Extreme functions. "Application of a derivative of functions research" - Presentation. Extreme function Download Presentation Extreme Functions 11 by Alimov
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P tinsel note
Presentation on mathematics on the topic: "Derivative. Points of extremum and inflection. Ascending and bulge function"It is intended for students of 1 course of institutions of ATO or students of 10 - 11 classes of secondary schools.
The purpose of using the presentation in the educational process:
Visual demonstration of presentation in the lesson with teacher's explanations
Independent study of material on the topic (with possible outline material)
Multiple possibility of using a presentation during remote learning
Fastening the material during the training, with independent formulation of the properties of the function schedule.
The presentation can be used in lessons as a visual manual, for self-studying the topic, to replenish gaps in students' knowledge as a result of passes of training sessions.
The presentation has a convenient interface, easy to handle, contains visibility and informativeness, uses hyperlinks and triggers.
04.10.2013 Lecturer of Mathematics File T.B.
Presentation Screenshots:
Slide 1.
GBOU SPO Petrozavodsky Forestry Technical School "Derivative. Points of extremum and inflection. Ascending and bulge function »Work algorithm: 1. Work with the presentation allows you to form basic concepts on the topic, get acquainted with the properties of the function from the position of the derivative. 2. The presentation contains definitions, graphs, properties and theorems, which, if necessary, can be completed by pressing the pause. 3. For the transition to the content -, management of the presentation - by clicking the Competition of the presentation "Interactive Mosaic" on the site The site Interactive manual performed teacher of mathematics of Petrozavodsky Forestry technical school Falina Tatyana Borisovna Petrozavodsk 2013
Slide 2.
Slide 3.
1. Incretion of the function O\u003e 0 O\u003e 0 The function y \u003d f (x) is called increasingly in the interval if the function y \u003d f (x) increases the function, the function y \u003d f (x) increases, if the greater value of the argument corresponds to the greater value of the function y \u003d F (x) U\u003e 0 Theorem: If the derivative is positive, then the function y \u003d f (x) increases at this gap.
Slide 4.
2. Disarl of the function function y \u003d f (x) is called decreasing in the interval, if with an increase in the argument, the function value decreases. Function decreases if the greater value of the argument corresponds to the smaller value of the function of the U< 0 у < 0 y=f(x) Теорема: Если производная на промежутке отрицательная, то функция y=f(x) на данном промежутке убывает.
Slide 5.
3. Points of maximum point x \u003d A is called the maximum point of the function y \u003d f (x) if the derivative at this point is 0, and during the transition through this point from left to right, the sign of the derivative changes with (+) to (-) max f (x ) U\u003e 0 U\u003e 0 + - XX U< 0 y=f(x) у < 0 у >0 0 Recognize a maximum point for graphics function is very simple. The graph of the function in the neighborhood of the maximum point look like a smooth "hill" x xma
Slide 6.
4. Minima points point x \u003d A is called a point of a minimum function y \u003d f (x) if the derivative at this point is 0, and during the transition through this point left to the direction of the derivative changes from (-) to (+) F (x) O3\u003e 0 U\u003e 0 U< 0 y=f(x) у < 0 у >0 - MI N + X X0 Recognize a point of a minimum on schedule a function is very simple. The graph of the function in the neighborhood of the minimum point looks like a smooth "sweep" point of the minimum and the maximum points are called extremum points. x xmin.
Lesson and presentation on the topic "Extreme function". Grade 11. Tutorial Alimova.
View the contents of the document
"8.12 Extremes Functions."
Topic: "Extremes of the function"
Tell me, and I will forget.
Show me, and I will remember.
Involve me, and I will learn.
Chinese wisdom.
Objectives lesson:
Educational:
Relying on the knowledge of students in a derivative function to help articulate and realize the definition of the concepts of critical, stationary points and extremum points; Test to the hypothesis: necessary and sufficient condition for the existence of the extremum function.
Create a condition for primary consolidation of learning skills analytically and graphically determine the presence of critical, stationary points and extremum points.
Prepare students for the delivery of the USE.
Developing:
Promote the development of educational and cognitive activity, logical thinking.
Educational:
To form the ability to observe, notice the patterns, to generalize, conduct reasoning by analogy.
Develop thinking, attention, student speech.
To form general labor skills in the conditions of the greatest responsibility and limited time.
Rise the ability to listen to another opinion and defend your point of view.
