Lesson type: learning new material.

Teaching methods: visual, partly exploratory.

The purpose of the lesson:

1. Introduce the concept of a tangent to the graph of a function at a point, find out what geometric sense derivative, derive the tangent equation and teach how to find it for specific functions.
2. Development of logical thinking, research skills, functional thinking, mathematical speech.
3. Develop communication skills at work, promote the development independent activity students.

Equipment: computer, multimedia projector, handouts.

Download:

Preview:

Lesson on the topic "Tangent. Tangent Equation"

Lesson type: learning new material.

Teaching methods:visual, partly exploratory.

The purpose of the lesson:

1. Introduce the concept of a tangent to the graph of a function at a point, find out what the geometric meaning of the derivative is, derive the tangent equation and teach how to find it for specific functions.
2. Development of logical thinking, research skills, functional thinking, mathematical speech.
3. Development of communication skills in work, to promote the development of independent activity of students.

Equipment: computer, multimedia projector, handouts.

Lesson plan

I organizational moment.
Checking students' readiness for the lesson. The message of the topic and motto of the lesson.

II Actualization of the material.
(Activate attention, show the lack of knowledge about the tangent, formulate the goals and objectives of the lesson.)

Let's discuss what is a tangent to a function graph? Do you agree with the statement that "A tangent is a straight line that has one common point with a given curve"?
There is a discussion. Statements of children (yes and why, no and why). During the discussion, we come to the conclusion that this statement is not true.

Examples.
1) The line x = 1 has one common point M(1; 1) with the parabola y = x2, but it is not tangent to the parabola. The straight line y = 2x – 1 passing through the same point is tangent to the given parabola.
2) Similarly, the line x = π is not tangent to the graph y = cos x , although it has the only common point K(π; 1) with it. On the other hand, the line y = - 1 passing through the same point is tangent to the graph, although it has infinitely many common points type;(π+2 πk; 1), where k is an integer, in each of which it touches the graph.

Picture 1

Figure 2

Setting goals and objectives for children in the lesson:find out what is the tangent to the graph of a function at a point, how to write an equation for a tangent?
What do we need for this?
Recall general form equations of a straight line, conditions for parallel lines, definition of a derivative, differentiation rules.

III Preparatory work for the study of new material.
Questioning material on cards: (tasks are completed on the board)
1 student: fill in the table of derivatives elementary functions

2 student: remember the rules of differentiation

3 student: write the equation of a straight line y = kx + 4 passing through the point A(3; -2).
(y=-2x+4)

4 student: make an equation of straight lines y=3x+b passing through the point С(4; 2).
(y = 3x - 2).

With the rest frontal work.

1. Formulate the definition of a derivative.
2. Which of the following lines are parallel? y = 0.5x; y \u003d - 0.5x; y \u003d - 0.5x + 2. Why?

Guess the scientist's name:

Key to answers

Who this scientist was, what his work is connected with, we will learn in the next lesson.
Check student answers on cards.
IV The study of new material.
To set the equation of a straight line on a plane, it is enough for us to know its angular
coefficient and coordinates of one point.

• Let's start with the slope

Figure 3

Consider the graph of the function y = f(x) differentiable at point A(x 0 , f(x 0 )) .
Pick a point on it M (x 0 + Δх, f(x 0 + Δх)) and draw a secant AM .
Question: what is the slope of the secant? (∆f/∆x=tgβ)

We will approximate the point along the arc M to point A . In this case, straight AM will rotate around the point A , approaching (for smooth lines) some limiting position - a straight line AT . In other words, AT , which has this property, is called tangent to the graph of the function y \u003d f (x) at point A (x 0, f (x 0)).

Slope of the secant AM at AM → 0 tends to the slope of the tangent AT Δf/Δx → f "(x 0 ) . The value of the derivative at a point x 0 take for the slope of the tangent. They say thatthe tangent is the limit position of the secant at ∆х → 0.

Existence of a derivative of a function at a point x 0 is equivalent to the existence of a (non-vertical) tangent at (x 0 , f(x 0 )) graph, while the slope of the tangent is equal to f "(x 0) . This is geometric meaning of the derivative.

Definition of a tangent: Tangent to a graph differentiable at a point x 0 function f is a line passing through a point(x 0 , f(x 0 )) and having a slope f "(x 0) .
Let's draw the functions tangent to the graph y \u003d f (x) at points x 1, x 2, x 3 , and note the angles they form with the x-axis. (This is the angle measured in the positive direction from the positive direction of the axis to the straight line.)

Figure 4

We see that the angle α 1 is acute, angle α 3 is obtuse, and angle α 2 zero, since the straight line l is parallel to the Ox axis. Tangent acute angle positive, stupid - negative. That's why f "(x 1)> 0, f" (x 2) \u003d 0, f "(x 3)

• We now derive the tangent equationto the graph of the function f at point A(x 0 , f(x 0 ) ).

General view of the straight line equation y = kx + b .

