The tangent of the angle of inclination of a straight line to the axis. Function derivative. The geometric meaning of the derivative

In mathematics, one of the parameters describing the position of a straight line on the Cartesian coordinate plane is slope this straight line. This parameter characterizes the slope of the straight line to the x-axis. To understand how to find the slope, first recall the general form of the equation of a straight line in the XY coordinate system.

In general, any line can be represented by the expression ax+by=c, where a, b and c are arbitrary real numbers, but necessarily a 2 + b 2 ≠ 0.

With the help of simple transformations, such an equation can be brought to the form y=kx+d, in which k and d are real numbers. The number k is a slope, and the equation of a straight line of this kind is called an equation with a slope. It turns out that to find the slope, you just need to bring the original equation to the above form. For a better understanding, consider a specific example:

Task: Find the slope of the line given by the equation 36x - 18y = 108

Solution: Let's transform the original equation.

Answer: The desired slope of this line is 2.

If, during the transformation of the equation, we obtained an expression of the type x = const and as a result we cannot represent y as a function of x, then we are dealing with a straight line parallel to the X axis. The slope of such a straight line is equal to infinity.

For lines that are expressed by an equation like y = const, the slope is zero. This is typical for straight lines parallel to the x-axis. For example:

Task: Find the slope of the line given by the equation 24x + 12y - 4(3y + 7) = 4

Solution: We bring the original equation to a general form

24x + 12y - 12y + 28 = 4

It is impossible to express y from the resulting expression, therefore the slope of this line is equal to infinity, and the line itself will be parallel to the Y axis.

geometric sense

For a better understanding, let's look at the picture:

In the figure, we see a graph of a function of the type y = kx. To simplify, we take the coefficient c = 0. In the triangle OAB, the ratio of the side BA to AO will be equal to the slope k. At the same time, the ratio VA / AO is the tangent acute angleα in right triangle OAV. It turns out that the slope of a straight line is equal to the tangent of the angle that this straight line makes with the x-axis of the coordinate grid.

Solving the problem of how to find the slope of a straight line, we find the tangent of the angle between it and the x-axis of the coordinate grid. The boundary cases, when the line under consideration is parallel to the coordinate axes, confirm the above. Indeed, for a straight line described by the equation y=const, the angle between it and the abscissa axis zero. The tangent of the zero angle is also zero and the slope is also zero.

For straight lines perpendicular to the x-axis and described by the equation x=const, the angle between them and the x-axis is 90 degrees. Tangent right angle is equal to infinity, and the slope of similar straight lines is equal to infinity, which confirms what was written above.

Tangent Slope

A common, often encountered in practice, task is also to find the slope of the tangent to the function graph at some point. The tangent is a straight line, therefore the concept of slope is also applicable to it.

To figure out how to find the slope of a tangent, we will need to recall the concept of a derivative. The derivative of any function at some point is a constant numerically equal to the tangent of the angle that forms between the tangent at the specified point to the graph of this function and the abscissa axis. It turns out that to determine the slope of the tangent at the point x 0, we need to calculate the value of the derivative of the original function at this point k \u003d f "(x 0). Let's consider an example:

Task: Find the slope of the line tangent to the function y = 12x 2 + 2xe x at x = 0.1.

Solution: Find the derivative of the original function in general form

y "(0,1) = 24 . 0.1 + 2 . 0.1 . e 0.1 + 2 . e 0.1

Answer: The desired slope at the point x \u003d 0.1 is 4.831

The continuation of the topic of the equation of a straight line on a plane is based on the study of a straight line from algebra lessons. This article gives generalized information on the topic of the equation of a straight line with a slope. Consider the definitions, get the equation itself, reveal the relationship with other types of equations. Everything will be discussed on examples of problem solving.

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Before writing such an equation, it is necessary to define the angle of inclination of a straight line to the O x axis with their slope. Let us assume that a Cartesian coordinate system O x is given on the plane.

Definition 1

The angle of inclination of the straight line to the axis O x, located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the straight line counterclockwise.

