Senior 1 Mathematics Problem of Evaluating Range!!! Before 6 o clock!!

-1 y 2 at 2 x 3

-2 y at 0 x 5

There are three scenarios.

a 0 hours 3 y 12-3a

06:12-3a y 3

Find the range of y=x 2-2x-1, in the first case: 2 x 3 and in the second case: 0 x 5

Solve 11 x 2 and draw the graph of the function which is a parabola, and then take x to find y2 x 14 and analyze it as 1

y=x 2-ax+3 0 x 3 Finding the Y Range This is to discuss the value of a, and to do this kind of problem, you should first turn the function into a vertex formula, i.e., y=(x-a 2) 2+3-(a 2) 2, and then solve it according to the value of x and discuss the value of a

y=x^2-2x-1

y'=2x-2 requires'=0 x=1 then at 2 x 3 Monotonically decreasing f(2) f f(3).

0 x 5 The biggest is f(1) and compare f(0) hef(5) to see who is smaller.

y=x^2-ax+3 y'=2x-a require'=0 x=a 2 discussion a 2 0 3 a 2 and 0 a 2 3 when a 2 0 decrements when 3 a 2 increments when 0 a 2 3 f(a 2) max Compare f(0) and f(3) to see who is smaller.

The first mentions, the first case (-1, 2).

The second case (-2, 14).

Question 2, A 6, is (12-3A, 3).

a 0, (3, 12-3a).

The minimum value of 0 a 6 is 3-a2 4, and the maximum value is the maximum of 0 and 12-3a.

Hope mine is helpful to you.

This is simple, look directly at the axis of symmetry, the equation is: b (2a), such as the first one is 1, and a 0, for the opening upward parabola, the axis of symmetry takes the minimum value, the farther away from the axis of symmetry the greater the value, the problem is solved, the second one is the same, but to discuss the size of a is OK, the function must be basically used in the future, the college entrance examination and function-related questions have 70 80 points, so, come on!

The discriminant formula is simpler, organized as (y-2)x 2+(y+1)x+(y-2)=0, the denominator is evergrandly higher than 0, and the defined domain x belongs to the real number r, then the discriminant formula =(y+1) 2-4(y-2) 2 is greater than or equal to zero, that is, y 2-6y 5 is less than or equal to zero, and 1 is less than or equal to x less than or equal to 5

The first step is to disassemble the numerators of the formula so that 3 x+2 roots are under x.

In the second step, take (1 root number x) as a whole, turn it into a one-dimensional quadratic polynomial, or one-dimensional quadratic function, and turn it into a vertex formula, yield: 3*-1 3

The third step is that you should know the basic properties of the quadratic function, so I won't say much about that. Because the whole (x under 1 root number) is greater than zero, that is, (-1 3) cannot be taken, and (x under 1 root number) is much larger than 0, the farther away from the axis of symmetry -1 3, the larger the result will be, so even if (x under 1 root number) is taken to 0, the result of the whole value is 0So the range is greater than 0

y = x parts (3 + 2 times the root number x).

To process parsing: y=(3+2 x) x

3/x+2/√x

3(1/√x+1/3)^2-1/3

When x +, 3(1 x+1 3) 2 1 3 y>0

or y'=-(3+ x) x 2<0, the function y is monotonically reduced within the defined domain (0,+.

When x +, 3 x +2 x 0

y>0

y=(3 + 2√x)/x= 3/x + 2/√x= 3(1/√x + 1/3)^2 - 1/3

Since x>0, 1 x>0,y cannot take the minimum value -1 3;Suppose 1 x = 0 and there is y = 0

So, the range of the function is y>0

The process uses software, I'm here for your share, please reward.

1. First find the definition field, x in the root number x is greater than or equal to 0, and x is the denominator, so x is greater than 0

2. Use the commutation method: let t be equal to the root number x, t is greater than 0, and then simplify the square of 3, y=3 (1/t 1) + 2 (1 t t) 4, the value range of t 1 is also greater than 0,5, and only need to solve the value range of a one-dimensional quadratic equation.

y=(3+2 root number x) x

3 x+2 root number x

3 (1 root number x+1 3) 2-1 3

Because 1 root number x>0

So y>0

y=(3+2 times the root number x) x

y=3 x+2 root number x

y=3(1 root number x+1 3) 2-1 3

1 root number x>0

y>0

First of all, the denominator is a real number other than o, but the root number is generally greater than or equal to o, and there is a denominator limit, so the inner must also be greater than 0, and the result is x>0, and it is not equal to 0!

Let x=t divide the numerator and denominator by t using the basic inequality.

1. By the title: x>=-1, so y>0

When x=-1, y=0, when x>-1, 1 y=[under the root(x+1)]+1 [under the root(x+1)], 1 y>=2, that is, y<=1 2. (Think of x+2 as x+1 + 1).

So the range is: [0,1 2].

