Given the function y = 4 ^ x-3 * 2 ^ x + 3, when its range is [1,7], the value range of X is To solve the problem thinking, detailed process and each step of thinking

Given the function y = 4 ^ x-3 * 2 ^ x + 3, when its range is [1,7], the value range of X is To solve the problem thinking, detailed process and each step of thinking

Because: y = 4 ^ x-3 × 2 ^ x + 3, y ∈ [1,7]. So: 1 ≤ 4 ^ x-3 × 2 ^ x + 3 ≤ 7-2 ≤ 4 ^ x-3 × 2 ^ x ≤ 4-2 ≤ (2 ^ 2) ^ x-3 × 2 ^ x ≤ 4-2 ≤ (2 ^ x) ^ 2-3 × 2 ^ x ≤ 4, that is: (2 ^ x) ^ 2-3 (2 ^ x) + 2 ≥ 0 (1) And: (2 ^ x) ^ 2-3 (2 ^ x) - 4 ≤ 0 (2) By (...)
Given the function y = - 2x & # 178; + 3, X ∈ {- 2, - 1,0,1,2}, then its range is []
Please explain the process, thank you!
solution
When x = - 2, y = - 5
When x = - 1, y = 1
When x = 0, y = 3
When x = 1, y = 1
When x = 2, y = - 5
The value range is: {1,3, - 5}
Given that the value range of function y = x2-3x + 3 (x > 0) is [1,7], then the value range of X is ()
A. （0，4]B. [1，4]C. [1，2]D. （0，1]∪[2，4]
∵ y = x2-3x + 3 = (x − 32) & nbsp; the symmetry axis of 2 + 34 is x = 32 and ∵ x > 0. The range of function is [1,7]. When x2-3x + 3 = 1, we can get x = 1 or x = 2. When x2-3x + 3 = 7, we can get x = 4 or x = - 1 (rounding) function. Y = x2-3x + 3 decreases monotonically in (0,1) and increases monotonically in [2,4]. Therefore, when the range of function is [1
Given the odd function y = f (x), when x > 0, f (x) = x & # 179; + 2x & # 178; - 1, if
Less than 0 is equivalent to f (- x)
=（-x）^3+2(-x)^2-1
=-x³+2x²-1
　　f(x)=-x³+2x²-1
When the formula of X0
That is, f (- x) = - x ^ 3 + 2x ^ 2-1
From the odd function, we can get f (x) = - f (- x) = - (- x ^ 3 + 2x ^ 2-1) = x ^ 3-2x ^ 2 + 1
From the definition of odd function, f (x) = - f (- x) = - (- x ^ 3 + 2x ^ 2-1) = x ^ 3-2x ^ 2 + 1
Given that the range of function y = 4 ^ X - (3 * 2 ^ x) + 3 is [1,7], find the value range of X
Let 2 ^ x = t
y=t^2-3t+3
The range is [1,7],
t^2-3t+3=1
T = 1 or 2
t^2-3t+3=7
T = 4 or - 1
T>0
So - 1 shed
2^x=4
X=2
T = 1 or 2
2 ^ x = 1 or 2
X = 0 or 1
Y = T ^ 2-3T + 3 lowest point t = 3 / 2
So t ≠ 1
Otherwise, the minimum value is not 1
t∈[2,4]
The value range of X should be: [1,2]
Let f (x) = ax ^ + BX + C (a > 0 and C ≠ 0), and f (1) = - A / 2, prove that f (x) has at least one zero point in the interval (0,2)
f(0)=c f(1)=-a\2=a+b+c f(2)=4a+2b+c=a-c
1. When C > 0, f (0) = C > 0, f (1) = - A-2
In the exponential function, how to find the definition field for the known value range, such as: known 4 ^ x-3 * 2 ^ x + 3; when the value range is [1,7], find the value range of X
Let f (x) = 4 ^ x-3 * 2 ^ x + 3, let t = 2 ^ x > 0, then f (T) = T ^ 2-3T + 3 = (T-3 / 2) ^ 2 + 3 / 4
So 1 ≤ (T-3 / 2) ^ 2 + 3 / 4 ≤ 7, so 1 / 4 ≤ (T-3 / 2) ^ 2 ≤ 25 / 4,
So - 5 / 2 ≤ T-3 / 2 ≤ - 1 / 2 or 1 / 2 ≤ T-3 / 2 ≤ 5 / 2,
So - 1 ≤ t ≤ 1, or 2 ≤ t ≤ 4, so t ∈ (0,1] ∪ [2,4]
So x ∈ (- ∞, 0] ∪ [1,2]
If a, B, C are greater than or equal to 1, f (x) = ax ^ 2 + BX + C has two different zeros in the interval (0,1), then the minimum value of F (1)
If f (x) = ax ^ 2 + BX + C has two different zeros in the interval (0,1), then f (1)
Minimum value of
If a, B, C are greater than or equal to 1, and the function f (x) = ax ^ 2 + BX + C has two different zeros in the interval (- 1,0), then the minimum value of F (1)
There are only two zeros in a quadratic function, and the smaller one is [- B - √ (B & # 178; - 4ac)] / (2a). When a and B are both greater than 0, the zeros mentioned above are less than 0, so it is impossible for the two zeros to be in the (0,1) interval;
The range of function y = 2tan (x - π / 3), X ∈ [0,2 π / 3]
TaNx is singly increased on (- π, π)
So 2tan (x - π / 3) is increasing on (- 2 π / 3,4 π / 3)
So 2tan0 = - 2 √ 3 / 3
2tanπ/3=2√3/3
So the range is [- 2 √ 3 / 3,2 √ 3 / 3]
I'm very happy to answer for you. I wish you progress in your study! If you don't know, you can ask!
For the function f (x) = lnx-ax ^ 2-bx, there are two zeros x1, x2
The students upstairs have questions to answer
Because the zero point is the zero point of the original function, not the zero point of the derivative function