Function y = (SiNx + 1) / (1 + X & # 178;) why is this a bounded function? How to judge

Because | SiNx + 1 | ≤ 2 (1)
1+x²≥1
0

Increasing interval of function y = - X & # 178; - 1

The function y = - 1 / 3 (x + 3) &# 178
1. The opening direction of parabola_ 2. The vertex coordinates are: ;
3. The axis of symmetry is: ；
4. On the left side of the axis of symmetry, the curve goes from left to right
Namely:_
On the right side of the axis of symmetry, the curve goes from left to right
Namely:_____
5. The graph of the function has the most_ The point is ；
6. When x The function has the most The value is .
7. Function y = - 1 / 3 (x + 3) & # 178; image can be regarded as function y = - 1 / 3x & # 178; image orientation Translation_ It's a unit

The function y = - 1 / 3 (x + 3) &# 178
1. The opening direction of parabola__ Next___ 2. The vertex coordinates are:___ (-3,0)___ ;
3. The axis of symmetry is:__ x=-3______ ；
4. On the left side of the axis of symmetry, the curve goes from left to right__ Incremental____
Namely:______ Y increases with the increase of X_______________________________________
On the right side of the axis of symmetry, the curve goes from left to right_ Decreasing_____
Namely:_________ Y decreases with the increase of X____________________________________
5. The graph of the function has the most_ High____ The point is___ （-3,0)___ ；
6. When x_ =-3___ The function has the most_ Big___ The value is_ 0___ .
7. Function y = - 1 / 3 (x + 3) & # 178; image can be regarded as function y = - 1 / 3x & # 178; image orientation_ Left___ Translation__ 3___ It's a unit

Let f (x) = ax ^ 2 + BX + C (a ≠ 0), if | f (0) | ≤ 1, | f (- 1) | ≤ 1, | f (1) | ≤ 1, try to prove that for any - 1

In this paper, we find that \124124124124124 (f (0) is the result of | f (x) | = 124; = | [(x ^ 2 + x) f (1) = A-B + A-B + a = [f (1) + F (- 1-1-1) - 2F (0)] / 2B = [f (1) + [f (f (f (1) + (f (f (1) + (f (f (- 1) - 1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-2-x-2) - f (0 (2) (2-2-x ^ 2-x ^ 2-x ^ 2) f (f (0) f (0) f (- 1) |

a> 0, a ^ (x1 + x2)

a^(2x1)+a^(2x2)=(a^x1)^2+(a^x2)^2>=2(a^x1)(a^x2)=2a^(x1+x2)>=a^(x1+x2)

1. What is the fixed mode of thinking?
2. What should we pay attention to?
3. The choice of the end of the interval?

1. Solve the inequality
2. Pay attention to greater than or equal to and less than or equal to (i.e. equal to); "0" question, whether the algebraic formula is meaningful

F (x) = xlnx, G (x) = x ^ 3 + ax ^ 2-x + 2 (1) if the monotone decreasing interval of function g (x) is (- 1 / 3,1), find the analytic expression of function g (x) (2) under the condition of (1), find the tangent equation of the image of function y = g (x) passing through point P (1,1) (3) belongs to (0, + infinity) for all x, 2f (x) is less than or equal to G (x) + 2, and find the value range of real number a

G (x) '= 3x ^ 2 + 2ax-1, G (1)' = 2 + 2A = 0, a = - 1, G (x) = x ^ 3-x ^ 2-x + 2, slope k = g (1) '= 0, tangent equation is y = 1,3,2f (x) ≤ G (x) + 2 2xlnx ≤ x ^ 3-ax ^ 2-x + 2 ax ^ 2 ≤ x ^ 3-2xlnx + 2 (x > 0) a ≤ x ^ 3-2xlnx + 2 / x ^ 2, Let f (x) = x ^ 3-2xlnx + 2, f (x)' = 3x ^

What should we do if we can't separate the variables into the form of a ≥ f (x)?
Give the general solution routine, Sheila

If we can't get the form of F (a) > = g (x)
Let f (x) = contain a, X formula > = 0, find out the necessary and sufficient condition (i.e. the range of a) under which it defines constant greater than or equal to 0
That is, Fmin (x) > = 0 is the problem of finding the minimum value
The problem of substantial constancy is the problem of seeking the maximum value

If the definition field of a quadratic function is always r, what are the requirements for △ then?

Why is the domain of quadratic function R
If the definition field of quadratic function is r, then it also includes 0? If x takes 0, then the analytic expression of quadratic function is meaningless?

The general formula y = ax ^ 2 + BX + C, right
When x = 0, y = C
Where is no point?

Function y = (SiNx + 1) / (1 + X & # 178;) why is this a bounded function? How to judge