-x) f (x) > 0 I hope to be a little more detailed. What method is used? (image or derivative)

one
E ^ x is an increasing function
At X0
The slope of e ^ x is greater than 1
So it's an increasing function
Minus interval (- infinite, 0)
Increasing interval (0, + infinite)
two
f'(x)=(e^xx-e^x)/x^2
Inequality is
k(1-x)(e^x)/x>e^x/x^2-e^x/x
e^x>0
k(1-x)/x>1/x^2-1/x
Simultaneous * x ^ 2
k(1-x)x>1-x
kx(1-x)>1-x
When x > 1 / K
The equation holds
So the solution is (1 / K, + infinity)