Find the definition and range of the following functions (1). Y = 2 ^ x + 3

Find the definition and range of the following functions (1). Y = 2 ^ x + 3

).y=2^x+3
The domain is r
2 ^ x > 0, then Y > 3
The range is (3, + infinity)
X belongs to all real numbers, Y > 3
For a function f (x) whose domain of definition is all real numbers, if there is a real number x0 such that f (x0) = x0 holds, then x0 is called the fixed point of F (x0)
Given the function f (x) = AX2 + (B + 1) x + (B-1) (a ≠ 0), if any real number B, f (x) always has two different fixed points, find the value range of real number a
This problem uses the construction method, transforms into the quadratic function to have two zeros
Let g (x) = f (x) - x = AX2 + BX + (B-1)
Let g (x) = 0
So AX2 + BX + (B-1) = 0 has two unequal real roots
△>0
b²-4a（b-1）＞0
Parameter separation
When b > 1, a < B & # 178; / (4b-4) holds
That is, the minimum value of a < B & # 178; / (4b-4)
I don't know if the LZ derivative will be zero. Here I use the derivative to get the minimum value of 1 when B = 2
a＜1
When B = 1, △ = 1 holds
When B ＜ 1, a ＞ B & # 178; / (4b-4) is constant
The maximum of a ＞ B & # 178; / (4b-4)
Similarly, it can be seen from the derivative that B & # 178; / (4b-4) increases monotonically at (- ∞, 0) and decreases monotonically at (0,1)
So the maximum value is 0 when it is 0
a＞0
All three conditions must be met at the same time
So we take the intersection
0＜a＜1
The second solution is to treat B & # 178; - 4a (B-1) > 0 as a new quadratic equation with one variable
Consider a as a known quantity
Use △ again
That is, f (x) = AX2 + (B + 1) x + (B-1) = 0 has two different roots
The discriminant = (B + 1) ^ 2-4a (B-1) = B ^ 2-2 (2a-1) B + 4A + 1 = (b-2a + 1) ^ 2-4a ^ 2 + 8a is always greater than zero
That is - 4A ^ 2 + 8A > 0
The solution is 0
Y = f (x) + 1-2f (x), where f (x) ∈ [3 / 8,4 / 9], find the range of function
Let t = root 1 / 2-x, and the final answer is [7 / 8,7 / 9]
(2011 Jiading District the first mock exam) two times function y=ax2+bx+c, a C < 0, the zero point of the function is ()
A. 1b. 2C. 0d. Not sure
∵ AC ＜ 0, ∵ △ = b2-4ac ＞ 0, ∵ the corresponding equation AX2 + BX + C = 0 has two unequal real roots, so the quadratic function and X-axis have two intersections
Given f (2x + 1) = 3x-2, find the analytic expression of F (x),
Let 2x + 1 = t, then x = (t-1) / 2
f(t)=3[(t-1)/2]-2=3t/2-7/2
The analytic expression of F (x) is obtained by replacing T with X
f(x)=3x/2-7/2
Let f (x) = AX2 + BX + C (a is not equal to 0) for any real number, f (2 + T) = f (2-T) holds, then f (- 1). What is the size of F (1) f (2)
Who will?
I want a specific process,
If f (2 + T) = f (2-T) holds, then f (x) is symmetric with respect to x = 2, then x = 2 is the axis of symmetry of F (x)
Discuss the positive and negative of A
If a ＞ 0, the opening of F (x) is upward, f (2) is the minimum, and - 1 and 1 are on the left side of the symmetry axis, decreasing, and the distance between - 1 and 2 is greater than that between 1 and 2, then f (- 1) ＞ f (1) ＞ f (2)
If a ＜ 0, then the opening of F (x) is downward, f (2) takes the maximum, and - 1 and 1 are on the left side of the symmetry axis, increasing, and the distance between - 1 and 2 is greater than that between 1 and 2, then f (- 1) ＜ f (1) ＜ f (2)
Given f (3x) = 2x ^ 2-1, find the analytic expression of F (x)
Please speed up the process~~~
Let a = 3x
x=a/3
f(a)=2(a/3)²-1=2a²/9-1
f(x)=2x²/9-1
Let y = 3x, that is, x = Y / 3, and substitute y in.
Solution: Let f (x) = ax ^ 2 + BX + C, (a, B, C are constants), because f (3x) = 9ax ^ 2 + 3bx + C = 2x ^ 2-1, so 9A = 2, 3b = 0, C = - 1, so a = 2 / 9, B = 0, C = - 1, so f (x) = 2 / 9x ^ 2-1.
If u = 3x, then x = u / 3
So f (U) = 2 * (U / 3) ^ 2-1 = 2U ^ 2 / 9-1
The above formula holds for any u belonging to R
Therefore, the expression of F (x) is f (x) = 2x ^ 2 / 9-1
Let t = 3x, x = t / 3, f (T) = 2 (T / 3) ^ 2-1 = 2 / 9t ^ 2-1, so f (x) = 2 / 9x ^ 2-1
The quadratic function f (x) = ax ^ 2 + BX + C (a, B, C ∈ R, a ≠ 0) is known and satisfies the following conditions: 1. F (- 1) = 0.2. For any real number x, f (x) - x ≥ 0
3. When x ∈ (0,2), f (x) ≤ ((x + 1) / 2) ^ 2
(1) . find the value of F (1)
(2) Find the value of a, B, C
(3) If G (x) = f (x) - MX (M is a real number) is a monotone function when x ∈ [- 1,1], the value range of M is obtained
From F (- 1) = 0, A-B + C = 0 is obtained. ① for any real number x, if f (x) - x ≥ 0, then f (1) ≥ 1. And the discriminant of equation AX ^ 2 + BX + C = x △ = (B-1) ^ 2-4ac ≤ 0. ③ when x ∈ (0,2), then f (1) ≤ ((1 + 1) / 2) ^ 2 = 1
It is the most intuitive to use the image method to solve the problem.
(1) f (x) = 3x ^ 2-2, find the analytic formula of F (2x-1)
f(x)=3x^2-2,
f(2x-1)=3（2x-1)²-2
=3(4x²-4x+1)-2
=12x²-12x+1
Just take it in and find it out
12X²－12X+1
As shown in the figure, the image of quadratic function y = AX2 + BX + C intersects with X axis at a and B, and intersects with y axis at point C, and ∠ ACB = 90 °, AC = 12, BC = 16
∵∠ ACB = 90 °, AC = 12, BC = 16, ∵ AB = ac2 + BC2 = 20; ∵ AC ⊥ BC, OC ⊥ AB, ∵ Ao = ac2 △ AB = 7.2, ∵ Bo = 12.8; ∵ a (- 7.2, 0), B (12.8, 0); ∵ CO2 = ob · OA, ∵ OC = 9.6, ∵ C (0, 9.6), substituting a (- 7.2, 0), B (12.8, 0), C (0, 9.6) into the function y = a (x + 7.2) (x-12.8), 9.6 = - 92.16, a = - 548 The relation of image is y = - 548 (x + 7.2) (x-12.8)  - 548x2 + 712x + 9.6