Type of lesson: An educational lesson with new material.
During the classes:
I. . Organizing time(Method information and reporting)
Actualization of knowledge. "Brainstorm"
1. Add a derivative function: (The task is performed independently, with further self-test, the number of correct tasks is noted in the Self Control Sheet)
| f (x) \u003d 3x 2 - 4 x + 5 | ||
| f (x) \u003d sin x - cos x | ||
| f (x) \u003d e x + ln x | ||
| f (x) \u003d e 2x - 6e x + 7 | ||
| f (x) \u003d - x 3 + 3x 2 + 9 x - 29 |
2. Solve inequality: (at the blackboard)
3. Representation of the intervals of the function: (at the board two student)
A) f (x) \u003d 3x - 9 (1 point)
B) f (x) \u003d x 2 + 6x - 9 (2 points)
II. . Research.(on millimeter paper)
Answer the questions:
IV . Advance hypothesis(Partially search (heuristic method))
(Students put forward hypothesis)
If the derivative changes the sign from "-" to "+", and at the very point is 0, then this point will be a point of a minimum function. (for the hypothesis nomination - 4 points)
Answer the questions:
Name the gaps of increasing and descending the obtained schedule.
How does a derivative behave near this point when switching through this point? And in this very point?
Working with a textbook.
P. 265 - 266. Find a hypothesis formulated in the text.
Read it.
Minimum points and maximum points are called extremum points.
What will we do in today's lesson?
(learn to find the points of extremum functions)
What is the topic of our lesson?
Extreme function. Signed theme lesson.
Post Posts(method of stimulating schoolchildren's learning activities)
The French mathematician Pierre farm 4th century proved by the hypothesis of the hypothesis proved by the hypothesis.
(historical reference)
Pierre Ferma(1601-1665) - French mathematician, one of the creators of analytical geometry and the theory of numbers (farm theorems). Proceedings on probability theory, calculus of infinitely small and optics (Farm principle).
(students read the formulation of theorem )
Working with a book page 267
Find what points are called stationary, critical.
(Points in which the derived function is zero, called stationary
Points in which the function has a derivative equal to zero or undifferentiated, called critical points of this function )
Signal Card Works.
If the statement is true - "yes", if not, "no" (the game "Yes, no"
For the correct answer 1 point
P. 268 Theorem. . (Students read it and give explanations as they understand it)
A sufficient sign of Extremum.
At the board: for the correct execution - 5 points.
Make an algorithm for finding extremum points functions.
1. Find the field definition area.
2. Find F "( x.).
x.) \u003d 0 or f "( x.) does not exist.
(The derivative is 0 in zeros of the numerator, the derivative does not exist in zeros of the denominator)
4. Place the definition area and these points on the coordinate direct.
5. Determine the signs of the derivative on each of the intervals
6. Apply signs.
7. Record the answer.
(Practical method)
Work with EGE materials
The function y \u003d f (x) is determined on the interval (-4; 5). The figure shows the graph of its derivative. Find the point of the minimum function y \u003d f (x) 
The function y \u003d f (x) is determined on the interval (- 6; 6). The figure shows the graph of its derivative. Find the dots in which the derivative function is zero (answer : x \u003d - 4; x \u003d - 2; x \u003d 1; x \u003d 5).
Total lesson: raising estimates (for self-control sheets)
student reflection
I would like to learn better ...
I like …
I do not like …
At the lesson, I felt ...
With homework, I ...
View the contents of the presentation
"8.12 Extreme Functions"
Tell me, and I will forget. Show me, and I will remember. Involve me, and I will learn.
Chinese wisdom.
f (x) \u003d 3x 2 - 4 x + 5
f (x) \u003d sin x - cos x
f (x) \u003d e x + ln x
f (x) \u003d e 2x - 6e x + 7
f (x) \u003d - x 3 + 3x 2 + 9 x - 29
cOS X + SIN X
2e. 2x - 6 E. x.
-3 X. 2 + 6 x + 9
Build a function graph: y \u003d x 2 -6x + 8;
Answer the questions:
- Name the gaps of increasing and descending the obtained schedule.
- Name the point of the minimum function.
- Answer the questions:
- Name the gaps of increasing and descending the obtained schedule.
- Name the point of the maximum function.
- How does a derivative behave near this point when switching through this point? And in this very point?
