Corner point of the graph. Tangent to a graph of a function at a point

Job type: 7

Condition

The line y=3x+2 is tangent to the graph of the function y=-12x^2+bx-10. Find b , given that the abscissa of the touch point is less than zero.

Show Solution

Decision

Let x_0 be the abscissa of the point on the graph of the function y=-12x^2+bx-10 through which the tangent to this graph passes.

The value of the derivative at the point x_0 is equal to the slope of the tangent, i.e. y"(x_0)=-24x_0+b=3. On the other hand, the tangent point belongs to both the graph of the function and the tangent, i.e. -12x_0^2+bx_0-10= 3x_0 + 2. We get a system of equations \begin(cases) -24x_0+b=3,\\-12x_0^2+bx_0-10=3x_0+2. \end(cases)

Solving this system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the condition of the abscissa, the touch points are less than zero, therefore x_0=-1, then b=3+24x_0=-21.

Answer

Job type: 7
Subject: geometric sense derivative. Tangent to function graph

Condition

The line y=-3x+4 is parallel to the tangent to the graph of the function y=-x^2+5x-7. Find the abscissa of the point of contact.

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Decision

The slope of the line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is y"(x_0). But y"=-2x+5, so y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is -3.Parallel lines have the same slopes.Therefore, we find such a value x_0 that =-2x_0 +5=-3.

We get: x_0 = 4.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

Show Solution

Decision

From the figure, we determine that the tangent passes through the points A(-6; 2) and B(-1; 1). Denote by C(-6; 1) the point of intersection of the lines x=-6 and y=1, and by \alpha the angle ABC (it can be seen in the figure that it is sharp). Then the line AB forms an obtuse angle \pi -\alpha with the positive direction of the Ox axis.

As you know, tg(\pi -\alpha) will be the value of the derivative of the function f(x) at the point x_0. notice, that tg \alpha =\frac(AC)(CB)=\frac(2-1)(-1-(-6))=\frac15. From here, by the reduction formulas, we obtain: tg(\pi -\alpha) =-tg \alpha =-\frac15=-0.2.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The line y=-2x-4 is tangent to the graph of the function y=16x^2+bx+12. Find b , given that the abscissa of the touch point is greater than zero.

Show Solution

Decision

Let x_0 be the abscissa of the point on the graph of the function y=16x^2+bx+12 through which

is tangent to this graph.

The value of the derivative at the point x_0 is equal to the slope of the tangent, i.e. y "(x_0)=32x_0+b=-2. On the other hand, the tangent point belongs to both the graph of the function and the tangent, i.e. 16x_0^2+bx_0+12=- 2x_0-4 We get a system of equations \begin(cases) 32x_0+b=-2,\\16x_0^2+bx_0+12=-2x_0-4. \end(cases)

Solving the system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the condition of the abscissa, the touch points are greater than zero, therefore x_0=1, then b=-2-32x_0=-34.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The figure shows a graph of the function y=f(x) defined on the interval (-2; 8). Determine the number of points where the tangent to the graph of the function is parallel to the straight line y=6.

Show Solution

Decision

The line y=6 is parallel to the Ox axis. Therefore, we find such points at which the tangent to the function graph is parallel to the Ox axis. On this chart, such points are extremum points (maximum or minimum points). As you can see, there are 4 extremum points.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The line y=4x-6 is parallel to the tangent to the graph of the function y=x^2-4x+9. Find the abscissa of the point of contact.

Show Solution

Decision

The slope of the tangent to the graph of the function y \u003d x ^ 2-4x + 9 at an arbitrary point x_0 is y "(x_0). But y" \u003d 2x-4, which means y "(x_0) \u003d 2x_0-4. The slope of the tangent y \u003d 4x-7 specified in the condition is equal to 4. Parallel lines have the same slopes. Therefore, we find such a value x_0 that 2x_0-4 \u003d 4. We get: x_0 \u003d 4.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The figure shows the graph of the function y=f(x) and the tangent to it at the point with the abscissa x_0. Find the value of the derivative of the function f(x) at the point x_0.

Show Solution

Decision

From the figure, we determine that the tangent passes through the points A(1; 1) and B(5; 4). Denote by C(5; 1) the point of intersection of the lines x=5 and y=1, and by \alpha the angle BAC (it can be seen in the figure that it is acute). Then the line AB forms an angle \alpha with the positive direction of the Ox axis.

In this article, we will analyze all types of problems for finding

Let's remember geometric meaning of the derivative: if a tangent is drawn to the graph of a function at a point, then the slope of the tangent (equal to the tangent of the angle between the tangent and the positive direction of the axis ) is equal to the derivative of the function at the point .