1. Let's find the angular coefficient k \u003d f "(x 0), we get y \u003d f "(x0) ∙ x + b, f (x) \u003d f "(x 0 )∙x + b
2. Let's find b. b \u003d f (x 0) - f "(x 0) ∙ x 0.
3. Substitute the obtained values k and b into the equation of a straight line: y \u003d f "(x 0) ∙x + f (x 0) - f "(x 0) ∙x 0 or y \u003d f (x 0) + f "(x 0) (x - x 0)

• Generalization of lecture material.

- formulate an algorithm for finding the tangent equation at a point?

1. The value of the function at the point of contact
2. Common derivative of a function
3. The value of the derivative at the point of contact
4. Substitute the found values ​​into general equation tangent.

V Consolidation of the studied material.

1. Oral work:
1) B what points of the graph is tangent to it
a) horizontal;
b) forms an acute angle with the x-axis;
c) forms an obtuse angle with the x-axis?
2) For what values ​​of the argument is the derivative of the function given by the graph
a) equal to 0;
b) more than 0;
c) less than 0?

Figure 5

Figure 6

3) The figure shows the graph of the function f(x) and a tangent to it at a point with an abscissa x0 . Find the value of the derivative of a function f "(x) at the point x 0 .

Figure 7

2. Written work.
No. 253 (a, b), No. 254 (a, b). (fieldwork, with commentary)

3. Solution of reference problems.
Let's consider four types of tasks. Children read the condition of the problem, offer a solution algorithm, one of the students draws it up on the board, the rest write it down in a notebook.
1. If a touch point is given
Write an equation for a tangent to a function graph f(x) = x 3 - 3x - 1 at point M with abscissa -2.
Solution:

1. Let's calculate the value of the function: f(-2) =(-2) 3 - 3(-2) - 1 = -3;
2. find the derivative of the function: f "(x) \u003d 3x 2 - 3;
3. calculate the value of the derivative: f "(-2) \u003d - 9 .;
4. let's substitute these values ​​into the tangent equation: y = 9(x + 2) - 3 = 9x + 15.

Answer: y = 9x + 15.

2. By the ordinate of the point of contact.
Write an equation for a tangent at a point on a graph with ordinate y 0 = 1.
Solution:

Answer: y \u003d -x + 2.

3. Preset direction.
Write tangent equations to the graph y \u003d x 3 - 2x + 7 , parallel to the line y = x .
Solution.
The desired tangent is parallel to the straight line y=x . So they have the same slope k \u003d 1, y "(x) \u003d 3x2 - 2. Abscissa x 0 points of contact satisfies the equation 3x 2 - 2 \u003d 1, from where x 0 = ±1.
Now we can write the tangent equations: y = x + 5 and y = x + 9 .
Answer: y = x + 5 , y = x + 9 .

4. Conditions for touching the graph and the straight line.
A task. At what b straight y = 0.5x + b is tangent to the graph of the function f(x) = ?
Solution.
Recall that the slope of a tangent is the value of the derivative at the point of tangency. The slope of this straight line is k = 0.5. From here we get the equation for determining the abscissa x of the point of contact: f "(x) \u003d = 0.5. Obviously, its only root is x = 1. The value of this function at this point is y(1) = 1. So, the coordinates of the touch point are (1; 1). Now it remains to choose such a value of the parameter b, at which the line passes through this point, that is, the coordinates of the point satisfy the equation of the line: 1 = 0.5 1 + b, whence b = 0.5.

5. Independent work educational character.

Work in pairs.


Check: the results of the solution are entered in a table on the board (one answer from each pair), discussion of the answers.

6. Finding the angle of intersection of the graph of a function and a straight line.
The angle of intersection of the graph of the function y = f(x) and direct line l called the angle at which the tangent to the graph of the function intersects the line at the same point.
No. 259 (a, b), No. 260 (a) - disassemble at the board.

7. Independent work of a controlling nature.(differentiated work, the teacher checks for the next lesson)
1 option.

Option 2.

1. At what points is the tangent to the graph of the function f(x) = 3x 2 - 12x + 7 parallel to the x axis?
2. Equate the tangent to the graph of the function f(x)= x 2 - 4 at the point with the abscissa x 0 = - 2. Draw the pattern.
3. Find out if the line is y \u003d 12x - 10 tangent to the graph of the function y = 4x3.

3 option.

VI Summing up the lesson.
1. Answers to questions
- what is called a tangent to the graph of a function at a point?
What is the geometric meaning of the derivative?
- formulate an algorithm for finding the tangent equation at a point?
2. Remember the goals and objectives of the lesson, have we achieved this goal?
3. What were the difficulties in the lesson, what moments of the lesson did you like the most?
4. Marking for the lesson.
VII Commentary on homework: p. 19 (1, 2), No. 253 (c), No. 255 (d), No. 256 (d), No. 257 (d), No. 259 (d). Prepare a report on Leibniz.

Literature

1. Algebra and the beginning of analysis: a textbook for grade 10 educational institutions. Compilers:. M. Nikolsky, M. K. Potapov, N. N. Reshetnikov, A. V. Shevkin. - M.: Education, 2008.

