Master class "Derivative of a function in the exam" material for preparing for the exam (gia) in algebra (grade 11) on the topic. Derivative How to solve the graphs of the derivative of a function

This section contains the problems of the exam in mathematics on topics related to the study of functions and their derivatives.

Demo versions Unified State Exam 2020 year they can meet under the number 14 for basic level and under number 7 for the profile level.

Take a close look at these three function graphs.
Have you noticed that these functions are in some sense "related"?
For example, in those areas where the graph of the green function is located above zero, the red function increases. In those areas where the graph of the green function is below zero, the red function decreases.
Similar remarks can be made for the red and blue graphs.
You can also notice that the zeros of the green function (dots x = −1 and x = 3) coincide with the extremum points of the red graph: at x = −1 on the red chart we see a local maximum, at NS = 3 on the red chart, the local minimum.
It is easy to see that the local highs and lows of the blue chart are reached at the same points where the red chart passes through the value y = 0.
Several more conclusions can be drawn about the peculiarities of the behavior of these graphs, because they are really related to each other. Look at the function formulas located under each of the graphs, and by means of calculations, make sure that each previous one is a derivative for the next one and, accordingly, each next one is one of the pre-forms of the previous function.

φ 1 (x ) = φ" 2 (x ) φ 2 (x ) = Φ 1 (x )
φ 2 (x ) = φ" 3 (x ) φ 3 (x ) = Φ 2 (x )

Let's remember what we know about the derivative:

Derivative of a function y = f(x) at the point NS expresses the rate of change of the function at the point x.

The physical meaning of the derivative lies in the fact that the derivative expresses the rate of the process described by the dependence y = f (x).

The geometric meaning of the derivative lies in the fact that its value at the point under consideration is equal to the slope of the tangent drawn to the graph of the differentiable function at this point.

Now let there be no red graph in the picture. Let us assume that we do not know the function formulas either.

May I ask you something related to the behavior of a function φ 2 (x ) if it is known that it is the derivative of the function φ 3 (x ) and the antiderivative function φ 1 (x )?
Can. And many questions can be answered exactly, because we know that the derivative is a characteristic of the rate of change of a function, so we can judge some of the behavior of one of these functions by looking at the graph of the other.

Before answering the following questions, scroll up the page so that the top figure containing the red graph is hidden. When the answers are given, put it back in to check the result. And only after that see my solution.

Attention: To enhance the teaching effect answers and solutions are loaded separately for each task by sequential pressing of buttons on a yellow background. (When there are a lot of tasks, the buttons may appear with a delay. If the buttons are not visible at all, check if your browser is allowed JavaScript.)

1) Using the graph of the derivative φ" 2 (x ) (in our case, this is a green graph), determine which of the 2 values of the function is greater φ 2 (−3) or φ 2 (−2)?

The graph of the derivative shows that in the segment [−3; −2] its values are strictly positive, which means that the function in this segment only increases, therefore the value of the function at the left end x = −3 is less than its value at the right end x = −2.

Answer: φ 2 (−3) φ 2 (−2)

2) Using the antiderivative graph Φ 2 (x ) (in our case, this is a blue graph), determine which of the 2 values of the function is greater φ 2 (−1) or φ 2 (4)?

The antiderivative graph shows that the point x = −1 is in the increasing region, hence the value of the corresponding derivative is positive. Point x = 4 is in the decreasing region and the value of the corresponding derivative is negative. Since the positive value is greater than the negative one, we conclude that the value of the unknown function, which is precisely the derivative, at point 4 is less than at point −1.

Answer: φ 2 (−1) > φ 2 (4)

There are a lot of similar questions you can ask about the missing schedule, which leads to a large variety of problems with a short answer, built according to the same scheme. Try to solve some of them.

Tasks for determining the characteristics of a graph derivative of a function.

Picture 1.

Figure 2.

Problem 1

y = f (x ) defined on the interval (−10.5; 19). Determine the number of integer points at which the derivative of the function is positive.

