The topic of the lesson is “The set of function values in USE problems. Finding the set of function values Set of function values y 4 x
Today in the lesson we will turn to one of the basic concepts of mathematics - the concept of a function; Let's take a closer look at one of the properties of a function - the set of its values.
During the classes
Teacher. When solving problems, we notice that sometimes it is precisely finding the set of values of a function that puts us in difficult situations. Why? It would seem that studying the function from the 7th grade, we know a lot about it. Therefore, we have every reason to make a preemptive move. Let's "play" with a lot of function values today to solve many of the questions on this topic in the upcoming exam.
Sets of values of elementary functions
Teacher. To begin with, it is necessary to repeat the graphs, equations and sets of values of the basic elementary functions over the entire domain of definition.
Graphs of functions are projected onto the screen: linear, quadratic, fractional-rational, trigonometric, exponential and logarithmic, for each of them a set of values is verbally determined. Pay attention to the fact that the linear function E(f) = R or one number, for linear fractional
This is our alphabet. By adding to it our knowledge of graph transformations: parallel translation, stretching, compression, reflection, we can solve the problems of the first part USE and even a little more difficult. Let's check it out.
Independent work
At task words and coordinate systems printed for each student.
1. Find the set of function values on the entire domain of definition:
A) y= 3 sin X ;
b) y = 7 – 2 X
;
V) y= -arccos( x + 5):
G) y= | arctg x |;
e)
2. Find the set of function values y = x 2 in between J, If:
A) J = ;
b) J = [–1; 5).
3. Define a function analytically (by an equation) if the set of its values:
1) E(f(x)) = (–∞ ; 2] and f(x) - function
a) square
b) logarithmic,
c) demonstrative;
2) E(f(x)) = R \{7}.
When discussing a task 2independent work, draw students' attention to the fact that, in the case of monotonicity and continuity of the function y=f(x)at a given interval[a;b],the set of its meanings-gap,whose ends are the values f(a)and f(b).
Answer options for the task 3.
1.
A) y = –x 2 + 2 , y = –(x
+ 18) 2 + 2,
y= a(x – x c) 2 + 2 at A < 0.
b) y= -| log 8 x | + 2,
V) y = –| 3 x – 7 | + 2, y = –5 | x | + 3.
2.
a) b)
V) y = 12 – 5x, Where x ≠ 1 .
Finding the set of values of a function using the derivative
Teacher. In the 10th grade, we got acquainted with the algorithm for finding the extrema of a function continuous on a segment and finding its set of values without relying on the graph of the function. Remember how we did it? ( With the help of the derivative.) Let's recall this algorithm .
1. Make sure the function y = f(x) is defined and continuous on the segment J = [a; b]. 2. Find the function values at the ends of the segment: f(a) and f(b). Comment. If we know that a function is continuous and monotonic on J, then you can immediately answer: E(f) = [f(a); f(b)] or E(f) = [f(b); f(A)]. 3. Find the derivative and then the critical points x kJ. 4. Find function values at critical points f(x k). 5. Compare function values f(a), f(b) And f(x k), choose the largest and smallest values of the function and give an answer: E(f)= [f hire; f naib]. |
Problems for the application of this algorithm are found in USE options. For example, in 2008 such a task was proposed. You have to solve it Houses .
Task C1. Find the largest value of a function
f(x) = (0,5x + 1) 4 – 50(0,5x + 1) 2
at | x + 1| ≤ 3.
Homework conditions printed out for each student .
Finding the set of values of a complex function
Teacher. The main part of our lesson will be non-standard tasks containing complex functions, the derivatives of which are very complex expressions. And the graphs of these functions are unknown to us. Therefore, for the solution, we will use the definition of a complex function, that is, the dependence between the variables in the order of their nesting in this function, and the assessment of their range (the interval of change in their values). Problems of this type are found in the second part of the exam. Let's turn to examples.