When a line is parallel to Ox or coincidence occurs in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is defined on the interval [ 0 , π) .

Definition 2

Slope of a straight line is the tangent of the slope of the given line.

The standard notation is k. From the definition we get that k = t g α . When the line is parallel to Ox, the slope is said to not exist because it goes to infinity.

The slope is positive when the graph of the function is increasing and vice versa. The figure shows various variations of the location of the right angle relative to the coordinate system with the value of the coefficient.

To find this angle, it is necessary to apply the definition of the slope coefficient and calculate the tangent of the inclination angle in the plane.

Solution

From the condition we have that α = 120 °. By definition, you need to calculate the slope. Let's find it from the formula k = t g α = 120 = - 3 .

Answer: k = - 3 .

If the angular coefficient is known, but it is necessary to find the angle of inclination to the x-axis, then the value of the angular coefficient should be taken into account. If k > 0, then the right angle is acute and is found by the formula α = a r c t g k . If k< 0 , тогда угол тупой, что дает право определить его по формуле α = π - a r c t g k .

Example 2

Determine the angle of inclination of the given straight line to O x with a slope equal to 3.

Solution

From the condition we have that the slope is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made according to the formula α = a r c t g k = a r c t g 3 .

Answer: α = a r c t g 3 .

Example 3

Find the angle of inclination of the straight line to the O x axis, if the slope = - 1 3 .

Solution

If we take the letter k as the designation of the slope, then α is the angle of inclination to the given straight line in the positive direction O x. Hence k = - 1 3< 0 , тогда необходимо применить формулу α = π - a r c t g k При подстановке получим выражение:

α = π - a r c t g - 1 3 = π - a r c t g 1 3 = π - π 6 = 5 π 6 .

Answer: 5 pi 6 .

An equation of the form y \u003d k x + b, where k is a slope, and b is some real number, is called the equation of a straight line with a slope. The equation is typical for any straight line that is not parallel to the O y axis.

If we consider in detail a straight line on a plane in a fixed coordinate system, which is given by an equation with a slope that looks like y \u003d k x + b. In this case, it means that the coordinates of any point on the line correspond to the equation. If we substitute the coordinates of the point M, M 1 (x 1, y 1), into the equation y \u003d k x + b, then in this case the line will pass through this point, otherwise the point does not belong to the line.

Example 4

Given a straight line with slope y = 1 3 x - 1 . Calculate whether the points M 1 (3 , 0) and M 2 (2 , - 2) belong to the given line.

Solution

It is necessary to substitute the coordinates of the point M 1 (3, 0) into the given equation, then we get 0 = 1 3 3 - 1 ⇔ 0 = 0 . The equality is true, so the point belongs to the line.

If we substitute the coordinates of the point M 2 (2, - 2), then we get an incorrect equality of the form - 2 = 1 3 · 2 - 1 ⇔ - 2 = - 1 3 . We can conclude that the point M 2 does not belong to the line.

Answer: M 1 belongs to the line, but M 2 does not.

It is known that the straight line is defined by the equation y = k · x + b passing through M 1 (0 , b) , substitution yielded an equality of the form b = k · 0 + b ⇔ b = b . From this we can conclude that the equation of a straight line with a slope y = k · x + b on the plane defines a straight line that passes through the point 0, b. It forms an angle α with the positive direction of the O x axis, where k = t g α .

Consider, for example, a straight line defined using a slope given by the form y = 3 · x - 1 . We get that the straight line will pass through the point with coordinate 0, - 1 with a slope of α = a r c t g 3 = π 3 radians along the positive direction of the O x axis. From this it can be seen that the coefficient is 3.

The equation of a straight line with a slope passing through a given point

It is necessary to solve a problem where it is necessary to obtain the equation of a straight line with a given slope passing through the point M 1 (x 1, y 1) .

