Municipal budgetary educational institution
average comprehensive school №13
"Piecewise functions"
Sapogova Valentina and
Donskaya Alexandra
Head Consultant:
Berdsk
1. Determination of main goals and objectives.
2. Questionnaire.
2.1. Determining the relevance of the work
2.2. Practical significance.
3. History of functions.
4. General characteristics.
5. Methods for specifying functions.
6. Construction algorithm.
8. Literature used.
1. Determination of main goals and objectives.
Target:
Find out a way to solve piecewise functions and, based on this, create an algorithm for their construction.
Tasks:
Get to know general concept about piecewise functions;
Find out the history of the term “function”;
Conduct a survey;
Identify ways to specify piecewise functions;
Create an algorithm for their construction;
2. Questionnaire.
A survey was conducted among high school students on their ability to construct piecewise functions. The total number of respondents was 54 people. Among them, 6% completed the work completely. 28% were able to complete the work, but with certain errors. 62% were unable to complete the work, although they made some attempts, and the remaining 4% did not start work at all.
From this survey we can conclude that the students of our school who are taking the program do not have a sufficient knowledge base, because this author does not pay attention to special attention for tasks of this kind. It is from this that the relevance and practical significance our work.
2.1. Determining the relevance of the work.
Relevance:
Piecewise functions are found both in the GIA and in the Unified State Exam; tasks that contain functions of this kind are scored 2 or more points. And, therefore, your assessment may depend on their decision.
2.2. Practical significance.
The result of our work will be an algorithm for solving piecewise functions, which will help to understand their construction. And it will increase your chances of getting the grade you want in the exam.
3. History of functions.
“Algebra 9th grade”, etc.;
Continuity and graphing of piecewise defined functions – complex topic. It is better to learn how to build graphs directly in a practical lesson. This is mainly a continuity study.
It is known that elementary function(see p. 16) is continuous at all points at which it is defined. Therefore, the discontinuity in elementary functions possible only at two types of points:
a) at points where the function is “redefined”;
b) at points where the function does not exist.
Accordingly, only such points are checked for continuity during the study, as shown in the examples.
For non-elementary functions the study is more complicated. For example, a function (the integer part of a number) is defined on the entire number line, but suffers a break at each integer x. Such questions are beyond the scope of the manual.
Before studying the material, you should repeat from the lecture or textbook what (what kind) break points there are.
Investigation of piecewise defined functions for continuity
Function set piecewise if she's on different areas domain of definition is given different formulas.
The main idea when examining such functions is to find out whether and how the function is defined at the points at which it is redefined. It then checks whether the function values to the left and right of such points are the same.
Example 1. Let us show that the function 
continuous.
Function 
is elementary and therefore continuous at the points at which it is defined. But, obviously, it is defined at all points. Consequently, it is continuous at all points, including at 
, as required by the condition.
The same is true for the function 
, and at 
it is continuous.
In such cases, continuity may only be broken where the function is overridden. In our example this is a point 
. Let's check it, for which we find the limits on the left and right:
The limits on the left and right are the same. It remains to be seen:
a) is the function defined at the point itself? 
;
b) if yes, does it match 
with limit values on the left and right.
By condition, if 
, That 
. That's why 
.
We see that (all are equal to the number 2). This means that at the point 
the function is continuous. So, the function is continuous along the entire axis, including the point 
.
Comments on the decision
a) It did not play a role in the calculations, substitute we have a specific number formula 
or 
. This is usually important when dividing by an infinitesimal because it affects the sign of infinity. Right here 
And 
are responsible only for function selection;
b) as a rule, notations 
And 
are equal, the same applies to the designations 
And 
(and is valid for any point, not just for 
). Below, for brevity, we use notation of the form 
;
c) when the limits on the left and right are equal, to check for continuity it actually remains to see whether one of the inequalities will be not strict. In the example, this turned out to be the 2nd inequality.
Example 2. We examine the function for continuity 
.
For the same reasons as in example 1, continuity can only be broken at the point 
. Let's check:
The limits on the left and right are equal, but at the very point 
the function is not defined (the inequalities are strict). It means that 
- dot repairable gap.
“Removable gap” means that it is enough either to make any of the inequalities non-strict, or to invent one for a separate point 
a function whose value at 
equals –5, or simply indicate that 
so that the entire function 
became continuous.
Answer: dot 
– removable break point.
Note 1. In the literature, a removable gap is usually considered a special case of a type 1 gap, but is more often understood by students as separate type rupture. To avoid discrepancies, we will adhere to the 1st point of view, and specially stipulate the “irremovable” gap of the 1st kind.