Take an arbitrary point on the tangent with coordinates :

And consider a right triangle:

In this triangle

From here

This is the equation of the tangent drawn to the graph of the function at the point.

To write the equation of the tangent, we only need to know the equation of the function and the point where the tangent is drawn. Then we can find and .

There are three main types of tangent equation problems.

1. Given a point of contact

2. Given the coefficient of slope of the tangent, that is, the value of the derivative of the function at the point.

3. Given the coordinates of the point through which the tangent is drawn, but which is not a tangent point.

Let's look at each type of problem.

one . Write the equation of the tangent to the graph of the function at the point .

.

b) Find the value of the derivative at the point . First we find the derivative of the function

Substitute the found values ​​into the tangent equation:

Let's open the brackets on the right side of the equation. We get:

Answer: .

2. Find the abscissas of the points at which the functions tangent to the graph parallel to the x-axis.

If the tangent is parallel to the x-axis, then the angle between the tangent and the positive direction of the axis zero, therefore, the tangent of the slope of the tangent is zero. So the value of the derivative of the function at the points of contact is zero.

a) Find the derivative of the function .

b) Equate the derivative to zero and find the values ​​in which the tangent is parallel to the axis:

We equate each factor to zero, we get:

Answer: 0;3;5

3 . Write equations of tangents to the graph of a function , parallel straight .

The tangent is parallel to the line. The slope of this straight line is -1. Since the tangent is parallel to this line, therefore, the slope of the tangent is also -1. I.e we know the slope of the tangent, and thus the value of the derivative at the point of contact.

This is the second type of problem for finding the tangent equation.

So, we are given a function and the value of the derivative at the point of contact.

a) Find the points at which the derivative of the function is equal to -1.

First, let's find the derivative equation.

Let's equate the derivative to the number -1.

Find the value of the function at the point .

(by condition)

.

b) Find the equation of the tangent to the graph of the function at the point .

Find the value of the function at the point .

(by condition).

Substitute these values ​​into the tangent equation:

.

Answer:

4 . Write an equation for a tangent to a curve , passing through a point

First, check if the point is not a touch point. If the point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Substitute the coordinates of the point in the equation of the function.

Title="(!LANG:1sqrt(8-3^2)">. Мы получили под корнем отрицательное число, равенство не верно, и точка не принадлежит графику функции и !} is not a point of contact.

This is the last type of problem for finding the tangent equation. First thing we need to find the abscissa of the point of contact.

Let's find the value.

Let be the point of contact. The point belongs to the tangent to the graph of the function . If we substitute the coordinates of this point into the tangent equation, we get the correct equality:

.

The value of the function at the point is .

Find the value of the derivative of the function at the point .

Let's find the derivative of the function first. This is .

The derivative at a point is .

Let us substitute the expressions for and into the equation of the tangent. We get the equation for:

Let's solve this equation.

Reduce the numerator and denominator of the fraction by 2:

We bring the right side of the equation to a common denominator. We get:

Simplify the numerator of the fraction and multiply both parts by - this expression is strictly greater than zero.

We get the equation

Let's solve it. To do this, we square both parts and go to the system.

Title="(!LANG:delim(lbrace)(matrix(2)(1)((64-48(x_0)+9(x_0)^2=8-(x_0)^2) (8-3x_0>=0 ) ))( )">!}

Let's solve the first equation.

We will decide quadratic equation, we get

The second root does not satisfy the condition title="(!LANG:8-3x_0>=0">, следовательно, у нас только одна точка касания и её абсцисса равна .!}

Let's write the equation of the tangent to the curve at the point . To do this, we substitute the value in the equation We already recorded it.

Answer:
.

Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the line passing through the point (x 0 ; f (x 0)), having slope f '(x 0), is called a tangent.

But what happens if the derivative at the point x 0 does not exist? There are two options:

1. The tangent to the graph also does not exist. The classic example is the function y = |x | at the point (0; 0).
2. The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).

Tangent equation

Any non-vertical straight line is given by an equation of the form y = kx + b, where k is the slope. The tangent is no exception, and in order to compose its equation at some point x 0, it is enough to know the value of the function and the derivative at this point.

So, let a function be given y \u003d f (x), which has a derivative y \u003d f '(x) on the segment. Then at any point x 0 ∈ (a; b) a tangent can be drawn to the graph of this function, which is given by the equation:

y \u003d f '(x 0) (x - x 0) + f (x 0)

Here f ’(x 0) is the value of the derivative at the point x 0, and f (x 0) is the value of the function itself.

Task. Given a function y = x 3 . Write an equation for the tangent to the graph of this function at the point x 0 = 2.

Tangent equation: y \u003d f '(x 0) (x - x 0) + f (x 0). The point x 0 = 2 is given to us, but the values ​​f (x 0) and f '(x 0) will have to be calculated.

