Find the range of the function f x 2x 3 x 1 2 x 2 and x 1 .

Find the range of the function f(x) 2x 3 x 1 ( 2 x 2 and x 1).

Solution: I suggest you abandon your current method and use the following method to solve:

From y=(2x-3) (x+1), y(x+1)=2x-3, (y-2)x=-y-3, so x=-(y+3) (y-2), we can know y≠2

and from -2 -(y+3) (y-2) 2:

y+3)/(y-2)≧-2...1)

y+3)/(y-2)≦2...2)

From (1) (y+3) (y-2)-2=[(y+3)-2(y-2)] (y-2)=(-y+7) (y-2)=-(y-7) (y-2) 0, (y-7) (y-2) 0, so y7 or y<2....3)

From (2) we get 2+(y+3) (y-2)=[2(y-2)+(y+3)] (y-2)=(3y-1) (y-2)=3(y-1 3) (y-2) 0, we get y1 3 or y>2....4)

3) (4)={y - asymptote.

Since x lim(2x-3) (x+1)=x lim(2-3 x) (1+1 x)=2

So y=2 is the horizontal asymptote of the function.

x -1+lim(2x-3) (x+1)=x -1+lim[2-5 (x+1)]=- x -1+ means x is close to -1 from the right side of -1).

x -1-lim(2x-3) (x+1)=x -1-lim[2-5 (x+1)]=+ x -1- means x is close to -1 from the left of -1).

f(-2)=(-4-3)/(-2+1)=7

f(2)=(4-3)/(2+1)=1/3.

Using these discussions, you can also draw an image of the function.

Uh......Because you don't make t=x+1, f(x)=2+(-5 x+1)= -5 t+2

And because -1 t 3 and t ≠ 0, the image of y=-5 t is drawn first, and then the image is shifted upwards by two unit lengths, so that the value range is obtained.

After seeking the derivation, I slowly found that f verified that (x)>0(x>1 2) is constant.

So the minimum value obtained at x = 1 2 is f(1 2).

f(x) can be arbitrarily large, so the value of the focal mode domain is (f(1 2),

First, the domain of the function is defined as all real numbers that make the denominator unequal and older than zero. Because the denominator is 2x + 3, the domain is defined as x ≠ 3 2.

Next, ask for the range of the function. You can write a function in the form of:

f(x) =x-1)/(2x+3) =1/2 - 5/4)/(2x+3)

As you can see, when x takes any real number, 2x + 3 can take any real number and zero. So (5 4) (2x+3) can take any real number (excluding 0), and 1 2 is a constant.

Thus, it can be concluded that the range of the function is all real numbers that are not equal to 1 2. That is, the range is (-1 2) (1 2,+.)

Solution: F(x)=(2x-1) (x+1).

f(x)=1+(x-2) (x+1) because the domain is defined as [3,5].

And (x-2) (x+1) is an increasing function on [3,5], so it is an increasing function in the defined domain [3,5], so when x=3, the minimum value f(x)=5 4, x=5, and the maximum value f(x)=3 2, so the value range is [5 4,3 2].

Please remember to adopt it, thank you!

Solution: Others have already done it, I only talk about the method.

1. First of all, we must prove that this function is an increasing function.

There are two ways to prove that this function is an increasing function: one is to prove that the derivative of f(x) is always positive at [3,5]. The second is to deform the function f(x) to f(x)=1-3 (x+2) with a definition, which can prove that this function is an increasing function.

2. Since f(x) is an increasing function, we can find the maximum and minimum values of f(x) in [3,5], that is, the maximum value = f(5) = 4 7

Min = f(3) = 2 5

This is the range of f(x).

Yes [2 5, 4 7].

f（x）=x-1/x+2=1-3/x+2

x∈［3，5］

f(x) is a monotonically increasing function, x=3 is the minimum, f(x)=2 5x=5 is the maximum, and f(x)=4 7

The range of f(x) is 2 5, 4 7

Hello! Solution:

x-1) (x+2)=1-3 (x+2), substituting x=3 and x=5.

When x=3. Formula = 2 5

When x=5, the equation = 4 7

The range of f(x) is [2 5, 4 7].

If there are other questions after this question, send and click on my avatar to ask me for help, it is not easy to answer the question, please understand, thank you.

Yours is the driving force of my service.

Good luck with your studies!

The top is the increment function, and the bottom is the increment function, so this is still a monotonic function.

Just calculate the values of the function at x=3 and x=5 and the range is determined.

Solution: f(x)=x 2-3x-4

The axis of symmetry of f(x) is -3 (-2)=3 2, that is, f(x) has a minimum value at 3 2, and the minimum value is -25 4 |-1-3/2|>|3-3/2|

f(x) has a maximum value at -1, and the solution gives a maximum value of 0 f(x) and a value range of [-25 4,0) at (-1,3).

Solution: f(x)=x 2-3x-4=(x-3 2) 2-25 4 3 2 (

When x=3 2, f(x) has a minimum value of -25 4 and f(x)=x 2-3x-4=(x-4)(x+1) then when x=-1, f(x)=0

x again (-1,3).

f（x）∈[9/4，0）

Is this a standard of gray?

Give the power to adopt put. Thanks and have a nice day.

1. Formula.

f(x)=[x (3 2)] 25 4)2, make this quadratic function image, mark the axis of symmetry and the known interval;

3. Combine the image and the known interval to determine the maximum and minimum values by using the monotonicity of the function;

4. Give the value range.

Answer: [ 25 4, 4).

Solution: f(x)=x 2-3x-4=(x-3 2) 2-25 4 3 2 (

When x=3 2, f(x) has a minimum value of -25 4 and f(x)=x 2-3x-4=(x-4)(x+1) then when x=-1, f(x)=0

x again (-1,3).

f（x）∈[9/4，0）

Is this a standard of gray?

Give the power to adopt put. Thanks and have a nice day.

1. Formula.

f(x)=[x (3 2)] 25 4)2, make this quadratic function image, mark the axis of symmetry and the known interval;

3. Combine the image and the known interval to determine the maximum and minimum values by using the monotonicity of the function;

4. Give the value range.

Answer: [ 25 4, 4).

Solution: f(x)=x 2-3x-4

The axis of symmetry of f(x) is -3 (-2)=3 2, that is, f(x) has a minimum value at 3 2, and the minimum value is -25 4 |-1-3/2|>|3-3/2|

f(x) has a maximum value at -1, and the solution gives a maximum value of 0 f(x) and a value range of [-25 4,0) at (-1,3).

1 0 and x (-1,3

When x = 3 2, f(x) is minimum, f(x = -25 4), and when x = 3, f(x) is maximum: f(x = -4

The function f(x)=x -3x-4 has a range of -25 4,-4 in x (-1,3).

f(x)=(x-3)+1 (x-3)+3>=2*root number((x-3)*1 (x-3))+3=5The equal sign holds if and only if x-3=1 (x-3). At this point x=4

So the value range is [5, infinite).

f(x)=

The range is [-4,4] =[-4,4].

So the range of the original function is [-4,4].

Hope it helps!