Answer the questions:
- Name the gaps of increasing and descending the obtained schedule.
- Name the point of the maximum function.
- How does a derivative behave near the neighborhood of this point when switching through this point? And in this very point?
Pierre Farm (1601-1665) - French mathematician, one of the creators of analytic geometry and the theory of numbers (farm theorems). Proceedings on probability theory, calculus of infinitely small and optics (Farm principle).
Pierre Farm opened the methods of finding extremums and tangents, which, from a modern point of view, are reduced to finding a derivative.
Required sign of Extremum .
Algorithm finding extremum points functions
1. Find the field definition area.
2. Find F "( x. ).
3. Find critical points, i.e. points where f "( x. ) \u003d 0 or f "( x. ) does not exist. (The derivative is 0 in zeros of the numerator, the derivative does not exist in zeros of the denominator)
4. Place the definition area and these points on the coordinate direct.
5. Determine the signs of the derivative on each of the intervals
6. Apply signs.
7. Record the answer.
d / s: paragraph 50, No. 912 (2.4),
913(2,4), 914(2,4)
- I can …
- I know …
- I would like to learn better ...
- I like …
- I do not like …
- At the lesson, I felt ...
- With homework, I ...
The great philosopher of Confucius once said: "Three paths lead to knowledge: the way of reflection is the path is the most noble, the path of imitation is the path the easiest and way of experience is the path is the wrongty." Performing homework, each of you will pass your way to know.
- Confucius, Kun Tzu (born approximately 551 - died 479 BC. Er), Ancient Chinese thinker, founder of Confucianism.
Point x1 is called a minimum point
Functions
f (x),
if a
in
Some
neighborhood point x1 is performed
inequality
f (x) f (x1)
Functions values \u200b\u200bat points x0 and x1
called respectively
maximum and minimum function.
Maximum and minimum function called
Extreme function.
Y f (x)
x1 x2.
x3.
X. On one interval, the function may have
several extremes, and it may be that
at least one point is more maximum in
other.
Maximum or minimum functions on some
the gap is not in the general case
The greatest and smallest functions of the function.
If at some point x0 differentiable
F (X) function has an extremum, then in some
The neighborhood of this point is performed the theorem
Farm and derived function at this point
equal to zero:
F (x0) 0 However, the function can have an extremum at the point,
In which it is not differentiable.
For example, a function
Y X.
has a minimum at point
x 0
But it does not differ at this point. In order for the Y \u003d F (X) function
extremum at point x0, it is necessary to
Its derivative at this point was equal
zero or did not exist. Points in which the necessary
The condition of extremum is called
Critical or stationary.
Tob, if at any point there is an extremum,
then this point is critical.
But the critical point is not necessarily
Point of extremum. Find Critical Points and Extremes
Functions:
1
Y X.
2
y (x) 2 x
Y 2 x 0 at x 0
2
x 0
Y 0
- critical point y.
x 0
Y X.
2
X. 2
y x 1.
3Apply the necessary extremum condition:
Y (x 1) 3x
2
Y 3x 0 at x 0
3
x 0
Y 1.
2
- critical point y.
Y X.
2
Y 1.
X. If during the transition through the point x0 derivative
Differential function y \u003d f (x) changes
sign from a plus on minus, then x0 is a point
maximum, and if with a minus on plus, then x0
There is a minimum point. Let the derivative change the sign from the plus to minus,
those. At some interval
a; X.
0
F (X) 0
and at some interval
x; B.
0
F (X) 0
Then the function y \u003d f (x) will increase on
a; X.
0and will decrease on
x; B.
0
By definition of an increasing function
f (x0) f (x) for all
x a; x0.
For decreasing function
f (x0) f (x) for all
x0.
x x0; B.
- Maximum maximum.
Similarly proved for a minimum. 1
Find a derivative function
Y f (x)
2
Find critical points function, in
which derivative is zero or
does not exist. 3
Explore the sign derived from the left and
to the right of each critical
Points.
4
Find an extremum function. Explore the function on the extremum:
y x (x 1)
3Apply a schema
extremum:
1
Research
Functions
on the
Find a derived function:
y (x (x 1)) (x 1) 3x (x 1)
3
3
2
(x 1) (x 1 3x) (x 1) (4 x 1)
2
22
We find critical points:
(x 1) (4 x 1) 0
2
x1 1.