The equality y 1 = k · x + b can be considered valid, since the line passes through the point M 1 (x 1 , y 1) . To remove the number b, it is necessary to subtract the equation with the slope coefficient from the left and right sides. It follows from this that y - y 1 = k · (x - x 1) . This equality is called the equation of a straight line with a given slope k, passing through the coordinates of the point M 1 (x 1, y 1) .

Example 5

Compose the equation of a straight line passing through the point M 1 with coordinates (4, - 1), with a slope equal to - 2.

Solution

By condition, we have that x 1 \u003d 4, y 1 \u003d - 1, k \u003d - 2. From here, the equation of the straight line will be written in this way y - y 1 = k · (x - x 1) ⇔ y - (- 1) = - 2 · (x - 4) ⇔ y = - 2 x + 7.

Answer: y = - 2 x + 7 .

Example 6

Write the equation of a straight line with a slope that passes through the point M 1 with coordinates (3, 5) parallel to the straight line y \u003d 2 x - 2.

Solution

By condition, we have that parallel lines have coinciding angles of inclination, hence the slope coefficients are equal. To find the slope from given equation, it is necessary to recall its basic formula y = 2 x - 2, hence it follows that k = 2 . We compose an equation with a slope coefficient and get:

y - y 1 = k (x - x 1) ⇔ y - 5 = 2 (x - 3) ⇔ y = 2 x - 1

Answer: y = 2 x - 1 .

The transition from the equation of a straight line with a slope to other types of equations of a straight line and vice versa

Such an equation is not always applicable for solving problems, since it has a not very convenient notation. To do this, it must be presented in a different form. For example, an equation of the form y = k · x + b does not allow you to write down the coordinates of the direction vector of the straight line or the coordinates of the normal vector. To do this, you need to learn how to represent equations of a different kind.

We can get the canonical equation of a straight line in a plane using the equation of a straight line with a slope. We get x - x 1 a x = y - y 1 a y . It is necessary to move the term b to the left side and divide by the expression of the resulting inequality. Then we get an equation of the form y = k x + b ⇔ y - b = k x ⇔ k x k = y - b k ⇔ x 1 = y - b k .

The equation of a straight line with a slope has become the canonical equation of a given straight line.

Example 7

Bring the equation of a straight line with slope y = - 3 x + 12 to canonical form.

Solution

We calculate and represent in the form of a canonical equation of a straight line. We get an equation of the form:

y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3

Answer: x 1 = y - 12 - 3.

The general equation of a straight line is easiest to obtain from y = k x + b, but this requires transformations: y = k x + b ⇔ k x - y + b = 0. The transition is made from general equation direct to equations of another kind.

Example 8

An equation of a straight line of the form y = 1 7 x - 2 is given. Find out if the vector with coordinates a → = (- 1 , 7) is a normal straight line vector?

Solution

To solve it, it is necessary to switch to another form of this equation, for this we write:

y = 1 7 x - 2 ⇔ 1 7 x - y - 2 = 0

The coefficients in front of the variables are the coordinates of the normal vector of the straight line. Let's write it like this n → = 1 7 , - 1 , hence 1 7 x - y - 2 = 0 . It is clear that the vector a → = (- 1 , 7) is collinear to the vector n → = 1 7 , - 1 , since we have a fair relation a → = - 7 · n → . It follows that the original vector a → = - 1 , 7 is a normal vector of the line 1 7 x - y - 2 = 0 , which means that it is considered a normal vector for the line y = 1 7 x - 2 .

Answer: Is

Let's solve the problem inverse to this one.

Need to move from general view equation A x + B y + C = 0 , where B ≠ 0 , to the slope equation. To do this, we solve the equation for y. We get A x + B y + C = 0 ⇔ - A B · x - C B .

The result is an equation with a slope equal to - A B .

Example 9

An equation of a straight line of the form 2 3 x - 4 y + 1 = 0 is given. Get the equation of a given line with a slope.

Solution

Based on the condition, it is necessary to solve for y, then we get an equation of the form:

2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 2 3 x + 1 ⇔ y = 1 6 x + 1 4 .

Answer: y = 1 6 x + 1 4 .