First, let's find the value of the function. Everything is easy here: f (x 0) = f (2) = 2 3 = 8;
Now let's find the derivative: f '(x) \u003d (x 3) ' \u003d 3x 2;
Substitute in the derivative x 0 = 2: f '(x 0) = f '(2) = 3 2 2 = 12;
So we get: y = 12 (x - 2) + 8 = 12x - 24 + 8 = 12x - 16.
This is the tangent equation.

Task. Compose the equation of the tangent to the graph of the function f (x) \u003d 2sin x + 5 at the point x 0 \u003d π / 2.

This time we will not describe in detail each action - we will only indicate the key steps. We have:

f (x 0) \u003d f (π / 2) \u003d 2sin (π / 2) + 5 \u003d 2 + 5 \u003d 7;
f '(x) \u003d (2sin x + 5) ' \u003d 2cos x;
f '(x 0) \u003d f '(π / 2) \u003d 2cos (π / 2) \u003d 0;

Tangent equation:

y = 0 (x − π /2) + 7 ⇒ y = 7

In the latter case, the line turned out to be horizontal, because its slope k = 0. There is nothing wrong with that - we just stumbled upon an extremum point.

Y \u003d f (x) and if at this point a tangent can be drawn to the function graph that is not perpendicular to the x-axis, then the slope of the tangent is f "(a). We have already used this several times. For example, in § 33 it was established, that the graph of the function y \u003d sin x (sinusoid) at the origin forms an angle of 45 ° with the abscissa axis (more precisely, the tangent to the graph at the origin makes an angle of 45 ° with the positive direction of the x axis), and in example 5 of § 33 points were found on given schedule functions, in which the tangent is parallel to the x-axis. In example 2 of § 33, an equation was drawn up for the tangent to the graph of the function y \u003d x 2 at the point x \u003d 1 (more precisely, at the point (1; 1), but more often only the value of the abscissa is indicated, assuming that if the value of the abscissa is known, then the value of the ordinate can be found from the equation y = f(x)). In this section, we will develop an algorithm for compiling the equation of the tangent to the graph of any function.

Let the function y \u003d f (x) and the point M (a; f (a)) be given, and it is also known that f "(a) exists. Let us compose the equation of the tangent to the graph of the given function in given point. This equation, like the equation of any straight line not parallel to the y-axis, has the form y = kx + m, so the problem is to find the values ​​of the coefficients k and m.

There are no problems with the slope k: we know that k \u003d f "(a). To calculate the value of m, we use the fact that the desired line passes through the point M (a; f (a)). This means that if we substitute the coordinates points M into the equation of a straight line, we get the correct equality: f (a) \u003d ka + m, from where we find that m \u003d f (a) - ka.
It remains to substitute the found values ​​of the whale coefficients into the equation straight:

We have obtained the equation of the tangent to the graph of the function y \u003d f (x) at the point x \u003d a.
If, say,
Substituting in equation (1) the found values ​​a \u003d 1, f (a) \u003d 1 f "(a) \u003d 2, we get: y \u003d 1 + 2 (x-f), i.e. y \u003d 2x-1.
Compare this result with the one obtained in Example 2 of § 33. Naturally, the same thing happened.
Let us compose the equation of the tangent to the graph of the function y \u003d tg x at the origin. We have: hence cos x f "(0) = 1. Substituting the found values ​​a \u003d 0, f (a) \u003d 0, f "(a) \u003d 1 into equation (1), we get: y \u003d x.
That is why we drew the tangentoid in § 15 (see Fig. 62) through the origin of coordinates at an angle of 45 ° to the abscissa axis.
Solving these is enough simple examples, we actually used a certain algorithm, which is embedded in formula (1). Let's make this algorithm explicit.

ALGORITHM FOR COMPOSING THE EQUATION OF THE FUNCTION TANGENT TO THE GRAPH y \u003d f (x)

1) Designate the abscissa of the point of contact with the letter a.
2) Calculate 1 (a).
3) Find f "(x) and calculate f" (a).
4) Substitute the found numbers a, f(a), (a) into formula (1).

Example 1 Write an equation for the tangent to the graph of the function at the point x = 1.
Let's use the algorithm, taking into account that in this example

On fig. 126 shows a hyperbola, a straight line y \u003d 2x is built.
The drawing confirms the given calculations: indeed, the line y \u003d 2-x touches the hyperbola at the point (1; 1).