1
x2
43
Explore the sign derived from the left and
to the right of each critical
Points:
y.
y.
1
4
1
At point x \u003d 1, there is no extremum.
X. 4
We find extreme function:
27
1
f min.
256
4 If the first derivative is differentiable
functions y \u003d f (x) at point x0 is zero, and
Second derivative at this point
positive, then x0 is a point
minimum, and if the second derivative
Negative, x0 is a maximum point. Let be
F (x0) 0
F (x0) 0
hence
f (x) f (x) 0
and in some neighborhood of the point x0, i.e. function
F (X)
will increase by
a; B.
containing a point x0.
But
F (x0) 0
At the interval
a; X.
F (X) 0
And on the interval
x; B.
F (X) 0
0
0Thus, the function
F (X)
When switching through the point x0 changes the sign with
minus on plus, therefore this point
It is a minimum point.
Similarly
Proven
Maximum function.
happening
for Research Scheme for Extremum in
this case is similar to the previous one, but
The third item should be replaced by:
3
Find the second derivative and
Determine its sign in each
Critical point. From the second sufficient condition it follows that
If at a critical point the second derivative
functions are not equal to zero, then this point is
Point of extremum.
Reverse statement is not true: if in
Critical point Second derivative
functions are zero, then this point is also
It may be an extremum point.
IN
this case for research function
It is necessary to use the first sufficient
The condition of extremum.
Objectives of the lesson: Educational: - to systematize knowledge and create multi-level monitoring conditions (self-control, interconnection) of the learning of knowledge and ability to develop: - promote the formation of skills to apply the knowledge gained in a new situation, develop mathematical thinking, , mobility, ability to communicate
Memo. Interval method. The main provisions: 1. The mark of the work (private) is uniquely determined by the signs of the factors (divide and divider). 2. The work mark does not change (changes to the opposite), if you change the sign from the even (odd) number of factors. 3. A sign of a linear function with a nonzero angular coefficient and a sign of a quadratic function to the right of a large (or only one) root coincide with the sign of their older coefficient. 4. If a strictly increasing (decreasing) function has a root, it is positive to the right of the root (negative) and when moving through the root changes the sign. Remarks: 1. In the absence of the roots, the sign of the quadratic function coincides with the sign of its older coefficient throughout the field of determining this function. 2. Position 3 and Remark 1 are valid for a polynomial of either degree.
Work with a schedule. Work with a schedule. Consider the figure, which shows the graph of the function y \u003d x³-3x². Consider the neighborhood of the point x \u003d 0, i.e.nekto the interval containing this point. It can be seen from the figure that such a neighborhood exists and the greatest value takes the function at point x \u003d 0. This point is called a maximum point. Similarly, the point x \u003d 2 is called a minimum point, since the function at this point takes the value less than at any point of the neighborhood x \u003d 2. Consider the figure, which shows the graph of the function y \u003d x³-3x². Consider the neighborhood of the point x \u003d 0, i.e.nekto the interval containing this point. It can be seen from the figure that such a neighborhood exists and the greatest value takes the function at point x \u003d 0. This point is called a maximum point. Similarly, the point x \u003d 2 is called a minimum point, since the function at this point takes the value less than at any point of the neighborhood x \u003d 2.
It is necessary to remember: point x 0 is called the maximum point of the function f (x), if there is such a neighborhood of the point x 0, which is inequality from x 0 from this neighborhood, the point x 0 is called the maximum point F (X), if there is Such a neighborhood of the point X 0, which for all x different from x 0 from this neighborhood is carried out inequality f (x) f (x 0). (Figure 2) Maximum points and minimum points are called extremum points. Maximum points and minimum points are called extremum points.
A little from the history of mathematics: Pierre Farm. (1601 - 1665) The work of the adviser in the city parliament of Toulouse did not interfere with the farm to engage in mathematics. Gradually, he acquired the glory of one of the first mathematicians of France. He roldied with the French scientist R. Descartes in the creation of analytical geometry, general methods for solving tasks per maximum and minimum. His receptions of constructing tangents to curves, calculating the areas of curvilinear figures, calculating curvilinear lengths paved the road to the creation of differential and integral calculus. From the work of the farm began a new mathematical science - the theory of numbers.