The derivative of the function is positive in those areas where the function increases. The figure shows that these are the intervals (−10.5; −7.6), (−1; 8.2) and (15.7; 19). Let's list the whole points inside these intervals: "−10", "- 9", "−8", "0", "1", "2", "3", "4", "5", "6", "7", "8", "16", "17", "18". There are 15 points in total.

Answer: 15

Remarks.
1. When in problems about the graphs of functions it is required to name "points", as a rule, they mean only the values of the argument x , which are the abscissas of the corresponding points located on the graph. The ordinates of these points are the values of the function, they are dependent and can be easily calculated if necessary.
2. When listing the points, we did not take into account the edges of the intervals, since the function at these points does not increase or decrease, but "unfolds". The derivative at such points is neither positive nor negative, it is equal to zero, therefore they are called stationary points. In addition, we do not consider here the boundaries of the domain of definition, because the condition says that this is an interval.

Task 2

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Determine the number of integer points at which the derivative of the function f " (x ) is negative.

The derivative of the function is negative in those areas where the function decreases. The figure shows that these are the intervals (−7.6; −1) and (8.2; 15.7). Integer points within these intervals: "−7", "- 6", "−5", "- 4", "−3", "- 2", "9", "10", "11", "12 "," 13 "," 14 "," 15 ". There are 13 points in total.

Answer: 13

See notes for the previous task.

To solve the following problems, you need to remember one more definition.

The maximum and minimum points of the function are united by a common name - extremum points .

At these points, the derivative of the function is either zero or does not exist ( necessary extremum condition).
However, a necessary condition is a sign, but not a guarantee of the existence of an extremum of a function. A sufficient condition for an extremum is a change in the sign of the derivative: if the derivative at a point changes sign from "+" to "-", then this is the maximum point of the function; if the derivative at a point changes sign from "-" to "+", then this is the minimum point of the function; if the derivative of the function is equal to zero at a point, or does not exist, but the sign of the derivative does not change to the opposite when passing through this point, then the specified point is not the extremum point of the function. This can be an inflection point, a break point, or a break point in the graph of a function.

Problem 3

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Find the number of points at which the tangent to the graph of the function is parallel to the straight line y = 6 or matches it.

Recall that the equation of the line has the form y = kx + b , where k- the coefficient of inclination of this straight line to the axis Ox... In our case k= 0, i.e. straight y = 6 not tilted, but parallel to the axis Ox... This means that the required tangents must also be parallel to the axis Ox and must also have a slope coefficient of 0. Tangents have this property at the extremum points of functions. Therefore, to answer the question, you just need to calculate all the extreme points on the chart. There are 4 of them - two maximum points and two minimum points.

Answer: 4

Problem 4

Functions y = f (x ) defined on the interval (−11; 23). Find the sum of the extremum points of the function on the segment.

On the indicated segment, we see 2 extremum points. The maximum of the function is reached at the point x 1 = 4, minimum at point x 2 = 8.
x 1 + x 2 = 4 + 8 = 12.

Answer: 12

Problem 5

Figure 1 shows the graph of the function y = f (x ) defined on the interval (−10.5; 19). Find the number of points at which the derivative of the function f " (x ) is equal to 0.

The derivative of the function is equal to zero at the extremum points, of which 4 are visible on the graph:
2 points of maximum and 2 points of minimum.

Answer: 4

Tasks for determining the characteristics of a function according to the graph of its derivative.

Picture 1.

Figure 2.

Problem 6

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). At what point of the segment [−6; 2] the function f (x ) takes the largest value.

On the indicated interval, the derivative was nowhere positive, therefore the function did not increase. It decreased or passed through stationary points. Thus, the function reached its greatest value on the left border of the segment: x = −6.

Answer: −6

Comment: The graph of the derivative shows that on the segment [−6; 2] it is equal to zero three times: at the points x = −6, x = −2, x = 2. But at the point x = −2, it did not change sign, which means that there could not be an extremum of the function at this point. Most likely there was an inflection point in the original function graph.

Problem 7

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). At what point of the segment the function takes the smallest value.

On the segment, the derivative is strictly positive, therefore, the function on this segment only increased. Thus, the function reached the smallest value on the left border of the segment: x = 3.