Exercise 1. For functions y = f(x) And y = g(x) write a complex function y = f(g(x)) and find its set of values:
A) f(x) = –x 2 + 2x +
3, g(x) = sin x;
b) f(x) = –x 2 + 2x +
3, g(x) = log 7 x;
V)
g(x) = x 2 + 1;
G)
Solution. a) A complex function has the form: y= -sin 2 x+2sin x + 3.
Introducing an intermediate argument t, we can write this function like this:
y= –t 2 + 2t+ 3, where t= sin x.
At the inner function t= sin x the argument takes any value, and the set of its values is the segment [–1; 1].
So for the outer function y = –t 2 +2t+ 3 we have learned the interval of change of the values of its argument t: t[-1; 1]. Let's look at the graph of the function y = –t 2 +2t + 3.
Note that the quadratic function for t[-1; 1] takes the smallest and largest values at its ends: y hiring = y(–1) = 0 and y naib = y(1) = 4. And since this function is continuous on the interval [–1; 1], then it also takes on all values between them.
Answer: y .
b) The composition of these functions leads us to a complex function which, after introducing an intermediate argument, can be represented as follows:
y= –t 2 + 2t+ 3, where t= log 7 x,
Function t= log 7 x
x (0; +∞ ), t (–∞ ; +∞ ).
Function y = –t 2 + 2t+ 3 (see graph) argument t takes any value, and the quadratic function itself takes all values no greater than 4.
Answer: y (–∞ ; 4].
c) The complex function has the following form:
Introducing an intermediate argument, we get:
Where t = x 2 + 1.
Since for the inner function x R , A t .
Answer: y (0; 3].
d) The composition of these two functions gives us a complex function
which can be written as
notice, that
So, at
Where k Z , t [–1; 0) (0; 1].
Drawing a graph of a function we see that for these values t
y(–∞ ; –4] c ;
b) over the entire domain of definition.
Solution. First, we examine this function for monotonicity. Function t= arcctg x- continuous and decreasing on R and the set of its values (0; π). Function y= log 5 t is defined on the interval (0; π), is continuous and increases on it. This means that this complex function is decreasing on the set R . And it, as a composition of two continuous functions, will be continuous on R .
Let's solve problem "a".
Since the function is continuous on the entire number line, it is continuous on any part of it, in particular, on a given segment. And then it on this segment has the smallest and largest values and takes all the values between them:
f(4) = log 5 arcctg 4.
Which of the resulting values is greater? Why? And what will be the set of values?
Answer: 
Let's solve the problem "b".
Answer: at(–∞ ; log 5 π) throughout the domain of definition.
Task with parameter
Now let's try to compose and solve a simple equation with a parameter of the form f(x) = a, Where f(x) - the same function as in task 4.
Task 5. Determine the number of roots of the log 5 equation (arcctg x) = A for each parameter value A.
Solution. As we have already shown in task 4, the function at= log 5 (arctg x) is decreasing and continuous on R and takes values less than log 5 π. This information is enough to give an answer.
Answer: If A < log 5 π, то уравнение имеет единственный корень;
If A≥ log 5 π, then there are no roots.
Teacher. Today we have considered problems related to finding the set of function values. On this path, we discovered a new method for solving equations and inequalities - the method of estimation, so finding the set of values of a function has become a means of solving problems of a higher level. At the same time, we saw how such problems are constructed and how the monotonicity properties of a function facilitate their solution.
And I would like to hope that the logic that connected the tasks considered today surprised you, or at least surprised you. It cannot be otherwise: climbing a new peak leaves no one indifferent! We notice and appreciate beautiful paintings, sculptures, etc. But mathematics also has its own beauty, attractive and bewitching - the beauty of logic. Mathematicians say that nice solution- this is usually correct solution and it's not just a phrase. Now you yourself have to find such solutions, and we have indicated one of the ways to them today. Good luck to you! And remember: the road will be mastered by the walking one!
The function is the model. Let's define X as a set of values of an independent variable // independent means any.
A function is a rule by which, for each value of the independent variable from the set X, one can find the only value of the dependent variable. // i.e. for every x there is one y.
