Given the function f (x) = AX2 + BX + C, f (- 3) = f (1) = 0, f (0) = - 3, we can find the solution set of the equation f (x) = 2x

Given the function f (x) = AX2 + BX + C, f (- 3) = f (1) = 0, f (0) = - 3, we can find the solution set of the equation f (x) = 2x

From the known conditions: 9a − 3B + C = 0A + B + C = 0C = − 3; the solution is a = 1, B = 2, C = - 3; from F (x) = 2x, the solution is x2 + 2x-3 = 2x; the solution is x = ± 3, and the solution set is {− 3, 3}

Let the constant ∈ R, take the possible number of roots of equation | x + a | * 2 ^ x = 2011 as the set a of elements=

|X + a | * 2 ^ x = 2011, | x + a | = 2011 / 2 ^ x, | x + a | = 2011 (1 / 2) ^ x, the combination of number and shape y = | x + a | is the upturned broken line, y | = 2011 (1 / 2) ^ x is similar to the image of the exponential function y | = (1 / 2) ^ x, the number of possible intersections can be observed when the broken line is moved left and right,
A={1,2,3}

Using set to express the root of equation
If a bivariate linear equation has two identical roots, such as X1 = x2 = 1, which set is used to represent the solution, do you need to add "(2)" to the lower right corner of {1}, 1?
If you add, is this set a single element set?

Set has two characteristics: one is disorder, the other is mutual heterogeneity;
Mutual dissimilarity means that there cannot be two or more identical elements in a set, that is to say, there are no identical elements in the set
So just write: {1}
If you don't understand, you can ask

Let a, b > 0, if the equation LG (AX) LG (BX) + 1 = 0 about X has a solution, find the range of a / b

lg(ax)lg(bx)+1=0;
(lga+lgx)(lgb+lgx)+1=0;
(lgx)*(lgx)+(lga+lgb)*lgx+lga*lgb+1=0;
According to the formula of the solution of the quadratic equation of two variables, we can know that there is a solution if:
(lga+lgb)*(lga+lgb)-4*(lga*lgb+1)>=0;
That is, (LGA LGB) * (LGA LGB) > = 4;
That is LG (A / b) > = 2 or LG (A / b) = 100 or 0

LG (AX) = 2lg (x + 3) has two unequal real roots
Find out the answer
I figured it out to be (- 3,0) and (12, positive infinity)
Can a be negative?

Yes, the range of corresponding x is required
Did you work it out?
Actually want specific answer

Find the domain of definition, range of value and monotone interval of function y = log13 (x2-5x + 4)

From μ (x) = x2-5x + 4 ＞ 0, the solution is x ＞ 4 or X ＜ 1, so x ∈ (- ∞, 1) ∪ (4, + ∞). When x ∈ (- ∞, 1) ∪ (4, + ∞), {μ| μ = x2-5x + 4} = R +, the range of function y = log13 (x2-5x + 4) is (- ∞, + ∞). Because function y = log13 (x2-5x + 4) is composed of y = log13 μ (x) and μ (x) = x2-5x + 4, function y = log13 μ (x) is in its domain of definition The function μ (x) = x2-5x + 4 is a decreasing function on (- ∞, 52) and an increasing function on [52, + ∞]. Considering the monotonicity of the definition domain and composite function, the increasing interval of y = log13 (x2-5x + 4) is the interval in the definition domain where y = log13 μ (x) is a decreasing function and μ (x) = x2-5x + 4 is also a decreasing function, that is (- ∞, 1); the decreasing interval of y = log13 (x2-5x + 4) is a decreasing function In the domain of definition, let y = log13 μ (x) be the decreasing function and μ (x) = x2-5x + 4 be the increasing function, namely (4, + ∞)

Solution of definition field of logarithmic function (detailed)

Three kinds of methods for finding the domain of function definition
1、 Given the analytic expression of a function, the domain of its definition is usually solved by listing the inequality (Group) of restrictive conditions first
2、 In this paper, we give the definition domain of function and find the definition domain of function. The solution steps are as follows: if the definition domain of function is known, then the definition domain of composite function should be solved by inequality
3、 The solution steps are: if the known domain is, then the domain is the value range when

How to find the definition field of logarithmic function,
Concept and then a few examples to solve the domain of definition
（1）.y=lg2x
（2）.y=loga(1-x^2)

There are two key points to consider when calculating the definition domain of logarithmic function
First, the base must be greater than 0 and not equal to 1
Secondly, the true part must be greater than 0
(1) The base number of this problem is determined to be 10, so just list an equation: 2x > 0, and deduce the domain of definition as {x | x > 0}
Here, we should pay attention to that the domain is a set, which can also be written in the form of interval. We should pay attention to the representation method, which can not be written directly as x > 0
(2) The base number of this question is A. I don't know if the original question has explained the scope of A. if not, I must explain that a > 0 and a is not equal to 0
Second, 1-x ^ 2 > 0, = = > - 1

Finding the definition field of logarithmic function
y=√log1/2(x-1)
log1/2(x-1)≥0 ①
(x-1)＞0 ②
How can I calculate this result here?
X > 1 is obtained from 2
To sum up, 1 < x ≤ 2
Domain (1,2]

log1/2(x-1)≥0
The above formula can be converted to Log1 (/ 2) (x-1) ≥ log (1 / 2) 1 (because the logarithm of 1 is 0)
∵ the logarithmic function with the base of 1 / 2 is a decreasing function,
That is to say, X-1 ≤ 1 means x ≤ 2
Because the true number of logarithm must be greater than 0, X-1 > 0, x > 1,
To sum up, 1

Finding monotone decreasing interval of function,
Search for calculus steps

Given f (x) = (1 / 2) x ^ 2 + LNX
So, f '(x) = x + (1 / x)
Then, when x ∈ (1, e), f '(x) > 0
Then the function f (x) increases monotonically on (1, e) and decreases monotonically on (0,1) (E, positive infinity)