Add the math evaluation range of the first year of high school correct and immediately add 100 .

1. y= sinx cosx=tanx, but because x has a limit, the value range is a real number, but except for the y value corresponding to x with a limit.

If a=1 2, then we get: f(x)=x+1 (2x)+2, derivative of f(x) gives 1-1 (2x squared), and then makes it equal to 0, and the solution gives x=1 root number 2, so it can be proved that the function f(x) increases monotonically over the interval [1, + infinity], so that when x=1, the minimum value of y is 7 2

Therefore, the range y belongs to [7 2, +infinity).

2) If x belongs to [1,+infinity), f(x)>0 is constant immediately: such that g(x)=x squared +2x+a>0 is constant, because x is greater than 0

And for x belongs to [1,+infinity), g(x)>0 is constant, only 2 squared -4a<0, and the solution is a>1

1. y= y= sinx 2—cosx= 5 2*sin(x-m) where m=arctan2

It can be obtained: the range y belongs to [- 5 2, 5 2] 1) If a=1 2, then it can be obtained: f(x)=x+1 (2x)+2 can be proved: the function f(x) increases monotonically over the interval [1, + infinity].

So, when x = 1, the minimum value of y is 7 2

Therefore, the range y belongs to [7 2, +infinity).

2) If x belongs to [1,+infinity), f(x)>0 is constant, and immediately is: such that g(x)=x squared +2x+a>0 is constant.

For x to belong to [1,+infinity), g(x) increases monotonically over the interval [1,+infinity], so only the minimum value of g(x) g(1)>0 is constant.

So, get: 1+2+a>0

a>-3

1. y= sinx/cosx

y=tanx, so its range is (-

+2x+a) x x belongs to [1,+infinity)1)a=1 2 Find the minimum value of f(x).

f(x)=x+2+a x 2*x*a x+2=2a+2When a=1 2, the minimum value of f(x) is 2*1 2+2=32) If x belongs to [1,+infinity), f(x)>0 is constant to find the range of a.

f(x)=x+2+a x 2*x*a x+2=2a+2f(x)>0 is constant, i.e., 2a+2 0

a＞-1

2: (1) When a=1 2, f(x)=(x 2+2x+1 2) x=x+2+1 2x, the derivative of f(x) = 1-2 (4x 2) = (2x 2-1) 2x 2, because x [1, + 2x 2-1>0, 2x 2>0, then f(x) is an increasing function, f(x)min=f(1)=7 2

2) x [1,+ f(x)=(x 2+2x+a) x>0 is horizontally true, because x>1>0, to make f(x)>0 only x 2+2x+a>0,(x+1) 2-1+a>0, since the minimum value of x [1,+ x+1) 2 is 4,(x+1) 2>1-a, (1-a) is less than the minimum value of (x+1) 2 to make f(x)>0 horizontally true, that is, 1-a<4 then a>-3

1.Positive and negative infinity.

2.（1）f'(x)=x-x -1, let f'(x)=0, find the monotonicity and monotonic interval, and draw the graph.

Here's the train of thought.

1, What does it mean 4,a>9 16 5,a>=9 16 Other methods of bringing in. Replace x with y. The questions are not difficult to do in high school, but there are problems, and these questions are too easy for us to do in the third year of high school.

y=sin(1/2x+π/6)

x belongs to [0, 3].

6≤1/2x+π/6≤π/3

1 2 sin(1 2x+ 6) root number 3 2 value range [1 2, root number 3 2].

y=-cos(3x-π/3)

x belongs to [- 3, 3].

π3 ≤ 3x-π/3 ≤ 3

3x-3) covers the range from -- 3 to -- 3 to - Fuel 3 exactly 2.

The value range of y=-cos(3x- 3) is [-1,1].

y=[(x|+1)-1]/(x|+1)=1-1/(|x|+1)

x|0, so |x|+1≥1，-1/(|x|+1)≥-1，y≥0

Again Zhou Qin because of 1 (|x|+1) 0, y 1

(negative infinity, -1) and (-1, 1 2) and (1 2, positive infinity).

Greater than 0, less than or equal to 2

y=1+1/x²-x+1

x -x+1 must be greater than 0 and the maximum value is 1

The range of y is greater than 0 and less than or equal to 2

To find the range of y=1 2+x 2, we first need to understand what a range is.

Value range: a mathematical noun, in a function, the range of droplet values of the dependent variable is called the droplet range of the function, which is the set of all the values of the dependent variable in the defined domain of the function.

So let's first look at the x drop definition domain: y=1 2+x 2 Obviously, no matter what sub-value the x drop value is, o Therefore, we take a specific value 1, when x=1, you can draw a rough image of y(x) drops, when x=1 then y=1 2 when x=2 then y=2 when x=3 then y=9 2, then I didn't draw a picture, but through the numerical value, you can see that no matter what sub-value x takes, y is greater than or equal to 1 2, so the y drop range is (1 2, positive infinity) 1 2 is desirable, Then the x-drop range is x-drop squared, and a number of drops squared is greater than 0 no matter what value is taken, so the x-drop range is (0, positive infinity) to find the intersection (1 2, positive infinity) intersection (0, positive infinity) to obtain (1 2, positive infinity) in this problem x can take 0 However, if y=1 2x+x 2, then 0 cannot be taken to pull, because the denominator is 0 is meaningless.

1.Direct method: It is suitable for simple functions, that is, it is enough to correspond the definition field to the above.

2.Monotonicity method: Draw an image and judge the monotonicity according to the definition (generally used for quadratic functions, where the maximum value is not necessarily the endpoint value).

3.Commutation method: e.g. y=x-under the root number (x-1), x》1 to find the mouth straight: first change the yuan, remove the root number, then use the new yuan to represent the original element, and then assign the definition domain of the independent variable to the heart independent variable.

Wait a minute. 4.Where notation: ......

Hope to add clarity, what you want. Is it just a question with a solution process, or a method of how to evaluate the range. Or a combination of both.

Define the domain first, and then bring in the query. Note the axis of symmetry, the most valuable problem.

Image method.

It can be seen as the sum of the distances from the points to 1 and 2 on the number line, obviously the minimum is 1 and the maximum is positive infinity. ye[1,+infinity).