It follows from the definition that there are two concepts - an independent variable (which we denote by x and it can take any value) and a dependent variable (which we denote by y or f (x) and it is calculated from the function when we substitute x).
FOR EXAMPLE y=5+x
1. Independent is x, so we take any value, let x = 3
2. and now we calculate y, so y \u003d 5 + x \u003d 5 + 3 \u003d 8. (y is dependent on x, because what x we substitute, we get such y)
We say that the variable y is functionally dependent on the variable x and this is denoted as follows: y = f (x).
FOR EXAMPLE.
1.y=1/x. (called hyperbole)
2. y=x^2. (called parabola)
3.y=3x+7. (called straight line)
4. y \u003d √ x. (called the branch of the parabola)
The independent variable (which we denote by x) is called the argument of the function.
Function scope
The set of all values that a function argument takes is called the function's domain and is denoted by D(f) or D(y).
Consider D(y) for 1.,2.,3.,4.
1. D (y)= (∞; 0) and (0;+∞) //the whole set of real numbers except zero.
2. D (y) \u003d (∞; +∞) / / all the many real numbers
3. D (y) \u003d (∞; +∞) / / all the many real numbers
4. D (y) \u003d. Find the largest and smallest value of the function on this segment.
The derivative is positive for all x from interval (-1; 1)
, that is, the arcsine function increases over the entire domain of definition. Therefore, it takes the smallest value at x=-1, and the largest at x=1.
We got the range of the arcsine function 
.
Find the set of function values 
on the segment
.
Solution.
Find the largest and smallest value of the function on the given segment.
Let us determine the extremum points belonging to the segment
:
The dependence of one variable on another is called functional dependency. Variable dependency y from a variable x called function, if each value x matches a single value y.
Designation:
variable x called the independent variable or argument, and the variable y- dependent. They say that y is a function of x. Meaning y corresponding to the given value x, called function value.
All values it takes x, form function scope; all the values it takes y, form set of function values.
Designations: 
D(f)- argument values. E(f)- function values. If the function is given by a formula, then it is considered that the domain of definition consists of all values of the variable for which this formula makes sense.
Function Graph the set of all points on the coordinate plane is called, the abscissas of which are equal to the values of the argument, and the ordinates are equal to the corresponding values of the function. If some value x=x0 match multiple values (not just one) y, then such a correspondence is not a function. In order for the set of points of the coordinate plane to be a graph of some function, it is necessary and sufficient that any straight line parallel to the Oy axis intersects with the graph at no more than one point.
Ways to set a function
1) Function can be set analytically in the form of a formula. For example, 
2) The function can be defined by a table of many pairs (x; y).
3) The function can be set graphically. Value Pairs (x; y) displayed on the coordinate plane.
Function monotonicity
Function f(x) called increasing on a given numerical interval, if a larger value of the argument corresponds to a larger value of the function. Imagine that a certain point moves along the graph from left to right. Then the point will sort of "climb" up the chart.
Function f(x) called waning on a given numerical interval, if a larger value of the argument corresponds to a smaller value of the function. Imagine that a certain point moves along the graph from left to right. Then the point will, as it were, "roll" down the chart.
A function that is only increasing or only decreasing on a given numerical interval is called monotonous on this interval.
Function zeros and intervals of constancy
Values X, at which y=0, is called function zeros. These are the abscissas of the points of intersection of the graph of the function with the x-axis.
Such ranges of values x, on which the values of the function y either only positive or only negative are called intervals of sign constancy of the function.
Even and odd functions
Even function
1) The domain of definition is symmetrical with respect to the point (0; 0), that is, if the point a belongs to the domain of definition, then the point -a also belongs to the domain of definition.
2) For any value x f(-x)=f(x)
3) The graph of an even function is symmetrical about the Oy axis.
odd function has the following properties:
1) The domain of definition is symmetrical with respect to the point (0; 0).
2) for any value x, which belongs to the domain of definition, the equality f(-x)=-f(x)
3) The graph of an odd function is symmetrical with respect to the origin (0; 0).