Answer: 3

Problem 8

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of maximum points of the function f (x ) belonging to the segment [−5; 10].

According to the necessary condition for an extremum, the maximum of the function may be at the points where its derivative is zero. On a given segment, these are points: x = −2, x = 2, x = 6, x = 10. But according to the sufficient condition, it will definitely be only in those of them where the sign of the derivative changes from "+" to "-". On the graph of the derivative, we see that of the listed points, only the point is such x = 6.

Answer: 1

Problem 9

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of extremum points of the function f (x ) belonging to the segment.

The extrema of a function can be at those points where its derivative is 0. On a given segment of the derivative graph, we see 5 such points: x = 2, x = 6, x = 10, x = 14, x = 18. But at the point x = 14 the derivative has not changed its sign, therefore it must be excluded from consideration. This leaves 4 points.

Answer: 4

Problem 10

Figure 1 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−10.5; 19). Find the intervals of increasing function f (x ). In the answer, indicate the length of the longest of them.

The intervals of increase of the function coincide with the intervals of positiveness of the derivative. On the graph we see three of them - (−9; −7), (4; 12), (18; 19). The longest of them is the second. Its length l = 12 − 4 = 8.

Answer: 8

Assignment 11

Figure 2 shows the graph f " (x ) - the derivative of the function f (x ) defined on the interval (−11; 23). Find the number of points at which the tangent to the graph of the function f (x ) is parallel to the straight line y = −2x − 11 or matches it.

The slope (aka the tangent of the slope) of a given straight line k = −2. We are interested in parallel or coinciding tangents, i.e. straight lines with the same slope. Based on the geometric meaning of the derivative - the slope of the tangent at the considered point of the graph of the function, we recalculate the points at which the derivative is equal to −2. There are 9 such points in Figure 2. It is convenient to count them by the intersections of the graph and the grid line passing through the value −2 on the axis Oy.

Answer: 9

As you can see, using the same graph, you can ask a wide variety of questions about the behavior of a function and its derivative. Also, the same question can be attributed to the graphs of different functions. Be careful when solving this problem on the exam, and it will seem very easy to you. Other types of problems in this task - on the geometric meaning of the antiderivative - will be discussed in another section.

The derivative of the function $ y = f (x) $ at a given point $ x_0 $ is the limit of the ratio of the increment of the function to the corresponding increment of its argument, provided that the latter tends to zero:

$ f "(x_0) = (lim) ↙ (△ x → 0) (△ f (x_0)) / (△ x) $

Differentiation is the operation of finding a derivative.

Derivative table of some elementary functions

Function	Derivative
$ c $	$0$
$ x $	$1$
$ x ^ n $	$ nx ^ (n-1) $
$ (1) / (x) $	$ - (1) / (x ^ 2) $
$ √x $	$ (1) / (2√x) $
$ e ^ x $	$ e ^ x $
$ lnx $	$ (1) / (x) $
$ sinx $	$ cosx $
$ cosx $	$ -sinx $
$ tgx $	$ (1) / (cos ^ 2x) $
$ ctgx $	$ - (1) / (sin ^ 2x) $

Basic rules for differentiation

1. The derivative of the sum (difference) is equal to the sum (difference) of the derivatives

$ (f (x) ± g (x)) "= f" (x) ± g "(x) $

Find the Derivative of the Function $ f (x) = 3x ^ 5-cosx + (1) / (x) $

The derivative of the sum (difference) is equal to the sum (difference) of the derivatives.