2. Didactic materials on Algebra and Principles of Analysis for Grade 10 / B.M. Ivlev, S.M. Saakyan, S.I. Schwarzburd. - M.: Education, 2008.
3. Multimedia disk of the company "1C". 1C: Tutor. Mathematics (part 1) + USE options. 2006.
4. open bank assignments in mathematics/ http://mathege.ru/

Lessons 70-71. The equation of the tangent to the graph of the function

09.07.2015 5132 0

Target: get the equation of the tangent to the graph of the function.

I. Communication of the topic and objectives of the lessons

II. Repetition and consolidation of the material covered

1. Answers to questions on homework (analysis of unsolved problems).

2. Control of assimilation of material (test).

Option 1

1. Find the derivative of the function y \u003d 3x4 - 2 cos x .

Answer:

at the point x = π.

Answer:

3. Solve the equation y '(x) = 0 if

Answer:

Option 2

1. Find the derivative of the function y \u003d 5xb + 3 sin x .

Answer:

2. Calculate the value of the derivative of the function at the point x = π.

Answer:

3. Solve the equation y '(x) = 0 if

Answer:

III. Learning new material

Finally, let's move on to the final stage of studying the derivative and consider the use of the derivative in the remaining lessons. In this lesson, we will discuss the tangent to the graph of a function.

The concept of a tangent has already been considered earlier. It was shown that the graph of a function differentiable at the point a f (x) near a practically does not differ from the tangent graph, which means that it is close to the secant passing through the points (a; f (a)) and (a + Δx; f (a + Δx)). Any of these secants passes through the point M(a; f (a)). To write the equation of a tangent, you must specify its slope. Slope of the secant Δ f /Δx at Δх → 0 tends to a number f "(a), which is the slope of the tangent. Therefore, they say that the tangent is the limiting position of the secant at Δx→ 0.

Now we get the equation of the tangent to the graph of the function f (X). Since the tangent is a straight line and its slope f "(a), then we can write its equation y \u003d f "(a) x + b . Let's find the coefficient b from the condition that the tangent passes through the point M(a; f (a)). Substitute the coordinates of this point in the tangent equation and get: f (a) \u003d f "(a) a + b, whence b \u003d f (a) - f "(a) a. Now we substitute the found value b into the tangent equation and get: or This is the tangent equation. Let us discuss the application of the tangent equation.

Example 1

At what angle is the sinusoidintersects the x-axis at the origin?

The angle at which the graph of this function intersects the x-axis, equal to the angle slope a of the tangent drawn to the graph of the function f(x ) at this point. Let's find the derivative:Taking into account the geometric meaning of the derivative, we have: and a = 60°.

Example 2

Let's write the equation of the tangent graph of the function f (x) = -x2 + 4x at the point a = 1.

f "(x) and the function itself f (x) at point a = 1 and get: f "(a) = f "(1) = -2 1 + 4 = 2 and f (a) = f (1) = -12 + 4 1 = 3. Substitute these values ​​into the tangent equation. We have: y \u003d 2 (x - 1) + 3 or y \u003d 2x + 1.

For clarity, the figure shows a graph of the function f(x ) and tangent to this function. Touch occurs at a point M(1; 3).

Based on examples 1 and 2, we can formulate an algorithm for obtaining the equation of the tangent to the graph of the function y = f(x):

1) designate the abscissa of the point of contact with the letter a;

2) calculate f(a);

3) find f "(x) and calculate f "(a);

4) substitute the found numbers a, f (a), f "(a) into the formula y \u003d f '(a) (x - a) + f (a).

Note that initially the point a may be unknown and it has to be found from the conditions of the problem. Then in the algorithm in paragraphs 2 and 3 the word "calculate" must be replaced by the word "write" (which is illustrated by example 3).

In example 2, the abscissa a of the tangent point was specified directly. In many cases, the touch point is determined by various additional conditions.

Example 3

Let us write the equations of tangents drawn from the point A (0; 4) to the graph of the function f(x) \u003d - x 2 + 2x.

It is easy to check that point A does not lie on the parabola. At the same time, the points of contact of the parabola and the tangents are unknown, therefore, to find these points, an additional condition will be used - the passage of the tangents through point A.

Let us assume that the contact occurs at point a. Let's find the derivative of the function:Calculate the values ​​of the derivative f"(x ) and the function itself f (x) at the point of contact a and we get: f '(a) \u003d -2a + 2 and f (a ) = -a2 + 2a. Substitute these values ​​into the tangent equation. We have: or This is the tangent equation.

We write down the condition for the passage of the tangent through the point A, substituting the coordinates of this point. We get: 4or 4 = a2, whence a = ±2. Thus, the touch occurs at two points B(-2; -8) and C(2; 0). Therefore, there will be two such tangents. Let's find their equations. Let us substitute the values ​​a = ±2 into the tangent equation. We get: at a = 2 or yx \u003d -2x + 4; at a = -2 or y2 = 6x + 4. So, the equations of tangents are y1 = -2x + 4 and y2 = 6x + 4.

Example 4

Let's find the angle between the tangents using the conditions of the previous problem.