Example 3. Let's check if the function is continuous 
At the point 
The limits on the left and right are different: 
. Regardless of whether the function is defined at 
(yes) and if so, what is it equal to (equal to 2), point 
–point of irremovable discontinuity of the 1st kind.
At the point 
is happening final leap(from 1 to 2).
Answer: dot 
Note 2. Instead of 
And 
usually write 
And 
respectively.
Available question: how the functions differ
And 
,
and also their graphs? Correct answer:
a) The 2nd function is not defined at the point 
;
b) on the graph of the 1st function point 
“shaded”, on the 2nd graph – not (“punctured point”).
Dot 
, where the graph breaks off 
, is not shaded in both graphs.
It is more difficult to examine functions that are defined differently on three areas.
Example 4. Is the function continuous? 
?
Just as in examples 1 – 3, each of the functions 
,
And 
is continuous along the entire numerical axis, including the area in which it is specified. Breaking is possible only at the point 
and/or at the point 
, where the function is overridden.
The task is divided into 2 subtasks: examine the continuity of the function
And 
,
and period 
not of interest to the function 
, and point 
– for function 
.
1st step. Checking the point 
and function 
(we don’t write the index):
The limits are the same. By condition, 
(if the limits on the left and right are equal, then in fact the function is continuous when one of the inequalities is not strict). So, at the point 
the function is continuous.
2nd step. Checking the point 
and function 
:
Because the 
, dot 
– discontinuity point of the 1st kind, and the value 
(and whether it exists at all) no longer plays a role.
Answer: the function is continuous at all points except the point 
, where there is an irremovable discontinuity of the 1st kind - a jump from 6 to 4.
Example 5. Find function breakpoints 
.
We proceed according to the same scheme as in example 4.
1st step. Checking the point 
:
A) 
, since to the left of 
the function is constant and equal to 0;
b) ( 
– even function).
The limits are the same, but when 
the function is not defined by condition, and it turns out that 
– removable break point.
2nd step. Checking the point 
:
A) 
;
b) 
– the value of the function does not depend on the variable.
Limits vary: 
, dot 
– point of irremovable discontinuity of the 1st kind.
Answer:
– removable break point, 
is a point of irremovable discontinuity of the 1st kind; at other points the function is continuous.
Example 6. Is the function continuous? 
?
Function 
determined at 
, so the condition 
turns into a condition 
.
On the other hand, the function 
determined at 
, i.e. at 
. So the condition 
turns into a condition 
.
It turns out that the condition must be met 
, and the domain of definition of the entire function is a segment 
.
The functions themselves 
And 
are elementary and therefore continuous at all points at which they are defined - in particular, and at 
.
It remains to check what happens at the point 
:
A) 
;
Because the 
, see if the function is defined at the point 
. Yes, the 1st inequality is relatively weak 
, and that's enough.
Answer: the function is defined on the interval 
and is continuous on it.
More complex cases, when one of the component functions is non-elementary or not defined at any point in its segment, are beyond the scope of the manual.
NF1. Construct graphs of functions. Note whether the function is defined at the point at which it is being redefined, and if so, what the value of the function is (the word " If" is omitted from the function definition for brevity):
1) a) 
b) 
V) 
G) 
2) a) 
b) 
V) 
G) 
3) a) 
b) 
V) 
G) 
4) a) 
b) 
V) 
G) 
Example 7. Let 
. Then at the site 
build a horizontal line 
, and on the site 
build a horizontal line 
. In this case, the point with coordinates 
"punctured", and the point 
"painted over". At the point 
a discontinuity of the 1st kind (“jump”) is obtained, and 
.
NF2. Examine the continuity of functions defined differently on 3 intervals. Plot graphs:
1) a) 
b) 
V) 
G) 
d) 
e) 
2) a) 
b) 
V) 
G) 
d) 
e) 
3) a) 
b) 
V) 
G) 
d) 
e) 
Example 8. Let 
. Location on 
build a straight line 
, why we find 
And 
. Connecting the dots 
And 
segment. We do not include the points themselves, because when 
And 
the function is not defined by condition.
Location on 
And 
circle the OX axis (on it 
), however the points 
And 
"gouged out." At the point 
we obtain a removable gap, and at the point 
– discontinuity of the 1st kind (“jump”).