2. Let's talk about it later, let's go play first, if you have any questions, you can send me QQ on it, 471023334

1.Under the root number (x+1)]=t, and t>=0, then x=t 2-1, so y=t (t 2+1), then 1 y=t+1 t>=2(t!).=0), when t=0, y=0, thus 0<=y<=1 2

2.3x-1>=0, we get x>=1 3, and it increases on x>=1 3, so y>=(1 3) 3+0=1 27

3.First of all, the denominator (2-sinx) will not be 0, and the solution gives sinx=(2y-1) (y-1), and there is |sinx|< = 1 to get |(2y-1)/(y-1)|< = 1 launch 0< = y< = 2 3

4.First of all, the denominator (2-cosx) will not be 0, and it will be determined by y=(1-sinx) (2-cosx).

2y-1=ycosx-sinx=[root number under (y 2+1)]*sin(x+t), so |sin(x+t)|=|(2y-1) [under the root number (y 2+1)]|=1 gives 0<=y<=3 4

You can ask me any math and physics problems in the future.

I think the first question is 0 to 1/2, the second question is greater than or equal to 1/27, the third question is 0 to two-thirds, and the fourth question is 0 to 4/3 I don't know if it's right If you want specific steps Tell me I'll send it to you in the evening Is it okay.

Solution:1

Let u= (x 1), then u [0,

y＝u/(u²＋1)＝1/(u＋1/u)

According to the mean theorem:

When u 0, u 1 u 2

Again, apparently there is no maximum value for u 1 u.

u＋1/u∈[2,﹢∞

y＝1/(u＋1/u)

y∈(0,1/2]

When u 0, y u (u 1) 0

y∈[0,1/2]

2.The domain is obviously defined as: x [1 3,

In [1 3, on, x and (3x 1) are obviously both additive functions, and the sum of the additive functions is still an increasing function, then y is also an increasing function, and when x takes the minimum value, y takes the minimum value i.e. ymin y(1 3) 1 27

And obviously y does not have a maximum, then.

y∈[1/27,﹢∞

3.Let u=sinx, then u [ 1,1].

y＝(1－u)/(2－u)＝(u－2＋1)/(u－2)＝1＋1/(u－2)

u∈[﹣1,1]

u－2∈[﹣3,﹣1]

1/(u－2)∈[1,﹣1/3]

y∈[0,1/3]

Let u tan(x 2), then u ( y (u 1) (3u 1).

Let v u 1, then v (

y＝v²/[3(v＋1)²＋1]＝v²/(3v²＋6v＋4)＝1/3·(3v²＋6v＋4－6v－4)/(3v²＋6v＋4)＝1/3－2(3v＋2)/(9v²＋18v＋12)

Let w 3v 2, then w (

y＝1/3－2w/[9(w－2)²/9＋18(w－2)/3＋12]＝1/3－2w/(w²＋2w＋4)＝1/3－2/(w＋4/w＋2)

According to the mean theorem:

When w 0, w 4 w 4;When w 0, w 4 w 4 and obviously w 4 w have no maximum or minimum value.

w＋4/w∈(﹣4]∪[4,﹢∞w＋4/w＋2∈(﹣2]∪[6,﹢∞1/(w＋4/w＋2)∈[1/2,0)∪(0,1/6]∴﹣2/(w＋4/w＋2)∈[1/3,0)∪(0,1]∴y∈[0,1/3)∪(1/3,4/3]

When w 0, y 1 3 2w (w 2w 4) 1 3 y [0,4 3].

1=< y =<4

1=< (ax+b) (x 2+1) =<4 after decomposition.

ax+b >= -x^2-1

4x^2+4 >= ax+b

x^2+ax+b+1 >= 0

4x^2-ax-b+4 >= 0

x+a 2) 2+b+1-a 2 4 >= 0 gives b+1-a 2 4=0

2x+a 4) 2-b+4-a 2 16 >= 0 gives -b+4-a 2 16=0

Solve the system of binary quadratic equations.

a=4 b=3

or a=-4 b=3

Define domain: x≠-1

Range: Obviously, you need to determine the range of the exponent first.

If a>0 then the function has a minimum value, when x = - (b 2a), y takes the minimum value, and the minimum value is y = (4ac-b 2) 4a).

Let g(x)=(3x-1)(x+1)=3x +2x-1 have a minimum value, minimum value = (-12-4) 12=-4 3

and when x=-1 3.

The exponential function is monotonically increasing.

The original function also has a minimum value, which is f(-1 3)=2 (-3=)1 8, which is the value range: f(x) 1 8

Question 1. <>

Question 2: Stove Qixin Hidden Wheel.

Question 3. <>

Please order a side scum, thank you.

The evaluated range is a set of infinite numbers because the domain itself is not continuous.

You were taught math by a PE teacher, right? A simple one-stroke question will not be!

Let u 4x,v (8 - 2x), then u 8 v 8, y u v, substitute u 8 (y - u) 8, collate to get 9u -16yu 8y -64 0, discriminant (-16y) 36(8y -64) 0, solve -6 2 y 6 2, and since v 0, then -2 x 2, u 8, so y u v 8, and therefore the function range [-8,6 2].