The tangents drawn y1 = -2x + 4 and y2 = 6x + 4 make angles a1 and a2 with the positive direction of the abscissa axis (and tg a 1 = -2 and tg a 2 = 6) and between themselves the angle φ = a 1 - a2. We find, using the well-known formula,whence φ = arctan 8/11.

Example 5

Let's write the equation of the tangent to the graph of the functionparallel line y \u003d -x + 2.

Two lines are parallel to each other if they have the same slope. The slope of the straight line y \u003d -x + 2 is -1, the slope of the desired tangent is f '(a ), where a - abscissa of the point of contact. Therefore, to determine a, we have an additional condition f '(a) \u003d -1.

Using the formula for the derivative of private functions, we find the derivative:Find the value of the derivative at the point a and get:

We get the equationor (a - 2)2 = 4, or a - 2 = ±2, whence a = 4 and a = 0. Thus, there are two tangents that satisfy the condition of the problem. Let us substitute the values ​​a = 4 and a = 0 into the equation of the tangent y = f '(a)(x - a) + f (a). For a = 4 we have:and tangent y1 \u003d - (x - 4) + 3 or y1 \u003d -x + 7. With a \u003d 0 we get:and tangent y2 \u003d - (x - 0) - 1 or y2 \u003d -x - 1. So, the equations of tangents y1 \u003d -x + 7 and y2 \u003d -x - 1.

Note that if f "(a ) does not exist, then the tangent or does not exist (as in the function f (x) = |x| at the point (0; 0) - fig. a, or vertical (like the functionat the point (0; 0) - fig. b.

So the existence of a derivative of a function f (x) at the point a is equivalent to the existence of a non-vertical tangent at the point (a; f (a)) graphics. In this case, the slope of the tangent is equal to f "(a). This is the geometric meaning of the derivative.

The concept of a derivative allows one to carry out approximate calculations. It has been repeatedly noted that at Δх→ 0 function values f(x ) and its tangent y(x) practically coincide. Therefore, at Δx→ 0 function behavior f (x) in a neighborhood of the point x0 can be approximately described by the formula(actually the tangent equation). This formula is successfully used for approximate calculations.

Example 6

Calculate the value of the function at the point x = 2.03.

Find the derivative of this function: f "(x) \u003d 12x2 - 4x + 3. We will assume that x \u003d a + Δx, where a \u003d 2 and Δx \u003d 0.03. We calculate the values ​​​​of the function and its derivative at point a and get: and Now let's define the value of the function in given point x = 2.03. We have:

Of course, the above formula is convenient to use if the values f (a) and f "(a ) is easy to calculate.

Example 7

Compute

Consider the functionLet's find the derivative:We will assume that x = a + Δx, where a = 8 and Δx = 0.03. Let's calculate the values ​​of the function and its derivative at the point a and get:Now let's determine the value of the function at a given point x = 8.03. We have:

Example 8

Let us generalize the result obtained. Consider the power function f (x) = x n and we will assume that x = a + Δx and Δx→ 0. Find f "(x) = n x n -1 and calculate the values ​​of the function and its derivative at point a, we get: f (a) \u003d an and f ’(a) \u003d nan -1 . Now we have the formula f (x) = a n + nan -1 Δx. Let's use it to calculate the number 0.98-20. We will assume that a = 1, Δх = -0.02 and n = -20. Then we get:

Of course, the above formula can be used for any other functions, in particular trigonometric ones.

Example 9

Let's calculate tg 48°.

Consider the function f(x) = tg x and find the derivative:We will assume that x = a + Δ x, where a = 45° = π/4 and (Once again, note that in trigonometry, angles are usually measured in radians). Let's find the values ​​of the function and its derivative at the point a and get:Now let's calculate(taking into account that π = 3.14).

IV. test questions

1. The equation of the tangent to the graph of the function.

2. Algorithm for deriving the tangent equation.

3. The geometric meaning of the derivative.

4. Application of the tangent equation for approximate calculations.

V. Task in the lessons

§ 29, no. 1 (a); 2 (b); 5 (a, b); 6 (c, d); 9(a); 10 (b); 12 (d); 14(a); 17; 21(a); 22 (a, c); 24 (a, b); 25(a); 26.

VI. Homework

§ 29, no. 1 (b); 2 (c); 5 (c, d); 6 (a, b); 9 (b); 10(a); 12(b); 14 (b); eighteen; 21(c); 22 (b, d); 24 (c, d); 25 (b); 27.

VII. Creative tasks

1. At what points x are tangent to function graphs parallel?

Answer: x \u003d -1, x \u003d 3.

2. For what x are the tangents to the graphs of functions y \u003d 3 cos 5 x - 7 and y = 5 cos 3 x + 4 are parallel?

Answer:

3. At what angles do the curves y = x2 intersect and

Answer: π/2 and arctg 3/5.

4. At what angles do the curves intersect y = cos x and y = sin x?

Answer:

5. To the parabola y \u003d 4 - x2 at the point with the abscissa x \u003d 1, a tangent is drawn. Find the point of intersection of this tangent with the y-axis.