NF3. Graph the functions and make sure they are continuous:
1) a) 
b) 
V) 
G) 
d) 
e) 
2) a) 
b) 
V) 
G) 
d) 
e) 
NF4. Make sure the functions are continuous and graph them:
1) a) 
b) 
V) 
2 a) 
b) 
V) 
3) a) 
b) 
V) 
NF5. Construct graphs of functions. Note the continuity:
1) a) 
b) 
V) 
G) 
d) 
e) 
2) a) 
b) 
V) 
G) 
d) 
e) 
3) a) 
b) 
V) 
G) 
d) 
e) 
4) a) 
b) 
V) 
G) 
d) 
e) 
5) a) 
b) 
V) 
G) 
d) 
e) 
NF6. Construct graphs of discontinuous functions. Note the function value at the point where the function is overridden (and whether it exists):
1) a) 
b) 
V) 
G) 
d) 
e) 
2) a) 
b) 
V) 
G) 
d) 
e) 
3) a) 
b) 
V) 
G) 
d) 
e) 
4) a) 
b) 
V) 
G) 
d) 
e) 
5) a) 
b) 
V) 
G) 
d) 
e) 
NF7. The same task as in NF6:
1) a) 
b) 
V) 
G) 
d) 
e) 
2) a) 
b) 
V) 
G) 
d) 
e) 
3) a) 
b) 
V) 
G) 
d) 
e) 
4) a) 
b) 
V) 
G) 
d) 
e) 
Charts piecewise given functions
Murzalieva T.A. teacher mathematicians MBOU"Bor secondary school" Boksitogorsky district Leningrad region
Target:
- master the linear spline method for constructing graphs containing a module;
- learn to apply it in simple situations.
Under spline(from the English spline - plank, rail) is usually understood as a piecewise given function.
Such functions have been known to mathematicians for a long time, starting with Euler (1707-1783, Swiss, German and Russian mathematician), but their intensive study began, in fact, only in the middle of the 20th century.
In 1946, Isaac Schoenberg (1903-1990, Romanian and American mathematician) first time using this term. Since 1960 with development computer technology began using splines in computer graphics and modeling.
1 . Introduction
2. Definition of a linear spline
3. Module Definition
4. Graphing
One of the main purposes of functions is description real processes occurring in nature.
But for a long time, scientists - philosophers and natural scientists - have identified two types of processes: gradual ( continuous ) And spasmodic.
When a body falls to the ground, it first occurs continuous increase driving speed , and at the moment of collision with the surface of the earth speed changes abruptly , becoming equal to zero or changing the direction (sign) when the body “bounces” from the ground (for example, if the body is a ball).
But since there are discontinuous processes, then means of describing them are needed. For this purpose, functions are introduced that have ruptures .
a - by the formula y = h(x), and we will assume that each of the functions g(x) and h(x) is defined for all values of x and has no discontinuities. Then, if g(a) = h(a), then the function f(x) has a jump at x=a; if g(a) = h(a) = f(a), then the “combined” function f has no discontinuities. If both functions g and h are elementary, then f is called piecewise elementary. "width="640" - One way to introduce such discontinuities is next:
Let function y = f(x)
at x is defined by the formula y = g(x),
and when xa - formula y = h(x), and we will consider that each of the functions g(x) And h(x) is defined for all values of x and has no discontinuities.
Then , If g(a) = h(a), then the function f(x) has at x=a jump;
if g(a) = h(a) = f(a), then the "combined" function f has no breaks. If both functions g And h elementary, That f is called piecewise elementary.
Charts continuous functions
Graph the function:
Y = |X-1| + 1
X=1 – formula change point
Word "module" came from Latin word"modulus", which means "measure".
Modulus of numbers A called distance (in single segments) from the origin to point A ( A) .
This definition reveals geometric meaning module.
Module (absolute value ) real number A the same number is called A≥ 0, and opposite number -A, if a
0 or x=0 y = -3x -2 at x "width="640" Graph the function y = 3|x|-2.
By definition of the modulus, we have: 3x – 2 at x0 or x=0
-3x -2 at x
x n) "width="640" . Let x be given 1 X 2 X n – points of change of formulas in piecewise elementary functions.
A function f defined for all x is called piecewise linear if it is linear on each interval
and besides, the coordination conditions are met, that is, at the points of changing formulas, the function does not suffer a break.
Continuous piecewise linear function called linear spline . Her schedule There is polyline with two infinities extreme links – left (corresponding to the values x n ) and right ( corresponding values x x n )
A piecewise elementary function can be defined by more than two formulas
Schedule - broken line with two infinite extreme links - left (x1).