$ f "(x) = (3x ^ 5)" - (cos x) "+ ((1) / (x))" = 15x ^ 4 + sinx - (1) / (x ^ 2) $

2. Derivative of the work

$ (f (x) g (x)) "= f" (x) g (x) + f (x) g (x) "$

Find the Derivative $ f (x) = 4x cosx $

$ f "(x) = (4x)" cosx + 4x (cosx) "= 4 cosx-4x sinx $

3. Derivative of the quotient

$ ((f (x)) / (g (x))) "= (f" (x) g (x) -f (x) g (x) ") / (g ^ 2 (x)) $

Find the Derivative $ f (x) = (5x ^ 5) / (e ^ x) $

$ f "(x) = ((5x ^ 5)" e ^ x-5x ^ 5 (e ^ x) ") / ((e ^ x) ^ 2) = (25x ^ 4 e ^ x- 5x ^ 5 e ^ x) / ((e ^ x) ^ 2) $

4. The derivative of a complex function is equal to the product of the derivative of the outer function by the derivative of the inner function

$ f (g (x)) "= f" (g (x)) g "(x) $

$ f "(x) = cos" (5x) · (5x) "= - sin (5x) · 5 = -5sin (5x) $

The physical meaning of the derivative

If a material point moves rectilinearly and its coordinate changes depending on time according to the law $ x (t) $, then the instantaneous speed of this point is equal to the derivative of the function.

The point moves along the coordinate line according to the law $ x (t) = 1,5t ^ 2-3t + 7 $, where $ x (t) $ is the coordinate at the time $ t $. At what point in time will the speed of the point be equal to $ 12 $?

1. Velocity is the derivative of $ x (t) $, so we find the derivative of the given function

$ v (t) = x "(t) = 1.5 · 2t -3 = 3t -3 $

2. To find at what time moment $ t $ the speed was equal to $ 12 $, compose and solve the equation:

The geometric meaning of the derivative

Recall that the equation of a straight line not parallel to the coordinate axes can be written in the form $ y = kx + b $, where $ k $ is the slope of the straight line. The coefficient $ k $ is equal to the tangent of the inclination angle between the straight line and the positive direction of the $ Ox $ axis.

The derivative of the function $ f (x) $ at the point $ x_0 $ is equal to the slope $ k $ of the tangent to the graph at this point:

Therefore, we can draw up a general equality:

$ f "(x_0) = k = tgα $

In the figure, the tangent to the function $ f (x) $ increases, therefore, the coefficient $ k> 0 $. Since $ k> 0 $, then $ f "(x_0) = tgα> 0 $. The angle $ α $ between the tangent and the positive direction $ Ox $ is acute.

In the figure, the tangent to the function $ f (x) $ decreases; therefore, the coefficient $ k< 0$, следовательно, $f"(x_0) = tgα < 0$. Угол $α$ между касательной и положительным направлением оси $Ох$ тупой.

In the figure, the tangent to the function $ f (x) $ is parallel to the $ Ox $ axis, therefore, the coefficient $ k = 0 $, therefore, $ f "(x_0) = tan α = 0 $. The point $ x_0 $, at which $ f "(x_0) = 0 $, called extreme.

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

The tangent line to the graph increases, therefore, $ f "(x_0) = tg α> 0 $

In order to find $ f "(x_0) $, find the tangent of the inclination angle between the tangent and the positive direction of the $ Ox $ axis. To do this, add the tangent to the triangle $ ABC $.

Find the tangent of the angle $ BAC $. (The tangent of an acute angle in a right triangle is the ratio of the opposite leg to the adjacent leg.)

$ tg BAC = (BC) / (AC) = (3) / (12) = (1) / (4) = 0.25 $

$ f "(x_0) = tg BAC = 0.25 $

Answer: $ 0.25

The derivative is also used to find the intervals of increasing and decreasing functions:

If $ f "(x)> 0 $ in the interval, then the function $ f (x) $ increases in this interval.

If $ f "(x)< 0$ на промежутке, то функция $f(x)$ убывает на этом промежутке.

The figure shows the graph of the function $ y = f (x) $. Find among the points $ x_1, x_2, x_3… x_7 $ those points at which the derivative of the function is negative.

In response, write down the number of points given.

It's very easy to remember.

Well, let's not go far, we will immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:

In our case, the base is a number:

Such a logarithm (that is, a logarithm with a base) is called "natural", and we use a special notation for it: write instead.

What is equal to? Of course, .

The derivative of the natural logarithm is also very simple:

Examples:

Find the derivative of the function.
What is the derivative of the function?

Answers: The exponent and natural logarithm are uniquely simple functions from the point of view of the derivative. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.