An equation of the form x a + y b \u003d 1 is solved in a similar way, which is called the equation of a straight line in segments, or canonical form x - x 1 a x = y - y 1 a y . It is necessary to solve it with respect to y, only then we get an equation with a slope:

x a + y b = 1 ⇔ y b = 1 - x a ⇔ y = - b a x + b .

The canonical equation can be reduced to a form with a slope. For this:

x - x 1 a x = y - y 1 a y ⇔ a y (x - x 1) = a x (y - y 1) ⇔ ⇔ a x y = a y x - a y x 1 + a x y 1 ⇔ y = a y a x x - a y a x x 1 + y 1

Example 10

There is a straight line given by the equation x 2 + y - 3 = 1 . Bring to the form of an equation with a slope.

Solution.

Based on the condition, it is necessary to transform, then we get an equation of the form _formula_. Both sides of the equation should be multiplied by -3 to get the required slope equation. Transforming, we get:

y - 3 = 1 - x 2 ⇔ - 3 y - 3 = - 3 1 - x 2 ⇔ y = 3 2 x - 3 .

Answer: y = 3 2 x - 3 .

Example 11

The straight line equation of the form x - 2 2 \u003d y + 1 5 is brought to the form with a slope.

Solution

It is necessary to calculate the expression x - 2 2 = y + 1 5 as a proportion. We get that 5 · (x - 2) = 2 · (y + 1) . Now you need to fully enable it, for this:

5 (x - 2) = 2 (y + 1) ⇔ 5 x - 10 = 2 y + 2 ⇔ 2 y = 5 x - 12 ⇔ y = 5 2 x

Answer: y = 5 2 x - 6 .

To solve such tasks, parametric equations of the straight line of the form x \u003d x 1 + a x λ y \u003d y 1 + a y λ should be reduced to the canonical equation of the straight line, only after that you can proceed to the equation with the slope.

Example 12

Find the slope of the straight line if it is given by parametric equations x = λ y = - 1 + 2 · λ .

Solution

You need to transition from parametric view to slope. To do this, we find the canonical equation from the given parametric one:

x = λ y = - 1 + 2 λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2 .

Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a straight line with a slope. To do this, we write this way:

x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1

It follows that the slope of the straight line is equal to 2. This is written as k = 2 .

Answer: k = 2 .

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The slope coefficient is straight. In this article, we will consider tasks related to the coordinate plane included in the exam in mathematics. These are assignments for:

- determination of the slope of a straight line, when two points through which it passes are known;
- determination of the abscissa or ordinate of the point of intersection of two lines on the plane.

What is the abscissa and ordinate of a point was described in this section. In it, we have already considered several problems related to the coordinate plane. What needs to be understood for the type of tasks under consideration? A bit of theory.

The equation of a straight line on the coordinate plane has the form:

where k – this is the slope of the straight line.

Next moment! Slope of a straight line equal to tangent angle of inclination of a straight line. This is the angle between the given line and the axisoh.

It lies between 0 and 180 degrees.

That is, if we reduce the equation of a straight line to the form y = kx + b, then further we can always determine the coefficient k (slope coefficient).

Also, if we can determine the tangent of the slope of the straight line based on the condition, then we will thereby find its slope.

The next theoretical moment!Equation of a straight line passing through two given points.The formula looks like:

Consider problems (similar to those from open bank assignments):

Find the slope of the straight line passing through the points with coordinates (–6; 0) and (0; 6).

In this problem, the most rational way to solve this is to find the tangent of the angle between the x-axis and the given straight line. It is known that it is equal to the angular coefficient. Consider a right triangle formed by a straight line and the x and y axes:

The tangent of an angle in a right triangle is the ratio of the opposite leg to the adjacent leg:

* Both legs are equal to six (these are their lengths).

Of course, this task can be solved using the formula for finding the equation of a straight line passing through two given points. But it will be a longer solution path.

Answer: 1

Find the slope of the straight line passing through the points with coordinates (5;0) and (0;5).

