If f (x) = 3x-2 / 2x + 1, X ∈ [1,4], then the range of y = f (x) is

If f (x) = 3x-2 / 2x + 1, X ∈ [1,4], then the range of y = f (x) is

The range is the range obtained by taking the maximum and minimum values in the domain
The domain is [1,4]
Then when x = 1, f (x) = 3 / 2
When x = 4, f (x) = 9 / 4
So the answer is [3 / 2,9 / 4]
If f (x) = MX / 4x-3 (x ≠ 3 / 4) has f [f (x)] = x in the domain of definition, then M =?
F[F（X）]=m【(mx）/（4x-3）】÷[4(mx）/（4x-3）-3]
=m^2x/(4mx-12x+9)=x
m^2/(4mx-12x+9)=1
∵ for all x in the domain,
The X on the denominator must be eliminated
So 4mx-12x = 0
M=3
I don't understand from here
=m^2x/(4mx-12x+9)=x
m^2/(4mx-12x+9)=1
∵ for all x in the domain,
The X on the denominator must be eliminated
So 4mx-12x = 0
M=3
Start with what you don't understand
This expression is equivalent to
4mx-12x+9=m^2
(4m-12)*x+(9-m^2)=0
It holds for any X
If M ≠ 3, then the left is a function of degree about X, which cannot be equal to 0
So m = 3
Another explanation:
On both sides of the equal sign, the left side is related to x, and the right side is not related to X
Then the coefficient before x must be 0, otherwise when x changes, the left value changes and the right value remains unchanged
Given that the definition field of F (2x = 1) = x ^ 2-3x = 2 is [1,2], then the value field of F (x) is -- forget, please answer
1 ≤ 2x-1 ≤ 22 ≤ 2x ≤ 31 ≤ x ≤ 3 / 2F (2x-1) = x ^ 2-3x-2, let 2x-1 = t, x = (1 + T) / 2F (T) = [(1 + T) / 2] ^ 2-3 (1 + T) / 2-2 = 1 / 4T ^ 2-t-13 / 4 = 1 / 4 (T ^ 2-4t + 4-4) - 13 / 4 = 1 / 4 (X-2) ^ 2-7 / 21 ≤ x ≤ 3 / 2, it can be seen that f (x) is a decreasing function in this interval, so when x = 1, take the maximum value y = - 13 / 4, when x =
Given that the definition field of function f (x) = (MX ^ 2 + 4x + m + 2) ^ - 1 / 2 is real number set R, the value range of real number m is obtained
Greater than or equal to 0 under root sign
The denominator is not equal to 0
So MX ^ 2 + 4x + m + 2 > 0 holds
M=0
It is 4x + 2 > 0, not constant
M ≠ 0 is a quadratic function
When m > 0, the opening is upward
And the discriminant is less than 0
16-4m(m+2)0
m-1+√5
And M > 0
To sum up
m>-1+√5
The definition field of function f (x) = (3x-1) / (2x + 3) is? And the value field is?
2X + 3 is not equal to zero, X is not equal to - 1.5
The range 3Y + 1 / 3-2y is not equal to - 1,5
Y is not equal to 1.5
Let the domain of function f (x) be d. if there is a non-zero constant l such that f (x + L) ≥ f (x) for any X &; m (M &; d), then f (x) is called a high-key function on M and l is a high-key value. If f (x) = x ^ 2 + 2x is a high-key function on (- ∞, 1), then the range of high-key value L is
（x+L）^2+2（x+L）≥x^2+2x
That is, L ^ 2 + 2lx + 2L ≥ 0 holds in (- ∞, 1]
l＜0
How did L ^ + 4L ≥ 0 come out?
