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Inverse function of y = 4x + 3 / X-2 I want to be more detailed. I need to define the domain
y=4x+3/x-2
y(x-2)=4x+3
xy-2y=4x+3
xy-4x=3+2y
x(y-4)=3+2y
x=(3+2y)/(y-4)
So the inverse function of y = 4x + 3 / X-2 is
y=(3+2x)/(x-4) （x≠4）
(exchange X and y of the above formula)
Is there an inverse function for y = x ^ 2-4x-5
No, because the function is defined as: every value x takes in the domain has a unique y value corresponding to it
The inverse function of function f (x) = x2-4x-5, X ∈ [- 2,1) is F-1 (x) =? There is a process, 3Q~
f(x)=x2-4x-5=(x-2)^2 - 9
Let y = f (x) and (X-2) ^ 2 = y + 9
|X-2 | = radical (y + 9)
Because of X-2
Function y = x2-4x-2 (- 1)
y=(x-2)²-6
Axis of symmetry x = 2
Then the domain of definition is on the same side of the axis of symmetry and is a monotonic decreasing function
Existence of inverse function
x=-1,y=3
x=1,y=-5
So - 5
Domain of inverse function y = 4 / 1 + x ^ 2
The domain of inverse function is y = 4 / 1 + x ^ 2, so y = 4 / (1 + x ^ 2) gives YX ^ 2 + y-4 = 0
Δ = B ^ 2-4ac = - 4Y (y-4) ≥ 0 (∵ x ∈ R) of equation (1)
∴ 0≤y≤4
The definition field of inverse function of y = 4 / 1 + x ^ 2 is [0,4]
The value range of the function y = (1 / 2) to the power of x2-2x + 1 (x greater than - 1 less than 1) is?
y=1/2x²-2x+1
=1/2(x²-4x)+1
=1/2(x-2)²+1-2
=1/2(x-2)²-1
Because - 1
The range of the function y = log2 (x2-2kx + k) is r
Given that the value range of function y = log2 (x2-2kx + k) is r, then the value range of K is ()
A．0＜k＜1
B．0≤k＜1
C. K ≤ 0 or K ≥ 1
D. K = 0 or K ≥ 1
Ask for the reason!
A:
The domain of y = log2 (X & # 178; - 2kx + k) is r
It shows that the minimum value of G (x) = x & # 178; - 2kx + k is not greater than 0
So: there is at least one zero point of parabola g (x)
Discriminant = (- 2K) &# - 178; - 4K > = 0
So: 4K & # 178; - 4K > = 0
4k(k-1)>=0
k=1
So: choose C
Note: the title is value field, not definition field
Y = 1 + log2 ^ X / 2 + (log2 ^ x) ^ 2
Log2 ^ x = t inverse solution
Let log2 ^ x = t
y=1+t/2+t^2
=(t+1/4)^2+15/16
∵t∈R
∴y∈[15/16,+∞）
Y = log2 (x), x > 1 for the range, there must be a detailed process
For example, y = log2 (x), x > 1 to find the range, there must be a detailed process
Log2 (x) increases monotonically on (1 positive infinity), log2 (x) > log2 (1) > 0, so y = log2 (x), and the range of x > 1 is (0 positive infinity)
Find the range of y = log2 (x ^ 2 + 4)
x^2+4>=4
The logarithmic function with base 2 is monotone increasing function
So the range is 2 to positive infinity
The minimum value is 2
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