Y=|x| - |x – 1|
Formula change points: x=0 and x=1.
Y(0)=-1, y(1)=1.
It is convenient to plot the graph of a piecewise linear function, pointing on coordinate plane vertices of the broken line.
In addition to building n vertices should build Also two points : one to the left of the vertex A 1 ( x 1; y ( x 1)), the other - to the right of the top An ( xn ; y ( xn )).
Note that a discontinuous piecewise linear function cannot be represented as a linear combination of the moduli of binomials .
Graph the function y = x+ |x -2| - |X|.
A continuous piecewise linear function is called a linear spline
1.Points for changing formulas: X-2=0, X=2 ; X=0
2. Let's make a table:
U( 0 )= 0+|0-2|-|0|=0+2-0= 2 ;
y( 2 )=2+|2-2|-|2|=2+0-2= 0 ;
at (-1 )= -1+|-1-2| - |-1|= -1+3-1= 1 ;
y( 3 )=3+|3-2| - |3|=3+1-3= 1 .
Construct a graph of the function y = |x+1| +|x| – |x -2|.
1 .Points for changing formulas:
x+1=0, x=-1 ;
x=0 ; x-2=0, x=2.
2 . Let's make a table:
y(-2)=|-2+1|+|-2|-|-2-2|=1+2-4=-1;
y(-1)=|-1+1|+|-1|-|-1-2|=0+1-3=-2;
y(0)=1+0-2=-1;
y(2)=|2+1|+|2|-|2-2|=3+2-0=5;
y(3)=|3+1|+|3|-|3-2|=4+3-1=6.
|x – 1| = |x + 3|
Solve the equation:
Solution. Consider the function y = |x -1| - |x +3|
Let's build a graph of the function /using the linear spline method/
- Formula change points:
x -1 = 0, x = 1; x + 3 =0, x = - 3.
2. Let's make a table:
y(- 4) =|- 4–1| - |- 4+3| =|- 5| - | -1| = 5-1=4;
y( -3 )=|- 3-1| - |-3+3|=|-4| = 4;
y( 1 )=|1-1| - |1+3| = - 4 ;
y(-1) = 0.
y(2)=|2-1| - |2+3|=1 – 5 = - 4.
Answer: -1.
1. Construct graphs of piecewise linear functions using the linear spline method:
y = |x – 3| + |x|;
1). Formula change points:
2). Let's make a table:
2. Construct function graphs using the teaching aid “Live Mathematics” »
A) y = |2x – 4| + |x +1|
1) Formula change points:
2) y() =
B) Build function graphs, establish a pattern :
a) y = |x – 4| b) y = |x| +1
y = |x + 3| y = |x| - 3
y = |x – 3| y = |x| - 5
y = |x + 4| y = |x| + 4
Use the Point, Line, and Arrow tools on the toolbar.
1. “Charts” menu.
2. “Build a graph” tab.
.3. In the "Calculator" window, enter the formula.
Graph the function:
1) Y = 2x + 4
1. Kozina M.E. Mathematics. 8-9 grades: collection elective courses. – Volgograd: Teacher, 2006.
2. Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova. Algebra: textbook. For 7th grade. general education institutions / ed. S. A. Telyakovsky. – 17th ed. – M.: Education, 2011
3. Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova. Algebra: textbook. For 8th grade. general education institutions / ed. S. A. Telyakovsky. – 17th ed. – M.: Education, 2011
4. Wikipedia, the free encyclopedia
http://ru.wikipedia.org/wiki/Spline
Back forward
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Textbook: Algebra 8th grade, edited by A. G. Mordkovich.
Lesson type: Discovery of new knowledge.
Goals:
for the teacher goals are fixed at each stage of the lesson;
for the student:
Personal goals:
- Learn to clearly, accurately, competently express your thoughts verbally and writing, understand the meaning of the task;
- Learn to apply acquired knowledge and skills to solve new problems;
- Learn to control the process and results of your activities;
Meta-subject goals:
In cognitive activity:
- Development logical thinking and speech, the ability to logically substantiate one’s judgments and carry out simple systematizations;
- Learn to put forward hypotheses when problem solving, understand the need to check them;
- Apply knowledge in a standard situation, learn to perform tasks independently;
- Transfer knowledge to a changed situation, see the task in the context of the problem situation;
In information and communication activities:
- Learn to conduct a dialogue, recognize the right to a different opinion;
In reflective activity:
- Learn to anticipate possible consequences your actions;
- Learn to eliminate the causes of difficulties.