I'm going to do it, because L ^ 2 + 2lx + 2L ≥ 0 holds on (- ∞, 1)
Let the left = g (x) = 2lx + (L ^ 2 + 2L), which is a linear function,
To always be above the x-axis on (- ∞, 1], then 2L is necessary
Finding the domain of definition and value of y = - 2x + 1 / 3x + 1
y=(-2x+1)/(3x+1)
=(-2x-2/3+2/3+1)/(3x+1)
=-2/3+(5/3)/(3x+1)
The definition field is: {x | x ≠ - 1 / 3, X belongs to real number}
The range is: {y | y ≠ - 2 / 3, y belongs to real number}
If there is a constant M > 0 for any x ∈ R and | f (x) | ≤ m | x |, then f (x) is an odd function over R
If there is a constant M > 0 for any x ∈ R and | f (x) | ≤ m | x | then f (x) is an odd function on R and satisfies that | f (x1) - f (x2) | ≤ 2 | x1-x2 | is an F function for all x 1x2. Why? (function graph)
F is an odd function, so f (0) = 0
For any x ∈ R, | f (x) | = | f (x) - f (0) | ≤ 2 | x-0 | = 2 | x |. It is equivalent to the existence of M = 2, satisfying the definition of F function
For any x ∈ R, | f (x) | = | f (x) - f (0) | ≤ 2 | x-0 | = 2 | x |. It is equivalent to the existence of M = 2, which satisfies the definition of F function.
For any x ∈ R, | f (x) | = | f (x) - f (0) | ≤ 2 | x-0 | = 2 | x |. It is equivalent to the existence of M = 2, which satisfies the definition of F function
If f (x) is an odd function on R, then m does not exist
F (x) = x ^ 2-2x-3 of x ^ 2-3x-4?
The definition field is to make the function meaningful as long as the denominator of the function is not equal to 0, so try to find x ^ 2-3x-4 is not equal to 0 in the definition field, first find x ^ 2-3x-4 = 0, get x = 4 or x = - 1, so the definition field is {x | x is not equal to 4 and X is not equal to - 1, X belongs to real number r}
If the numerator and denominator drop x + 1 at the same time, f (x) = (x-3) / (x-4)
Segment: X
The first problem is y = ln (x 1) / radical (- x ^ 2-3x 4) domain, the second problem is y = x-radical 3, 2f (x) + F (4-x) = x 2; - 1, change 4-x into x, and solve the equation. 1, X
The function f (x) defined on R satisfies the following conditions: ① f (x + y) + F (X-Y) = 2F (x) cosy, ② f (0) = 0, f (π / 2) = 1
I have proved that f (x) is an odd function. I don't know how to evaluate f (x) P.S. please don't copy thx
Let y = π / 2, f (x + π / 2) + F (x - π / 2) = 2F (x) cos π / 2 = 0,
So f (x + π / 2) = - f (x - π / 2) = f (π / 2-x)
Let x = π / 2, f (π / 2 + y) + F (π / 2-y) = 2F (π / 2) cosy = 2cosy,
f(π/2+y)+f(y+π/2)=2cosy
So f (y + π / 2) = cosy,
Let y + π / 2 = X,
So f (x) = SiNx
∵f(x+y)+f(x-y)=2f(x)cosy
Let x = y = 0
f(0+0)+f(0-0)=2f(0)cos0==>f(0)=0
f(0+π/2)+f(0-π/2)=2f(0)cos(π/2)==>f(π/2)+f(-π/2)=0
f(π/2+0)+f(π/2-0)=2f(π/2)cos(0)==>f(π/2)=1
∴f(-π/2)=-1
Ψ f (x)... Expansion
∵f(x+y)+f(x-y)=2f(x)cosy
Let x = y = 0
f(0+0)+f(0-0)=2f(0)cos0==>f(0)=0
f(0+π/2)+f(0-π/2)=2f(0)cos(π/2)==>f(π/2)+f(-π/2)=0
f(π/2+0)+f(π/2-0)=2f(π/2)cos(0)==>f(π/2)=1
∴f(-π/2)=-1
∴f(x)=sinx
Obviously, sin (x + y) + sin (X-Y) = 2sinxcosy holds