Farm theorem. If x 0 is the extremum point of the differentiable function f (x), then f (x) \u003d 0. If x 0 is the extremum point of the differentiable function f (x), then f (x) \u003d 0. The farm theorem has a visual geometrical meaning: tangent to graphics function y \u003d f (x) at point (x 0; f (x 0)), where x 0 - the extremum point of the function y \u003d f (x), parallel to the abscissa axis, and therefore Its angular coefficient F (x) is zero. The farm theorem has a visual geometrical meaning: tangent to graphics function y \u003d f (x) at point (x 0; f (x 0)), where x 0 - the extremum point of the function y \u003d f (x), parallel to the abscissa axis, and therefore Its angular coefficient F (x) is zero.
Stationary and critical points of the point in which the derived function is zero, are called stationary, i.e. If f (x) \u003d 0, then this is not enough to assert that X is an extremum point. Points in which the function has a derivative equal to zero, or undifferentiated, is called critical points of this function. Consider the function f (x) \u003d x³. Its derivative F (x) \u003d 3x², F (x) \u003d 0. However, x \u003d 0 is not an extremum point, since the function increases on the entire numeric axis (Figure 1). Formulate a sufficient condition that the stationary point is an extremum point.
0 to the left of the point x 0 and f (x) "title \u003d" (! Lang: Theorem: Let the function f (x) differentiate on the interval (A; b), x 0 є (a; b), and f (x) \u003d 0. Then: 1) if when switching through the stationary point x 0 functions f (x), its derivative changes the sign from the "plus" to "minus", i.e. f (x)\u003e 0 to the left of the point x 0 and f (x)" class="link_thumb"> 10 !} Theorem: Let the function f (x) be differentiated on the interval (a; b), x 0 є (a; b), and f (x) \u003d 0. Then: 1) if when switching through the stationary point x 0 functions f (x), its derivative changes the sign from the "plus" to "minus", i.e. f (x)\u003e 0 to the left of the point x 0 and f (x) 0 to the left of the point x 0 and f (x) 0 to the left of the point x 0 and f (x) "\u003e 0 to the left of the point x 0 and f (x) 0 to the left of the point x 0 and f (x)"\u003e 0 to the left of the point x 0 and f (x) "Title \u003d "(! Lang: Theorem: Let the function f (x) differentiate on the interval (a; b), x 0 є (a; b), and f (x) \u003d 0. Then: 1) if during the transition through a stationary point x 0 functions f (x) its derivative changes the sign from the "plus" to "minus", i.e. f (x)\u003e 0 to the left of the point x 0 and f (x)"> title="Theorem: Let the function f (x) be differentiated on the interval (a; b), x 0 є (a; b), and f (x) \u003d 0. Then: 1) if when switching through the stationary point x 0 functions f (x), its derivative changes the sign from the "plus" to "minus", i.e. f (x)\u003e 0 to the left of the point x 0 and f (x)"> !}
The plan for finding an extremum function. 1. Find a derivative function. 2. Find stationary points of function, i.e. derived to equate to zero. 3. Using the interval method, find out how the signs of the derivative change. 4. According to the transition signs of the function to determine the minimum or maximum points.
Consider the task 1: Find the points of the extremum function f (x) \u003d 9x-3. Solution: 1) Find a derivative function: F '(x) \u003d 9 2) We find stationary points: there are no stationary points. 3) This function is linear and increases on the entire numeric axis, so the extremum points does not have a function. Answer: The function f (x) \u003d 9x-3 does not have extremum points.
Consider the task 2: Find the points of the extremum function f (x) \u003d x ² -2x. Solution: 1) Find a derivative function: f '(x) \u003d 2x-2 2) We will find stationary points: 2x-2 \u003d 0x \u003d 1. 3) Using the interval method, we will find how the sign of the derivative changes (see figure): 4) when switching through the point x \u003d 1, the sign of the derivative changes from the sign from "-" to "+", so x \u003d 1 is a minimum point. Answer: point x \u003d 1 is a point of a minimum function f (x) \u003d x ² -2x.
Consider the task 3: Find the points of the extremum function f (x) \u003d x -4x³. Solution: 1) Find a derivative function: F '(x) \u003d 4x³-12x² 2) We will find stationary points: 4x³-12x² \u003d 0 x1 \u003d 0, x2 \u003d 3. 3) Using the interval method, we find how the sign of the derivative changes (see figure): 4) when switching through the point x \u003d 0, the sign of the derivative does not change, then this point is not an extremum point, and when switching through the point x 1 \u003d 3, the derivative Changes the sign from "-" to "+", therefore x 2 \u003d 3 - is a point of a minimum. Answer: Point x \u003d 3 is a point of a minimum function f (x) \u003d x -4x³.