Our points have coordinates (5;0) and (0;5). Means,

Let's bring the formula to the form y = kx + b

We got that the angular coefficient k = – 1.

Answer: -1

Straight a passes through points with coordinates (0;6) and (8;0). Straight b passes through the point with coordinates (0;10) and is parallel to the line a b with axle ox.

In this problem, you can find the equation of a straight line a, determine the slope for it. Straight line b the slope will be the same since they are parallel. Next, you can find the equation of a straight line b. And then, substituting the value y = 0 into it, find the abscissa. BUT!

In this case, it is easier to use the triangle similarity property.

The right triangles formed by the given (parallel) lines of coordinates are similar, which means that the ratios of their respective sides are equal.

The desired abscissa is 40/3.

Answer: 40/3

Straight a passes through points with coordinates (0;8) and (–12;0). Straight b passes through the point with coordinates (0; -12) and is parallel to the line a. Find the abscissa of the point of intersection of the line b with axle ox.

For this problem, the most rational way to solve it is to use the similarity property of triangles. But we will solve it in a different way.

We know the points through which the line passes a. We can write the equation of a straight line. The formula for the equation of a straight line passing through two given points is:

By condition, the points have coordinates (0;8) and (–12;0). Means,

Let's bring to mind y = kx + b:

Got that corner k = 2/3.

*The angular coefficient could be found through the tangent of the angle in a right triangle with legs 8 and 12.

We know that parallel lines have equal slopes. So the equation of a straight line passing through the point (0;-12) has the form:

Find value b we can substitute the abscissa and ordinate into the equation:

So the line looks like:

Now, to find the desired abscissa of the point of intersection of the line with the x-axis, you need to substitute y \u003d 0:

Answer: 18

Find the ordinate of the point of intersection of the axis oy and a straight line passing through point B(10;12) and a parallel line passing through the origin and point A(10;24).

Let's find the equation of a straight line passing through the points with coordinates (0;0) and (10;24).

The formula for the equation of a straight line passing through two given points is:

Our points have coordinates (0;0) and (10;24). Means,

Let's bring to mind y = kx + b

The slopes of the parallel lines are equal. Hence, the equation of a straight line passing through the point B (10; 12) has the form:

Meaning b we find by substituting the coordinates of the point B (10; 12) into this equation:

We got the equation of a straight line:

To find the ordinate of the point of intersection of this line with the axis OU must be substituted into the found equation X= 0:

*Easiest solution. With the help of parallel translation, we shift this line down along the axis OU to the point (10;12). The shift occurs by 12 units, that is, point A(10;24) "passed" to point B(10;12), and point O(0;0) "passed" to point (0;–12). So the resulting line will intersect the axis OU at the point (0;–12).

The desired ordinate is -12.

Answer: -12

Find the ordinate of the point of intersection of the line given by the equation

3x + 2y = 6, with axis Oy.

Coordinate of the point of intersection of the given line with the axis OU has the form (0; at). Substitute the abscissa into the equation X= 0, and find the ordinate:

Ordinate of point of intersection of a line with an axis OU equals 3.

* The system is being solved:

Answer: 3

Find the ordinate of the point of intersection of the lines given by the equations

3x + 2y = 6 and y = - x.

When two lines are given, and the question is about finding the coordinates of the point of intersection of these lines, the system of these equations is solved:

In the first equation, we substitute - X instead of at:

The ordinate is minus six.

Answer: – 6

Find the slope of the straight line passing through the points with coordinates (–2; 0) and (0; 2).

Find the slope of the straight line passing through the points with coordinates (2;0) and (0;2).

The line a passes through the points with coordinates (0;4) and (6;0). Line b passes through the point with coordinates (0;8) and is parallel to line a. Find the abscissa of the point of intersection of line b with the x-axis.

Find the ordinate of the point of intersection of the y-axis and the line passing through point B (6;4) and the parallel line passing through the origin and point A (6;8).