Answer: (0; 5).

6. A tangent is drawn to the parabola y \u003d 4x - x2 at the point with the abscissa x \u003d 3. Find the point of intersection of this tangent with the x-axis.

Answer: (9/2; 0).

7. Find the angle between two tangents drawn from the point (0; -2) to the parabola y \u003d x2.

Answer:

8. To the graph of the function y \u003d 3x2 + 3x + 2, tangents are drawn with slopes k 1 = 0 and k 2 = 15. Write the equation of a straight line passing through the points of contact.

Answer: y \u003d 12x - 4.

9. Find the equations of lines touching simultaneously the parabolas y = x2 + x - 2 and y = -x2 + 7x - 11.

Answer: y \u003d 7x - 11 and y \u003d x - 2.

10. Write the equation of the common tangent to the parabolas y \u003d -3x2 + 4x + 4 and y \u003d -3x2 + 16x - 20.

Answer: y = -2x + 7.

11. The tangent to the graph of the function y \u003d x2 - 4x - 3 is drawn at the point x \u003d 0. Find the area of ​​\u200b\u200bthe triangle formed by the tangent and the coordinate axes.

Answer: 9/8.

12. Find the area of ​​the triangle bounded by the coordinate axes and the tangent to the graph of the functionat the point x = 2.

Answer: 1.

VIII. Summing up the lessons

Sections: Maths

Goals.

• Generalize and systematize the rules of differentiation;
• Repeat the algorithm for constructing a tangent to the function graph, the scheme for studying the function;
• Solving problems for the use of the largest and the smallest value functions.

Equipment. Poster “Derivative. Rules for calculating derivatives. Applications of the derivative”.

During the classes

According to the cards, the students repeat the theoretical material.

1. Define the derivative of a function at a point. What is called differentiation? Which function is called differentiable at a point?

(The derivative of the function f at the point x is the number to which the ratio tends

A function that has a derivative at a point x 0 is called differentiable at that point. Finding the derivative of f is called differentiation.)

2. Formulate the rules for finding the derivative.

(1. Derivative of the sum (u + v)"=u"+v";
2. About the constant factor (Cu)"=Cu";
3. Derivative of the product (uv)"=u"v+uv";
4. Derivative of a fraction (u / v) "= (u" v-uv ") / v 2;
5. Derivative power function(xn)"=nxn+1 .)

3. What are the derivatives of the following functions:

4. How to find the derivative of a complex function?

(We must consistently represent it in the form of elementary functions and take the derivative according to known rules).

5. What are the derivatives of the following functions:

6. What is the geometric meaning of the derivative?

(The existence of a derivative at a point is equivalent to the existence of a non-vertical tangent at the point (x 0, f (x 0)) of the graph of the function, and the slope of this tangent is f "(x 0)).

7. What is the equation of the tangent to the graph of the function at the point (x 0, f (x 0))?

(The tangent equation has the form y \u003d f (x 0) + f "(x 0) (x-x 0))

8. Formulate an algorithm for constructing a graph of a function using a derivative.

(1. Find OOF.
2. Investigate for parity.
3. Investigate for periodicity.
4. Find the intersection points of the graph with the coordinate axes.
5. Find the derivative of the function and its critical points.
6. Find intervals of monotonicity and extrema of the function.
7. Build a table based on the results of the study.
8. Graph the function.)

9. Formulate theorems with the help of which it is fashionable to plot a function graph.

(1. Sign of increase (decrease).
2. A necessary sign of an extremum.
3. Sign of maximum (minimum).)

10. What formulas exist for approximate calculations of functions?

Individual work.

Level A (three options), level B (one option).

Level A

Option 1.

1. Write down the equation of the tangent to the graph of the function

f (x) \u003d (x -1) 2 (x -3) 3 parallel lines y \u003d 5-24x.

2. Write the number 18 as the sum of three positive terms so that one term is twice the other, and the product of all three terms is the largest.

4. Find the intervals of increase and decrease of the function f(x)=(x-1) e x+1.

Option 2.

1. At what angle to the abscissa is the tangent to the graph of the function f (x) \u003d 0.x 2 + x-1.5 at the point with the abscissa x 0 \u003d - 2? Write the equation for this tangent and complete the drawing for this problem.

2. As in B. 1.

3. Find the derivative of the function:

Level B

1. Find the derivative of the function:

a) f (x) \u003d e -5x;
b) f(x) = log 3 (2x 2 -3x+1).

2. Write the equation of the tangent to the graph of the function at the point with the abscissa x 0 if f (x) \u003d e -x, x 0 \u003d 1.

3. Find the intervals of increase and decrease of the function f(x)=x·e 2х.

Summary of the lesson.

The work is checked, a mark is given for theory and practice.

Homework given individually:

a) repeat the derivatives of trigonometric functions;
b) interval method;
c) the mechanical meaning of the derivative.

2. A: No. 138, No. 142, B: No. 137 (a, b), No. 140 (a).

3. Take the derivative of the functions:

a) f(x)=x 4 -3x 2 -7;
b) f(x)=4x 3 -6x;
c) f(x)=-2sin(2x-4);
d) f(x)=cos(2x-4).

