Let the function y = 4 ^ x * a + 2 ^ x + 1 hold when y > 0 is constant when x belongs to (- infinity, 1], and find the range of A

Let the function y = 4 ^ x * a + 2 ^ x + 1 hold when y > 0 is constant when x belongs to (- infinity, 1], and find the range of A

Let t = 2 ^ x, because x0, a > - 3 / 4
So - 3 / 40 is 4A + 2 + 1 > 0, a > - 3 / 4
So a > 0
To sum up, a > - 3 / 4

If y = loga-1 ^ x is a decreasing function in (0, positive infinity), then the value range of a is
If y = log (A-1) x is a decreasing function within (0, positive infinity), then the value range of a is not loga-1 ^ X

If y = log (A-1) x is a decreasing function in (0, positive infinity), then 0

Let the odd function f (x) be a decreasing function on (0, + ∞) and f (1) = 0, then the solution set of the inequality f (x) - f (- x) x < 0 is ()
A. （-1，0）∪（1，+∞）B. （-∞，-1）∪（1，+∞）C. （-∞，-1）∪（0，1）D. （-1，0）∪（0，1）

According to the meaning of the question, draw the function image that meets the condition: ∵ function y = f (x) is an odd function, ∵ f (x) - f (- x) 2x ＜ 0 is transformed into: 2F (x) x ＜ 0, that is, XF (x) ＜ 0. From the graph, when x ＞ 0, f (x) ＜ 0, then x ＞ 1; when x ＜ 0, f (x) ＞ 0, then x ＜ - 1; in conclusion, the solution set of F (x) - f (- x) 2x ＜ 0 is: (- ∞, - 1) ∪ (1, + ∞), so select: B

Given that the odd function f (x) (- ∞, 0) is a decreasing function and f (2) = 0, then the solution set of inequality (x-1) f (x-1) > 0 is

∵ f (x) is an odd function
∴f（-2）=0
On (- 2,0) and (2, + ∞)
With F (x)

If the odd function f (x) is a decreasing function in (- 1,1), then the solution of the inequality f (x) less than 0 is zero___

Because f (x) is an odd function
There is f (0) = 0
So f (x)

If the function y = ax ^ 2 + BX + C is an increasing function on (negative infinity, - 1) and a decreasing function on [- 1, positive infinity), then
A. B is greater than zero and a is less than zero
B. B = 2A less than zero
C. B = 2A greater than zero
D. The sign of a and B is indefinite
No direct answer

On both sides of the axis of symmetry x = - 1: increase left and decrease right
So the opening of the image is down, that is, a

If and only if the function y = x square + BX + C is monotone on [0, infinity]
Urgent, thank you

-b/20
From zero to infinity, the function is monotone, only if the normal of the quadratic function is less than or equal to zero

Discuss the monotonicity of function FX = x + 9 / X (x > 0), and prove your conclusion

Monotonically increasing on [3, + infinity]
Monotonically decreasing on (0,3]
Let's complete the proof by ourselves. Take X1 in the domain

Prove the monotonicity of function y = x + 1 / X on the domain of definition
There should be classified discussions,

y'=1-1/x^2
Y '= 0 when x = 1 or - 1
So when x = (- infinite, - 1), y is increasing
When x = (- 1,0), y is minus
When x = (0,1), y is minus
When x = (1, + infinity), y is increasing

What is the domain of definition of inverse scale function y = 1 / x? Is monotonicity like this? Prove the conclusion
Such as the title

The domain is (- infinity, 0) and (0, + infinity)
y'=-1/x^2,y"=2/x^3
When x tends to 0, y tends to infinity, so y axis is asymptote. When x tends to infinity, y tends to 0, so x axis is asymptote
When x > 0, y'0, the image is monotonically decreasing and concave
When x