What is the interval of the range of the function y x 1 x 2 Please list the process

There are many ways to solve this problem.

Method 1: Use derivatives to solve.

y'=(1-x^2)/(1+x*x)^2

Lingy'=0, which yields x=-1 or x=1

When x=-1, y=-1 2

When x=-1, y=-1 2

So the range range is [-1 2,1 2].

Method 2: (Solve using mean inequality).

1) When x=0, y=0

2) When x>0.

y=x (1+x*x)=1 (x+1 x) when x>0, x+1 x>=2

03) When x<0.

y=x (1+x*x)=1 (x+1 x) when x>0, x+1 x>=2

1/2<=y<0

Synthesis (1), (2), and (3) can be seen.

The range is [-1 2,1 2].

Method 3: (Solve with trigonometric variables).

Let x=tan

then y=tan (1+tan 2).

sin2α=2tanα/(1+tanα^2)y=1/2sin2α

and sin2 [1,1].

y∈[-1/2，1/2]

Medium y=x-6x+7

x=-b/2a=3

and a guess the source 0

y=-1 at x=2

x=-1.

y=14, then the range is [-1,14].

Medium y=x-6x+7

x∈[1,4]

x=1 at the panicle state.

y=2 x=4.

y=-1 x=3.

The y=-2 range is -2,2

There are many ways to solve this problem.

Method 1: Use derivatives to solve.

y'=(1-x^2)/(1+x*x)^2

Lingy'=0, which yields x=-1 or x=1

When x=-1, y=-1 2

When x=-1, y=-1 2

So the range range is [-1 2,1 2].

Method 2: (Solve using mean inequality).

1) When x=0, y=0

2) When x>0.

y=x (1+x*x)=1 (x+1 buried x)When x> quietly0, x+1 x>=2

Qiling A 0

Summary. First of all, we need to note that the expression under the root number cannot be negative, i.e., 1 - x 0 or x + 3 + x - 1 0 can be simplified

x 1 or x 2 or x 1 Therefore, the domain of the function can be expressed as the interval [-2, 1] 1, +where [-2, 1] indicates that the range of x values is closed [-2, 1], and [1, + means that the range of x values is half-open [1, +

Find the domain of the function y= 1-x++ x+3+x-1 1 and express it as an interval.

Hello dear, first of all, we need to note that the expression under the root number cannot be negative, i.e.: 1 - x 0 or x + 3 + x - 1 0 simplification can be found by: x 1 or x 2 or x 1 Therefore, the definition domain of the function can be expressed as the interval [-2, 1] 1, +, where [-2, 1] indicates that the value range of x is closed interval [-2, 1], and [1, + means that the value range of x is in the half-open interval [1, +

There is another problem, the function f(x) = 1 of the square of x, then f(2) = 4 1. Right or wrong.

This answer is wrong. According to the title, the reciprocal of the square of the function f(x) = x can be expressed as f(x) = 1 x. f(2) is required, then 2 is substituted into the function, and f(2) = 1 2 = 1 4.

Therefore, the value deficit base of plex f(2) is 1 4 instead of 4 1.

The method is as follows, please comma circle for reference:

If there is help from the landslide, please celebrate.

The parabola opening is upward, the axis of symmetry x 0, the minimum value is 0 1 1, and the only thing that is shouted is the value range [ 1, the increasing interval (0, Zheng Suipei) family object.

y=-x+2x+3 =-x-1)+4 The axis of symmetry is x=1 Now the cherry blossom is shouting in the interval [0,3], then the maximum value is f(1), the missing beat = 4, the minimum value is f(3)=0, so the value range is [0 ,4].

Solution: Transforms the original function into :

y=2x²-4x-3

2x Socks Burn - 4x+2-2-3

2(x²-2x+1)-5

2(x-1)²-5

It can be seen that y is a parabola with an axis of symmetry x = 1 and an opening pointing upward.

And: the minimum value is at x=1.

Because x=1 belongs to [0,3].

Then, the minimum value of y at [0,3] is: y(x=1)=2(1-1) -5=-5

For the two endpoints of 0,3.

For x=0, the segment has :

y(x=0)=2(0-1)²-5=-3

For :x=3, there is:

y(x=3)=2(3-1)²-5=3

Because: 3>-3

Therefore, the maximum value of y at [0,3] is y=3

In summary, the value range of y=2x-4x-3 in [0,3] is: [Xunxun-5,3] is solved.

As can be seen from the question, y=

x²-2mx-3

x-m) -m -3, so the axis of symmetry is x=m;

When m", the function increases monotonically on [-1,3) in the range of [2m-2,6-6m);

When m 3, the function decreases monotonically on [-1,3) in the range of (6-6m,2m-2];

When -1 m<(-1+3) 2=1, the minimum value is -m -3 when there is x=m, and the maximum value is 6-6m when m=3, so the value range is [-m -3,6-6m).

When 1 m<3, the minimum value is -m -3 when there is x=m, and the maximum value is 2m-2 when m=1, so the value range is [-m -3, 2m-2]. In summary:

y=(x²+4)/x= x+4/x

x∈[1，3]

y 2 4 = 4, when xx = 4 x i.e. x = 2, take the equal sign.

f(1)=5，f(3)=13/3＜5

The value range is [4,5].

If you don't learn inequality, use the following.

Set 1 x1 x2 3

f(x2)-f(x1)

x2+4/x2)-(x1+4/x1)

x2-x1) -4(x2-x1) (x1x2)=(x2-x1)[1-4 (x1x2)]=(x2-x1)(x1x2- 4) (x1x2) When x1,x2 [1,2], the above equation is 0

At this point, the function is monotonically decreasing.

when x1, x2 (2, 3).

Eq. 0 increases monotonically at this point.

The original function decreases monotonically on [1,2] and increases monotonically on (2,3).

The minimum value is f(2)=4

f(1)=5, f(3)=13 3 f(1) The maximum value is 5

The value range is [4,5].

The function is the minimum when x=2.

x^2+4/x=4

x=1→y=5

x=3→y=13/3

So the range is [4,5].