4. Name the function research scheme.

Class: 10

Presentation for the lesson

Back forward

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested this work please download the full version.

Lesson type: learning new material.

Teaching methods: visual, partially search.

The purpose of the lesson.

1. Introduce the concept of a tangent to the graph of a function at a point, find out what the geometric meaning of the derivative is, derive the tangent equation and teach how to find it for specific functions.
2. Develop logical thinking, mathematical speech.
3. Cultivate the will and perseverance to achieve the final results.

Equipment: interactive board, computer.

Lesson plan

I. Organizational moment

Checking students' readiness for the lesson. Message about the topic of the lesson and objectives.

II. Knowledge update.

(Recall with students the geometric definition of a tangent to a graph of a function. Give examples showing that this statement is not complete.)

Recall what is a tangent?

“A tangent is a straight line that has one common point with a given curve.” (Slide #2)

Discussion of the correctness of this definition. (After discussion, students come to the conclusion that this definition is incorrect.) To illustrate their conclusion, we give the following example.

Consider an example. (Slide #3)

Let a parabola and two straight lines be given , which has one common point M (1; 1) with this parabola. There is a discussion why the first line is not tangent to this parabola (Fig. 1), but the second one is (Fig. 2).

In this lesson, we must find out what is the tangent to the graph of a function at a point, how to write an equation for a tangent?

Consider the main tasks for compiling the tangent equation.

To do this, recall the general form of the equation of a straight line, the conditions for parallel lines, the definition of a derivative and the rules of differentiation. (Slide number 4)

III. Preparatory work for the study of new material.

1. Formulate the definition of a derivative. (Slide number 5)
2. Fill in the table of arbitrary elementary functions. (Slide number 6)
3. Remember the rules of differentiation. (Slide number 7)
4. Which of the following lines are parallel and why? (Make sure visually) (Slide number 8)

IV The study of new material.

To set the equation of a straight line on a plane, it is enough for us to know the slope and the coordinates of one point.

Let the graph of the function be given. A point is selected on it, at this point a tangent is drawn to the graph of the function (we assume that it exists). Find the slope of the tangent.

Let's increment the argument and consider on the graph (Fig. 3) the point P with the abscissa . The slope of the secant MP, i.e. the tangent of the angle between the secant and the x-axis, is calculated by the formula .

If we now tend to zero, then the point P will begin to approach the point M along the curve. We have characterized the tangent as the limiting position of the secant in this approximation. So, it is natural to assume that the slope of the tangent will be calculated by the formula .

Consequently, .

If to the graph of the function y = f (x) at the point x = a you can draw a tangent, non-parallel to the axis at, then expresses the slope of the tangent. (Slide number 10)

Or in another way. Derivative at a point x = a equal to the slope of the tangent to the graph of the function y = f(x) at this point.

This is the geometric meaning of the derivative. (Slide number 11)

Moreover, if:

Let us find out the general form of the tangent equation.

Let the straight line be given by the equation . We know that . To calculate m, we use the fact that the line passes through the point . Let's put it into the equation. We get , i.e. . Substitute the found values k and m into the equation of a straight line:

is the equation of the tangent to the graph of the function. (Slide number 12)

Consider examples:

Let's make the equation of a tangent:

(Slide number 14)

Solving these examples, we used a very simple algorithm, which is as follows: (Slide No. 15)

Consider typical tasks and their solution.

№1 Write the equation of the tangent to the graph of the function at the point.

(Slide number 16)

Solution. Let's use the algorithm, given that in this example .

2)

3) ;

4) Substitute the found numbers ,, into the formula.

№2 Draw a tangent to the graph of the function so that it is parallel to the straight line. (Slide number 17)

Solution. Let us refine the formulation of the problem. The requirement to “draw a tangent” usually means “make an equation for a tangent”. Let's use the tangent drawing algorithm, given that in this example .

The desired tangent must be parallel to the straight line. Two lines are parallel if and only if their slopes are equal. So the slope of the tangent must be equal to the slope of the given straight line: .No. Consequently: ; ., i.e.

V. Problem solving.

1. Solving problems on finished drawings (Slide No. 18 and Slide No. 19)

2. Solving problems from the textbook: No. 29.3 (a, c), No. 29.12 (b, d), No. 29.18, No. 29.23 (a) (Slide No. 20)

VI. Summarizing.

1. Answer the questions:

• What is called a tangent to the graph of a function at a point?
• What is the geometric meaning of the derivative?
• Formulate an algorithm for finding the tangent equation?

