If the image of the function y = a ^ x + B-1 (a > 0 and a is not equal to 1) passes through the second and third quadrant, what must A.B satisfy

If the image of the function y = a ^ x + B-1 (a > 0 and a is not equal to 1) passes through the second and third quadrant, what must A.B satisfy

Drawing is very important!
When a & lt; 1 and a & gt; 1, the image is opposite
When a & lt; 1, such as the red line in the figure, the image must pass through the second quadrant, and B-1 & lt; - 1, that is B & lt; 0, must pass through the third quadrant
When a & gt; 1, such as the black line in the figure, the image must pass through the first quadrant, and then 0 & gt; B-1 & gt; - 1, namely - 1 & gt; B & gt; 0, if it passes through the second and third quadrants

Let f (x) = (x ^ 2) times (e ^ x square) expand into a power series of X at (- infinity, + infinity)

Sweat
e^x=Σx^n/n!
e^(x^2)=Σ(x^2)^n/n!=Σx^2n/n!
(x^2)e^(x^2)=Σx^(2n+2)/n!

Given f (x) = 8 + 2x-x2, G (x) = f (2-x2), try to find the monotone interval of G (x)

∵ f (x) = 8 + 2x-x2 ∵ g (x) = f (2-x2) = - X4 + 2x2 + 8g '(x) = - 4x3 + 4x when G' (x) > 0 & nbsp;, - 1 ＜ x ＜ 0 or X ＞ 1 when G '(x) ＜ 0, X ＜ - 1 or 0 ＜ x ＜ 1, so the increasing interval of function g (x) is: (- 1,0) and (1, + ∞) the decreasing interval is: (- ∞, - 1) and (0,1)

Monotone interval of square + X-1 of y = 2x

The quadratic function with the opening upward decreases on the left and increases on the right
The axis of symmetry is x = - 1 / 4
Therefore, the monotone decreasing interval is (- ∞, - 1 / 4)
Monotone increasing interval is (- 1 / 4, + ∞)
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Monotone interval of the square of y = | - X - 2x + 2 |

y=|-x²-2x+2|
For this problem, we can go to the absolute value first
That is y = - X & sup2; - 2x-1 + 3
=-（x+1）²+3
=-（x+1）²+3
We can see that it increases monotonically in the interval [- ∞, - 1],
Monotonically decreasing in the interval [- 1, ∞]
After adding absolute value sign, the original negative value becomes positive value
Let - X & sup2; - 2x + 2 < 0
The solution is x ＜ - 3 - 1 or X ＞ 3 - 1
It can be seen that the intersection points of y = - X & sup2; - 2x + 2 image and X axis are: (- √ 3 - 1,0) and (√ 3 - 1,0)
Draw the image of y = - X & sup2; - 2x + 2
Then the image below the x-axis is folded up along the x-axis to get the image of y = | - X & sup2; - 2x + 2 |
It can be seen that it decreases monotonically in the interval (- ∞, - √ 3-1] ∪ [- 1, √ 3-1]
In the interval [- √ 3-1, - 1] ∪ [√ 3-1, + ∞) monotonically increases

What is the monotone interval of y = x square + 2X-4
It's derivative

Y '= 2x + 2, x > - 1 monotonically increasing
x

If the function f (x) = x ^ 2-2mx + 3 is a decreasing function on (- infinity, 2), then the value range of M is

The image of quadratic function, opening up, symmetry axis is x = m, in order to make the function minus function on (- infinity, 2), we must make the symmetry axis X = m on the right side of 2, that is {m | m > = 2} is the range

The function y = x ^ 2-2mx-1 increases monotonically on [1, positive infinity]. Find the value range of M

Y = x ^ 2-2mx-1 open up thinking: find the value range of the axis of symmetry. The axis of symmetry = m increases monotonically on [1, positive infinity], so m ≤ 1 -------- hope it can help you

In order to make the function y = 1 + 2 ^ x + (4 ^ x) * a hold when x belongs to (- infinity, 1)] and Y is greater than 0, find the value range of A

a> 0 or - 3 / 40
When A0 = = > - 3 / 4

If the function y = 1 + 2x + 4xa holds for Y > 0 on X ∈ (- ∞, 1], then the value range of a is______ ．

From the meaning of the question, we get that 1 + 2x + 4xa ＞ 0 is constant on X ∈ (- ∞, 1]; a ＞ - 1 + 2x4x is constant on X ∈ (- ∞, 1]; and ∵ t = - 1 + 2x4x = - (12) 2x - (12) x = - [(12) x + 12] 2 + 14. When x ∈ (- ∞, 1], the range of T is (- ∞, - 34], ｜ a ＞ - 34; that is to say, the range of a is (- 34, + ∞); so the answer is: (- 34, + ∞)