1. It is necessary to clearly understand that the slope of the straight line is equal to the tangent of the slope of the straight line. This will help you in solving many problems of this type.

2. The formula for finding a straight line passing through two given points must be understood. With its help, you can always find the equation of a straight line if the coordinates of two of its points are given.

3. Remember that the slopes of parallel lines are equal.

4. As you understand, in some problems it is convenient to use the sign of similarity of triangles. Problems are solved practically orally.

5. Tasks in which two lines are given and it is required to find the abscissa or ordinate of their intersection point can be solved graphically. That is, build them on the coordinate plane (on a sheet in a cell) and determine the intersection point visually. *But this method is not always applicable.

6. And the last. If a straight line and the coordinates of the points of its intersection with the coordinate axes are given, then in such problems it is convenient to find the slope through finding the tangent of the angle in the formed right triangle. How to "see" this triangle for various arrangements of lines on the plane is schematically shown below:

>> Line inclination angle from 0 to 90 degrees<<

>> Straight line angle from 90 to 180 degrees<<

That's all. Good luck to you!

Sincerely, Alexander.

P.S: I would be grateful if you tell about the site in social networks.

The derivative of a function is one of the most difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.

This article simply and clearly explains what a derivative is and why it is needed.. We will not now strive for mathematical rigor of presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of the function.

The figure shows graphs of three functions. Which one do you think grows the fastest?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here is another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

You can see everything on the chart right away, right? Kostya's income has more than doubled in six months. And Grisha's income also increased, but just a little bit. And Matthew's income decreased to zero. The starting conditions are the same, but the rate of change of the function, i.e. derivative, - different. As for Matvey, the derivative of his income is generally negative.

Intuitively, we can easily estimate the rate of change of a function. But how do we do it?

What we are really looking at is how steeply the graph of the function goes up (or down). In other words, how fast y changes with x. Obviously, the same function at different points can have a different value of the derivative - that is, it can change faster or slower.

The derivative of a function is denoted by .

Let's show how to find using the graph.

A graph of some function is drawn. Take a point on it with an abscissa. Draw a tangent to the graph of the function at this point. We want to evaluate how steeply the graph of the function goes up. A handy value for this is tangent of the slope of the tangent.

The derivative of a function at a point is equal to the tangent of the slope of the tangent drawn to the graph of the function at that point.

Please note - as the angle of inclination of the tangent, we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what is the tangent to the graph of a function. This is a straight line that has the only common point with the graph in this section, moreover, as shown in our figure. It looks like a tangent to a circle.

Let's find . We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite leg to the adjacent one. From triangle:

We found the derivative using the graph without even knowing the formula of the function. Such tasks are often found in the exam in mathematics under the number.

There is another important correlation. Recall that the straight line is given by the equation

The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. It expresses the geometric meaning of the derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the slope of the tangent.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas, and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point, the function is increasing. The tangent to the graph, drawn at the point, forms an acute angle; with positive axis direction. So the derivative is positive at the point.

At the point, our function is decreasing. The tangent at this point forms an obtuse angle; with positive axis direction. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If a function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

And what will happen at the maximum and minimum points? We see that at (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the slope of the tangent at these points is zero, and the derivative is also zero.

The point is the maximum point. At this point, the increase of the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from "plus" to "minus".

At the point - the minimum point - the derivative is also equal to zero, but its sign changes from "minus" to "plus".

Conclusion: with the help of the derivative, you can find out everything that interests us about the behavior of the function.

If the derivative is positive, then the function is increasing.

If the derivative is negative, then the function is decreasing.

At the maximum point, the derivative is zero and changes sign from plus to minus.

At the minimum point, the derivative is also zero and changes sign from minus to plus.

We write these findings in the form of a table:

Let's make two small clarifications. You will need one of them when solving the problem. Another - in the first year, with a more serious study of functions and derivatives.

A case is possible when the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it has remained positive as it was.

It also happens that at the point of maximum or minimum, the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

But how to find the derivative if the function is given not by a graph, but by a formula? In this case, it applies