Not every function is even or odd. Functions general view are neither even nor odd.
Periodic functions
Function f is called periodic if there exists a number such that for any x from the domain of definition the equality f(x)=f(x-T)=f(x+T). T is the period of the function.
Every periodic function has an infinite number of periods. In practice, the smallest positive period is usually considered.
The values of the periodic function are repeated after an interval equal to the period. This is used when plotting graphs.
Page 1
Lesson 3
"function range"
Objectives: - Apply the concept of the range of values to the solution of a specific problem;
solving typical problems.
For several years, problems have regularly appeared in exams in which it is required to select from a given family of functions those whose sets of values satisfy the declared conditions.
Let's consider such tasks.
Knowledge update.
What do we mean by the set of function values?
What is the value set of a function?
From what data can we find the set of function values? (According to the analytical notation of the function or its graph)
(cm USE assignments, part A)
What function values do we know? (The main functions are listed with their writing on the board; for each of the functions, its set of values \u200b\u200bis written down). As a result, on the board and in the students' notebooks
|
Function |
Many values |
|
y = x 2 y = x 3 y=| x| y=
|
E( y) = E( y) = [- 1, 1] E( y) = (– ∞, + ∞) E( y) = (– ∞, + ∞) E( y) = (– ∞, + ∞) E( y) = (0, + ∞) |
Can we, using this knowledge, immediately find the sets of values of the functions written on the blackboard? (see table 2).
What can help answer this question? (Graphs of these functions).
How to plot the first function? (Lower the parabola 4 units down).
|
Function |
Many values |
|
y = x 2 – 4 |
E( y) = [-4, + ∞) |
|
y = + 5 |
E( y) = |
|
y = – 5cos x |
E( y) = [- 5, 5] |
|
y= tg( x + / 6) – 1 |
E( y) = (– ∞, + ∞) |
|
y= sin( x + / 3) – 2 |
E( y) = [- 3, - 1] |
|
y=| x – 1 | + 3 |
E( y) = |
|
y=| ctg x| |
E( y) = |
|
y =  = | cos(x + /4) | |
E( y) = |
|
y=(x- 5) 2 + 3 |
E( y) = . Find the set of function values:   . 
Introduction of an algorithm for solving problems for finding the set of values of trigonometric functions. Let's see how we can apply our experience to the various tasks included in the options for a single exam. 1. Finding the values of functions for a given value of the argument. Example. Find the value of the function y = 2 cos(π/2+ π/4 ) – 1, If x = -π/2. Solution. y(-π/2) = 2 cos(- π/2 – π/4 )- 1= 2 cos(π/2 + π/4 )- 1 = - 2 sinπ/4 – 1 = - 2  – 1 =
= – 2. Finding the range of trigonometric functions
1≤ sinX≤ 1 2 ≤ 2 sinX≤ 2 9 ≤ 11+2sinX≤ 13 3 ≤ Let us write out the integer values of the function on the interval . This number is 3. Answer: 3.
at= sin 2 X- 2 3 sinx + 3 2 - 3 2 + 8, at= (sinX- 3) 2 -1. E ( sinX) = [-1;1]; E ( sinX -3) = [-4;-2]; E ( sinX -3) 2 = ; E ( at) = . Answer: .
Can we find a set of values for this function? (No.) What should be done? (Reduced to one function.) How to do it? (Use formula cos 2 x= 1-sin 2 x.) So, at= 1-sin 2 x+2sin x –2, y= -sin 2 x+2sin x –1, at= -(sin x –1) 2 . Well, now we can find a set of values and choose the smallest of them. 1 ≤ sin x ≤ 1, 2 ≤ sin x – 1 ≤ 0, 0 ≤ (sin x – 1) 2 ≤ 4, 4 ≤ -(sin x -1) 2 ≤ 0. So the smallest value of the function at hire= -4. Answer: -4.