Alone to perform the following tasks: 1) for this figure, determine the maximum points and the minimum function y \u003d f (x). 2) find stationary points: a) y \u003d e ² -2e; b) y \u003d 2x³-15x ² + 36x; c) y \u003d sinx-cosx; d) y \u003d (2 + x ²) / x. 3) Find Extreme Functions: a) f (x) \u003d x³-x; b) f (x) \u003d x -8x² + 3; c) f (x) \u003d x + sinx; d) f (x) \u003d x-cos2x.
Fizkultminutka. For students, it is proposed to perform several physical exercises to remove fatigue and stress for long-term work on the computer. 1. Sitting on a chair: - Hands behind the head; - elbows divorce sewing, tilt the head back; - elbows forward, head forward; - hands relaxed down; - Exercise repeat 4 - 5 times. 2. Sitting on a chair: - the head smoothly take back; - Tilt smoothly head forward; - Exercise repeat 4 - 5 times. 3. Eye exercise: - Quickly frog; - close your eyes and sit calmly; - Slowly count up to five; - Exercise repeat 4 - 5 times. 4. Exercise for the eyes: - Snacking his eyes hard; - Slowly count up to five; - open your eyes and see in the distance; - Exercise repeat 4 - 5 times. 5. Exercise for the eyes: - look at the index finger of an elongated hand; - View in the distance; - Exercise repeat 4 - 5 times.
Testing: To perform the test, you must open the file that is in the "Function Extreme" folder on the disk with: called "Test 1". As a result of the work, you get an assessment for your knowledge. Also to systematize knowledge, you can perform the following tests for the repetition of the material studied earlier ("Test 2", "Test 3", "Test 4", "Test 5"). To perform the test, you must open the file that is located in the "Extreme Functions" folder on the disk with: called "Test 1". As a result of the work, you get an assessment for your knowledge. Also to systematize knowledge, you can perform the following tests for the repetition of the material studied earlier ("Test 2", "Test 3", "Test 4", "Test 5").
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Signatures for slides:
Extreme function
Points from the function of determining the function in which: f '(x) \u003d 0 or does not exist, are called critical points of this function. How can they be the points of extremum function. (Fig. 1 and 2). F '(x 1) \u003d 0 F' (x 2) \u003d 0
Points from the function of determining the functions in which: f '(x) \u003d 0 extremes are not extremes
Let x be about a point from the function of determining the function f (x) and f '(x o) \u003d 0, if the derivative of the function changes its sign from "+" to "-" at the point x o or vice versa, then this point is an extremum. X 1 x 2 x 1 Max x 2 min
Extremes of the function X 0 is a maximum point (max) functions, if there is such a neighborhood of the point x 0, which for all x ≠ x 0 of this vicinity, the inequality f (x) ˂ f (x 0) is performed. X 0 - a minimum point (min) of the function, if there is such a neighborhood of the point x 0, which the laser of all x ≠ x 0 from this neighborhood is carried out inequality f (x) ˃ f (x 0).
Figure 1 Figure 2 According to the specified graphics of functions Y \u003d F (X), specify: -critical points; -stationar points; -Extremum function.
Algorithm for searching for extremum points Functions: 1. Find a derived function; 2. Having a derivative to zero - to find stationary points; 3. Explore the derivative on the "sign" - to conclude.
Perform a task 1. Go to the maximum function of the function 2. The point of the minimum function on (0;) on (0;)
In 8 2 9, the figure shows a graph of a function defined on the interval. Find the amount of extremum points features. 3. -2 1 4 5 8 10 -2 + 1 + 3 + 4 + 5 + 8 + 10 \u003d ...
The figure shows a graph of the derivative function f (x), determined on the interval (-9; 8). Find the extremum point of the function on the interval (-3; 3) -3 3 B8 - 2 + -
On the topic: Methodical development, presentations and abstracts
Presentation on the lessons of algebra in the 11th grade on the themes "Ascending and decrease of the function. Extremes function."
Presentation is compiled for three lessons. Part of the material I took from the presentations of other teachers, for which they are very much thanks. Bottomically already made material component at its discretion for this class ...