Subject goals:
- Find out what a piecewise function is;
- Learn to define a piecewise given function analytically from its graph;
During the classes
1. Self-determination educational activities
Purpose of the stage:
- include students in learning activities;
- determine the content of the lesson: we continue to repeat the topic of numerical functions.
Organization educational process at stage 1:
T: What did we do in previous lessons?
D: We repeated the topic of numerical functions.
U: Today we will continue to repeat the topic of previous lessons, and today we must find out what new things we can learn in this topic.
2. Updating knowledge and recording difficulties in activities
Purpose of the stage:
- update educational content, necessary and sufficient for the perception of new material: remember the formulas numerical functions, their properties and methods of construction;
- update mental operations, necessary and sufficient for the perception of new material: comparison, analysis, generalization;
- record an individual difficulty in an activity that demonstrates it personally significant level insufficiency of existing knowledge: specifying a piecewise given function analytically, as well as constructing its graph.
Organization of the educational process at stage 2:
T: The slide shows five numerical functions. Determine their type.
1) fractional-rational;
2) quadratic;
3) irrational;
4) function with module;
5) sedate.
T: Name the formulas corresponding to them.
3) 
;
4) 
;
U: Let's discuss what role each coefficient plays in these formulas?
D: Variables “l” and “m” are responsible for shifting the graphs of these functions left - right and up - down, respectively, the coefficient “k” in the first function determines the position of the branches of the hyperbola: k>0 - the branches are in the I and III quarters, k< 0 - во II и IV четвертях, а коэффициент “а” определяет направление ветвей параболы: а>0 - branches are directed upwards, and< 0 - вниз).
2. Slide 2
U: Define analytically the functions whose graphs are shown in the figures. (considering that they move y=x2). The teacher writes down the answers on the board.
D: 1) 
);
2)
;
3. Slide 3
U: Define analytically the functions whose graphs are shown in the figures. (considering that they are moving). The teacher writes down the answers on the board.
4. Slide 4
U: Using the previous results, define analytically the functions whose graphs are shown in the figures.
3. Identifying the causes of difficulties and setting goals for activities
Purpose of the stage:
- organize communicative interaction, during which the distinctive property a task that caused difficulty in learning activities;
- agree on the purpose and topic of the lesson.
Organization of the educational process at stage 3:
T: What is causing you difficulties?
D: Pieces of graphs are provided on the screen.
T: What is the purpose of our lesson?
D: Learn to define pieces of functions analytically.
T: Formulate the topic of the lesson. (Children try to formulate the topic independently. The teacher clarifies it. Topic: Piecewise given function.)
4. Construction of a project for getting out of a difficulty
Purpose of the stage:
- organize communicative interaction to build a new mode of action, eliminating the cause of the identified difficulty;
- fix new way actions.
Organization of the educational process at stage 4:
T: Let's read the task carefully again. What results are asked to be used as help?
D: Previous ones, i.e. those written on the board.
U: Maybe these formulas are already the answer to this task?
D: No, because these formulas define the quadratic and rational function, and the slide shows their pieces.
U: Let's discuss what intervals of the x-axis correspond to the pieces of the first function?
U: Then analytical method the assignment of the first function looks like: if
T: What needs to be done to complete a similar task?
D: Write down the formula and determine which intervals of the abscissa axis correspond to the pieces of this function.
5. Primary consolidation in external speech
Purpose of the stage:
- record the studied educational content in external speech.
Organization of the educational process at stage 5:
7. Inclusion in the knowledge system and repetition
Purpose of the stage:
- train skills in using new content in conjunction with previously learned content.
Organization of the educational process at stage 7:
U: Define analytically the function whose graph is shown in the figure.
8. Reflection on activities in the lesson
Purpose of the stage:
- record new content learned in the lesson;
- evaluate your own activities in the lesson;
- thank your classmates who helped get the lesson results;
- record unresolved difficulties as directions for future educational activities;
- discuss and write down homework.
Organization of the educational process at stage 8:
T: What did we learn about in class today?
D: With a piecewise given function.
T: What work did we learn to do today?
D: Ask this type functions analytically.
T: Raise your hand, who understood the topic of today's lesson? (Discuss any problems that have arisen with the other children).
Homework
- No. 21.12(a, c);
- No. 21.13(a, c);
- №22.41;
- №22.44.