2. What were the difficulties in the lesson, what moments of the lesson did you like the most?

3. Marking.

VII. Comments on homework

No. 29.3 (b, d), No. 29.12 (a, c), No. 29.19, No. 29.23 (b) (Slide No. 22)

Literature. (Slide 23)

1. Algebra and the beginning of mathematical analysis: Proc. For 10-11 cells. for students of educational institutions (basic level) / Edited by A.G. Mordkovich. – M.: Mnemosyne, 2009.
2. Algebra and the beginning of mathematical analysis: Problem book, For 10-11 cells. for students of educational institutions (basic level) / Edited by A.G. Mordkovich. – M.: Mnemosyne, 2009.
3. Algebra and the beginnings of analysis. Independent and test papers for grades 10-11. / Ershova A.P., Goloborodko V.V. – M.: ILEKSA, 2010.
4. USE 2010. Mathematics. Task B8. Workbook/ Under the editorship of A.L. Semenov and I.V. Yashchenko - M .: MTsNMO Publishing House, 2010.

Open algebra lesson in grade 11 19.10. 2011

Teacher: Gorbunova S.V.

Lesson topic: The equation of the tangent to the graph of the function.

Lesson Objectives

1. 
   Clarify the concept of "tangent".
2. 
   Derive the tangent equation.
3. 
   Write an algorithm for "drawing up the equation of the tangent to the graph of the function

y = f(x)".

1. 
   Start practicing skills and abilities in drawing up a tangent equation in various mathematical situations.
2. 
   To form the ability to analyze, generalize, show, use the elements of research, develop mathematical speech.

Equipment: computer, presentation, projector, interactive whiteboard, reminder cards, reflection cards.

Lesson structure:

1. 
   HE. U.
2. 
   Lesson topic message
3. 
   Repetition of the studied material
4. 
   Formulation of the problem.
5. 
   Explanation of new material.
6. 
   Creation of an algorithm for "drawing up the equation of a tangent".
7. 
   History reference.
8. 
   Consolidation. Development of skills and abilities in drawing up the tangent equation.
9. 
   Homework.
10. 
    Independent work with self-test
11. 
    Summing up the lesson.
12. 
    Reflection

During the classes

1. O.N.U.

2. Posting the topic of the lesson

The topic of today's lesson is "The equation of the tangent to the graph of a function." Open your notebooks, write down the date and the topic of the lesson. (slide 1)

Let the words that you see on the screen become the motto of today's lesson. (slide 2)

• 
  There are no bad ideas
• 
  Think creatively
• 
  take risks
• 
  Don't criticize

To prepare for the lesson, we will repeat the previously studied material. Attention to the screen. Write the solution down in your notebook.

2. Repetition of the studied material (slide 3).

Purpose: to test knowledge of the basic rules of differentiation.

Find the derivative of a function:

Who has more than one mistake? Who has one?

3. Update

Purpose: To activate attention, show the lack of knowledge about the tangent, formulate the goals and objectives of the lesson. (Slide 4)

Let's discuss what is a tangent to a function graph?

Do you agree with the statement that "A tangent is a straight line that has one common point with a given curve"?
There is a discussion. Statements of children (yes and why, no and why). During the discussion, we come to the conclusion that this statement is not true.

Let's look at specific examples:

Examples.(slide 5)
1) The line x = 1 has one common point M(1; 1) with the parabola y = x 2, but it is not tangent to the parabola.

The straight line y = 2x – 1 passing through the same point is tangent to the given parabola.

The line x = π is not tangent to the graph y = cos x, although it has the only common point K(π; 1) with it. On the other hand, the line y = - 1 passing through the same point is tangent to the graph, although it has infinitely many common points of the form (π+2 πk; 1), where k is an integer, in each of which it concerns the chart.

^ 4. Setting goals and objectives for children in the lesson: (slide 6)

Try to formulate the purpose of the lesson yourself.

Find out what is the tangent to the graph of the function at a point, derive the equation of the tangent. Apply the formula to problem solving
^ 5. Learning new material

See how the position of the line x=1 differs from the position y=2x-1? (slide 7)

Conclude what is a tangent?

The tangent is the limiting position of the secant.

Since the tangent is a straight line, and we need to write the equation for the tangent, what do you think we need to remember?

Recall the general form of the straight line equation. (y \u003d kx + b)

What is another name for the number k? (slope or tangent of the angle between this line and the positive direction of the Ox axis) k \u003d tg α

What is the geometric meaning of the derivative?

The tangent of the angle of inclination between the tangent and the positive direction of the x-axis

That is, I can write tg α = yˈ(x). (slide 8)

Let's illustrate this with a drawing. (slide 9)

Let a function y = f (x) be given and a point M belonging to the graph of this function. Let's define its coordinates as follows: x=a, y=f(a), i.e. M (a, f (a)) and let there be a derivative f "(a), i.e. at a given point, the derivative is defined. Draw a tangent through the point M. The tangent equation is the equation of a straight line, so it looks like: y \u003d kx + b. Therefore, the task is to find k and b. Pay attention to the blackboard, from what is written there, is it possible to find k? (yes, k = f "(a).)

How to find b now? The desired line passes through the point M (a; f (a)), we substitute these coordinates into the equation of the line: f (a) \u003d ka + b, hence b \u003d f (a) - ka, since k \u003d tg α \u003d yˈ(x), then b = f(a) – f "(a)a

Substitute the value of b and k into the equation y = kx + b.

y \u003d f "(a) x + f (a) - f "(a) a, taking the common factor out of the bracket, we get:

y \u003d f (a) + f "(a) (x-a).