Solution. at= 1-cos 2 x+ cos x + 1,5, at= -cos 2 x+ 2∙0.5∙cos x - 0,25 + 2,75, at= -(cos x- 0,5) 2 + 2,75. E(cos x) = [-1;1], E(cos x – 0,5) = [-1,5;0,5], E(cos x – 0,5) 2 = , E(-(cos x-0,5) 2) = [-2,25;0], E( at) = . The largest value of the function at naib= 2.75; smallest value at hire= 0.5. Let's find the product of the largest and smallest value of the function: at naib ∙at hire = 0,5∙2,75 = 1,375. Answer: 1.375. Solution. Let's rewrite the function in the form at =, at = Let us now find the set of values of the function. E(sin x) = [-1, 1], E(6sin x) = [-6, 6], E(6sin x + 1) = [-5, 7], E((6sin x + 1) 2) = , E(– (6sin x + 1) 2) = [-49, 0], E(– (6sin x + 1) 2 + 64) = , E( y) = [ Let's find the sum of integer values of the function: 4 + 5 + 6 + 7 + 8 = 30. Answer: 30.  Solution. 1) 2) Therefore, 2 X belong to the second quarter. 3) In the second quarter, the sine function decreases and is continuous. Means, given function 4) Calculate these values:
Answer :
 Solution. 1) Since a sine takes values from -1 to 1, then the set of difference values 2) The arccosine is a monotonically decreasing and continuous function. Hence, the set of values of the expression is a segment 3) When multiplying this segment by Answer:  Solution. Since the arc tangent is an increasing function, then 2) When increasing X from 3) When increasing from 4) Using the formula expressing the sine in terms of the tangent of a half angle, we find that
Hence, the desired set of values is the union of segments Answer: at= a sin x + b cos x or at= a sin(Rx) + bcos (Rx).
Solution. Let's find the value Let's transform the expression 15 sin 2x + 20 cos 2x = 25 ( 25 sin (2x + The set of function values y \u003d sin (2x + Then the set of values of the original function -25 Answer:
[-25; 25].
Function at= ctg X is decreasing on the segment [π/4; π/2], therefore, the function will take the smallest value at x =π/2, that is at(π/2) = сtg π/2 = 0; and the largest value is at x=π/4, that is at(π/4) = сtg π/4 = 1. Answer: 1, 0.  . Solution. Separate in equality It follows that the graph of the function f(x) is either a hyperbola (а≠ 0) or a straight line without a point. Moreover, if a; 2a) and (2a; If a \u003d 0, then f (x) \u003d -2 over the entire domain of definition x ≠ 0. Therefore, it is obvious that the desired values of the parameter are not equal to zero. Since we are only interested in the values of the function on the segment [-1; 1], then the classification of situations is determined by the fact that the asymptote x = 2a of the hyperbola (a≠0) is located relative to this segment. Case 1. All points of the interval [-1; 1] are to the right of the vertical asymptote x = 2a, that is, when 2a Case 2. The vertical asymptote intersects the interval [-1; 1], and the function decreases (as in case 1), that is, when Case 3. The vertical asymptote intersects the interval [-1; 1] and the function is increasing, i.e. -1 Case 4. All points of the interval [-1; 1] are to the left of the vertical asymptote, that is, 1 a > . and second Reception 5. Simplification of the formula defining a fractional rational function Reception 6. Finding the set of values of quadratic functions (by finding the vertex of the parabola and establishing the nature of the behavior of its branches). Reception 7. Introduction of an auxiliary angle for finding the set of values of some trigonometric functions. |
– 1.
, i.e. E (y) = .
,
, 8].
that is X belongs to the first quarter.
before 
.
. When multiplied by 
this segment will go to the segment 
.
.
we get 
.
.
.
before 
argument 2 X increases from 
before 
. Since the sine on such an interval increases, the function 
up to 1.
argument 2 X increases from 
. Since the sine decreases on such an interval, the function 
up to 1.
.
And 
, that is, the segment 
= 
= 25.
) = 25 () =
), where cos 
sin (2x + 
) and, if a > 0, increases monotonically on these rays.
.