Answer: y \u003d 2-x.
Example 2 Draw a tangent to the graph of the function so that it is parallel to the straight line y \u003d 4x - 5.
Let us refine the formulation of the problem. The requirement to "draw a tangent" usually means "make an equation for a tangent". This is logical, because if a person was able to compose an equation for a tangent, then he is unlikely to experience difficulties in constructing a straight line on the coordinate plane according to its equation.
Let's use the algorithm for compiling the tangent equation, given that in this example, But, unlike the previous example, there is ambiguity here: the abscissa of the tangent point is not explicitly indicated.
Let's start talking like this. The desired tangent must be parallel to the straight line y \u003d 4x-5. Two lines are parallel if and only if their slopes are equal. This means that the slope of the tangent must be equal to the slope of the given straight line: Thus, we can find the value of a from the equation f "(a) \u003d 4.
We have:
From the equation So, there are two tangents that satisfy the conditions of the problem: one at the point with the abscissa 2, the other at the point with the abscissa -2.
Now you can act according to the algorithm.

Example 3 From the point (0; 1) draw a tangent to the graph of the function
Let's use the algorithm for compiling the tangent equation, considering that in this example Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we act according to the algorithm.

By condition, the tangent passes through the point (0; 1). Substituting into equation (2) the values ​​x = 0, y = 1, we get:
As you can see, in this example, only at the fourth step of the algorithm we managed to find the abscissa of the touch point. Substituting the value a \u003d 4 into equation (2), we get:

On fig. 127 shows a geometric illustration of the considered example: a graph of the function

In § 32 we noted that for a function y = f(x), which has a derivative at a fixed point x, the approximate equality holds:

For the convenience of further reasoning, we change the notation: instead of x we ​​will write a, instead we will write x, and accordingly we will write x-a instead. Then the approximate equality written above will take the form:

Now take a look at fig. 128. A tangent is drawn to the graph of the function y \u003d f (x) at the point M (a; f (a)). Marked point x on the x-axis close to a. It is clear that f(x) is the ordinate of the graph of the function at the specified point x. And what is f (a) + f "(a) (x-a)? This is the ordinate of the tangent corresponding to the same point x - see formula (1). What is the meaning of approximate equality (3)? That to calculate the approximate value of the function, the value of the tangent ordinate is taken.

Example 4 Find the approximate value of the numerical expression 1.02 7 .
It's about about finding the value of the function y \u003d x 7 at the point x \u003d 1.02. We use formula (3), taking into account that in this example
As a result, we get:

If we use a calculator, we get: 1.02 7 = 1.148685667...
As you can see, the approximation accuracy is quite acceptable.
Answer: 1,02 7 =1,14.

A.G. Mordkovich Algebra Grade 10

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Consider the following figure:

It shows some function y = f(x) that is differentiable at the point a. Marked point M with coordinates (a; f(a)). Through an arbitrary point P(a + ∆x; f(a + ∆x)) of the graph, a secant MP is drawn.

If now the point P is shifted along the graph to the point M, then the straight line MP will rotate around the point M. In this case, ∆x will tend to zero. From here we can formulate the definition of a tangent to the graph of a function.

Tangent to function graph

The tangent to the graph of the function is the limiting position of the secant when the increment of the argument tends to zero. It should be understood that the existence of the derivative of the function f at the point x0 means that at this point of the graph there is tangent to him.

In this case, the slope of the tangent will be equal to the derivative of this function at this point f’(x0). This is the geometric meaning of the derivative. The tangent to the graph of the function f differentiable at the point x0 is some straight line passing through the point (x0;f(x0)) and having a slope f’(x0).

Tangent equation

Let's try to get the equation of the tangent to the graph of some function f at the point A(x0; f(x0)). The equation of a straight line with a slope k has the following form:

Since our slope is equal to the derivative f'(x0), then the equation will take the following form: y = f'(x0)*x + b.

Now let's calculate the value of b. To do this, we use the fact that the function passes through point A.

f(x0) = f’(x0)*x0 + b, from here we express b and get b = f(x0) - f’(x0)*x0.

We substitute the resulting value into the tangent equation:

y = f'(x0)*x + b = f'(x0)*x + f(x0) - f'(x0)*x0 = f(x0) + f'(x0)*(x - x0).

y = f(x0) + f'(x0)*(x - x0).

Consider the following example: find the equation of the tangent to the graph of the function f (x) \u003d x 3 - 2 * x 2 + 1 at the point x \u003d 2.

2. f(x0) = f(2) = 2 2 - 2*2 2 + 1 = 1.

3. f'(x) = 3*x 2 - 4*x.

4. f'(x0) = f'(2) = 3*2 2 - 4*2 = 4.

5. Substitute the obtained values ​​into the tangent formula, we get: y = 1 + 4*(x - 2). Opening the brackets and bringing like terms, we get: y = 4*x - 7.

Answer: y = 4*x - 7.

General scheme for compiling the tangent equation to the graph of the function y = f(x):

1. Determine x0.

2. Calculate f(x0).

3. Calculate f'(x)