We have obtained the equation of the tangent to the graph of the function y = f(x) at the point x = a.

To confidently solve problems on a tangent, you need to clearly understand the meaning of each element in this equation. Let's dwell on this again: (slide 10)

1. 
   (a, f (a)) - point of contact
2. 
   f "(a) \u003d tg α \u003d k slope angle or slope
3. 
   (x, y) - any point of the tangent

And so we derived the equation of the tangent, analyzed the meaning of each element in this equation, let's now try to derive an algorithm for compiling the equation of the tangent to the graph of the function y = f (x)

6. Drawing up an algorithm (slide 11).

I propose to create an algorithm for the students themselves:

1. 
   We denote the abscissa of the point of contact with the letter a.
2. 
   Let's calculate f(a).
3. 
   Find f "(x) and calculate f "(a).
4. 
   Let us substitute the found values ​​of the number a, f (a), f "(a) into the tangent equation.
5. 
   y \u003d f (a) + f "(a) (x-a).

(I distribute the algorithm printed in advance to students as a reminder for later work.)

1. 
   Historical background (slide 12).

Attention to the screen. Unscramble the word

1	4/3	9	-4	-1	-3	5

Answer: FLUX (slide 13).

What is the origin story of this name? (slide 14.15)

The concept of a derivative arose in connection with the need to solve a number of problems in physics, mechanics and mathematics. The honor of discovering the basic laws of mathematical analysis belongs to the English scientist Newton and the German mathematician Leibniz. Leibniz considered the problem of drawing a tangent to an arbitrary curve.

The famous physicist Isaac Newton, who was born in the English village of Woolstrop, made a significant contribution to mathematics. Solving problems on drawing tangents to curves, calculating the areas of curvilinear figures, he created a general method for solving such problems - flux method (derivatives), and called the derivative itself fluent .

He calculated the derivative and integral of the power function. He writes about differential and integral calculus in his work "Method of Fluxions" (1665 - 1666), which served as one of the beginnings of mathematical analysis, differential and integral calculus, which the scientist developed independently of Leibniz.

Many scientists in different years were interested in the tangent. Occasionally, the concept of a tangent was encountered in the works of the Italian mathematician N. Tartaglia (c. 1500 - 1557) - here the tangent appeared in the course of studying the issue of the angle of inclination of the gun, which ensures the greatest givenness of the flight of the projectile. I. Keppler considered the tangent in the course of solving the problem of the largest volume of a parallelepiped inscribed in a ball of given radius.

In the 17th century, on the basis of G. Galileo's theory of motion, the kinematic concept of the derivative was actively developed. Various presentation options are found in R. Descartes, the French mathematician Roberval, the English scientist D. Gregory, in the works of I. Barrow.

8. Consolidation (slide 16-18).

1) Compose the equation of the tangent to the graph of the function f (x) \u003d x² - 3x + 5 at the point with the abscissa

Solution:

Let's make the equation of a tangent (according to algorithm). Call a strong student.

1. 
   a = -1;
2. 
   f(a) = f(-1) = 1 + 3 + 5 = 9;
3. 
   f "(x) \u003d 2x - 3,
   f "(a) \u003d f "(-1) \u003d -2 - 3 \u003d -5;
4. 
   y \u003d 9 - 5 (x + 1),

y = 4 - 5x.

Answer: y = 4 - 5x.

USE assignments 2011 B-8

1. The function y \u003d f (x) is defined on the interval (-3; 4). The figure shows its graph and the tangent to this graph at the point with the abscissa a \u003d 1. Calculate the value of the derivative f "(x) at the point a \u003d 1.

Solution: to solve it, it is necessary to remember that if the coordinates of any two points A and B lying on a given line are known, then its slope can be calculated by the formula: k \u003d, where (x 1; y 1), (x 2; y 2) are the coordinates of points A, B, respectively. The graph shows that this tangent passes through points with coordinates (1; -2) and (3; -1), which means k=(-1-(-2))/(3-1)= 0.5.

2. The function y \u003d f (x) is defined on the interval (-3; 4). The figure shows its graph and the tangent to this graph at the point with the abscissa a = -2. Calculate the value of the derivative f "(x) at the point a \u003d -2.

Solution: the graph passes through the points (-2;1) (0;-1) . fˈ(-2)= -2

8. Homework (slide 19).

Preparation for the exam B-8 No. 3 - 10

^ 9. Independent work

Write the equation of the tangent to the graph of the function y \u003d f (x) at the point with the abscissa a.
option 1 option 2

f(x) = x²+ x+1, a=1 f(x)= x-3x², a=2

answers: 1st option: y=3x; Option 2: y \u003d -11x + 12

10. Summing up.

• 
  What is called a tangent to the graph of a function at a point?
• 
  What is the geometric meaning of the derivative?
• 
  Formulate an algorithm for finding the tangent equation at a point?

11. Reflection:

Choose an emoticon that matches your mood and state after the lesson. Thank you for the lesson.