Real processes occurring in nature can be described using functions. Thus, we can distinguish two main types of processes that are opposite to each other - these are gradual or continuous And spasmodic(an example would be a ball falling and bouncing). But if there are discontinuous processes, then there are special means to describe them. For this purpose, functions are introduced that have discontinuities and jumps, that is, in different parts of the number line, the function behaves according to different laws and, accordingly, is specified by different formulas. The concepts of discontinuity points and removable discontinuity are introduced.
Surely you have already come across functions defined by several formulas, depending on the values of the argument, for example:
y = (x – 3, for x > -3;
(-(x – 3), at x< -3.
Such functions are called piecewise or piecewise specified. Sections of the number line with various formulas tasks, let's call them components domain. The union of all components is the domain of definition of the piecewise function. Those points that divide the domain of definition of a function into components are called boundary points. Formulas that define a piecewise function on each component of the domain of definition are called incoming functions. Graphs of piecewise given functions are obtained by combining parts of graphs constructed on each of the partition intervals.
Exercises.
Construct graphs of piecewise functions:
1)
(-3, at -4 ≤ x< 0,
f(x) = (0, for x = 0,
(1, at 0< x ≤ 5.
The graph of the first function is a straight line passing through the point y = -3. It originates at a point with coordinates (-4; -3), runs parallel to the x-axis to a point with coordinates (0; -3). The graph of the second function is a point with coordinates (0; 0). The third graph is similar to the first - it is a straight line passing through the point y = 1, but already in the area from 0 to 5 along the Ox axis.
Answer: Figure 1.
2)
(3 if x ≤ -4,
f(x) = (|x 2 – 4|x| + 3|, if -4< x ≤ 4,
(3 – (x – 4) 2 if x > 4.
Let's consider each function separately and build its graph.
So, f(x) = 3 is a straight line, parallel to the axis Oh, but you only need to depict it in the area where x ≤ -4.
Graph of the function f(x) = |x 2 – 4|x| + 3| can be obtained from the parabola y = x 2 – 4x + 3. Having constructed its graph, the part of the figure that lies above the Ox axis must be left unchanged, and the part that lies under the abscissa axis must be symmetrically displayed relative to the Ox axis. Then symmetrically display the part of the graph where
x ≥ 0 relative to the Oy axis for negative x. We leave the graph obtained as a result of all transformations only in the area from -4 to 4 along the abscissa axis.
The graph of the third function is a parabola, the branches of which are directed downward, and the vertex is at the point with coordinates (4; 3). We depict the drawing only in the area where x > 4.
Answer: Figure 2.
3)
(8 – (x + 6) 2, if x ≤ -6,
f(x) = (|x 2 – 6|x| + 8|, if -6 ≤ x< 5,
(3 if x ≥ 5.
Construction of the proposed piecewise-specified function similarly previous point. Here the graphs of the first two functions are obtained from the transformations of the parabola, and the graph of the third is a straight line parallel to Ox.
Answer: Figure 3.
4) Graph the function y = x – |x| + (x – 1 – |x|/x) 2 .
Solution. The scope of this function is all real numbers, except zero. Let's expand the module. To do this, consider two cases:
1) For x > 0 we get y = x – x + (x – 1 – 1) 2 = (x – 2) 2.
2) At x< 0 получим y = x + x + (x – 1 + 1) 2 = 2x + x 2 .
Thus, we have a piecewise defined function:
y = ((x – 2) 2, for x > 0;
( x 2 + 2x, at x< 0.
The graphs of both functions are parabolas, the branches of which are directed upward.
Answer: Figure 4.
5) Draw a graph of the function y = (x + |x|/x – 1) 2.
Solution.
It is easy to see that the domain of the function is all real numbers except zero. After expanding the module, we obtain a piecewise given function:
1) For x > 0 we get y = (x + 1 – 1) 2 = x 2 .
2) At x< 0 получим y = (x – 1 – 1) 2 = (x – 2) 2 .
Let's rewrite it.
y = (x 2, for x > 0;
((x – 2) 2 , at x< 0.
The graphs of these functions are parabolas.
Answer: Figure 5.
6) Is there a function whose graph on the coordinate plane has common point from any straight line?
Solution.
Yes, it exists.
An example would be the function f(x) = x 3 . Indeed, the graph of a cubic parabola intersects with the vertical line x = a at point (a; a 3). Let now the straight line be given by the equation y = kx + b. Then the equation
x 3 – kx – b = 0 has real root x 0 (since a polynomial of odd degree always has at least one real root). Consequently, the graph of the function intersects with the straight line y = kx + b, for example, at the point (x 0; x 0 3).
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