Differentiation rules

The rules of what? Again a new term, again ?! ...

Differentiation is the process of finding a derivative.

That's all. How else to call this process in one word? Not a derivation ... The differential of mathematics is called the same increment of a function at. This term comes from the Latin differentia - difference. Here.

When deriving all these rules, we will use two functions, for example, and. We also need formulas for their increments:

There are 5 rules in total.

The constant is moved outside the derivative sign.

If is some constant number (constant), then.

Obviously, this rule also works for the difference:.

Let's prove it. Let, or easier.

Examples.

Find the derivatives of the functions:

at the point;
at the point;
at the point;
at the point.

Solutions:

(the derivative is the same at all points, since it is a linear function, remember?);

Derivative of a work

Everything is the same here: we introduce a new function and find its increment:

Derivative:

Examples:

Find the derivatives of the functions and;
Find the derivative of the function at the point.

Solutions:

Derivative of the exponential function

Now your knowledge is enough to learn how to find the derivative of any exponential function, not just the exponent (have you forgotten what it is?).

So, where is some number.

We already know the derivative of the function, so let's try to cast our function to a new radix:

To do this, we will use a simple rule:. Then:

Well, it worked. Now try to find the derivative, and don't forget that this function is tricky.

Happened?

Here, check yourself:

The formula turned out to be very similar to the derivative of the exponent: as it was, it remains, only a multiplier appeared, which is just a number, but not a variable.

Examples:
Find the derivatives of the functions:

Answers:

This is just a number that cannot be calculated without a calculator, that is, it cannot be written in a simpler form. Therefore, in the answer we leave it in this form.

Note that here is the quotient of two functions, so we apply the corresponding rule of differentiation:

In this example, the product of two functions:

Derivative of a logarithmic function

Here it is similar: you already know the derivative of the natural logarithm:

Therefore, to find an arbitrary one of the logarithm with a different base, for example:

You need to bring this logarithm to the base. How do you change the base of the logarithm? I hope you remember this formula:

Only now, instead of we will write:

The denominator is just a constant (constant number, no variable). The derivative is very simple:

The derivatives of exponential and logarithmic functions are almost never found in the exam, but it will not be superfluous to know them.

Derivative of a complex function.

What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if the logarithm seems difficult to you, read the topic "Logarithms" and everything will pass), but from the point of view of mathematics, the word "difficult" does not mean "difficult".

Imagine a small conveyor belt: two people are sitting and doing some kind of action with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second ties it with a ribbon. It turns out such a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.

Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then we will square the resulting number. So, we are given a number (chocolate bar), I find its cosine (wrapper), and then you square what I have (you tie it with a ribbon). What happened? Function. This is an example of a complex function: when, to find its value, we do the first action directly with the variable, and then another second action with the result of the first.

In other words, a complex function is a function whose argument is another function: .

For our example,.

We may well do the same actions in reverse order: first you square, and then I look for the cosine of the resulting number:. It is easy to guess that the result will almost always be different. An important feature of complex functions: when you change the order of actions, the function changes.

Second example: (same). ...

The action that we do last will be called "External" function, and the action taken first - respectively "Internal" function(these are informal names, I use them only to explain the material in simple language).

Try to determine for yourself which function is external and which is internal:

Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function

What is the first action to take? First, we will calculate the sine, and only then will we raise it to a cube. This means that it is an internal function, but an external one.
And the original function is their composition:.
Internal:; external:.
Examination: .
Internal:; external:.
Examination: .
Internal:; external:.
Examination: .
Internal:; external:.
Examination: .

we change variables and get a function.

Well, now we will extract our chocolate bar - look for a derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:

Another example:

So, let us finally formulate an official rule:

Algorithm for finding the derivative of a complex function:

Everything seems to be simple, right?

Let's check with examples:

Solutions:

1) Internal:;

External:;

2) Internal:;

(just do not try to reduce by now! Nothing can be taken out from under the cosine, remember?)

3) Internal:;

External:;

It is immediately clear that this is a three-level complex function: after all, this is already a complex function in itself, and from it we also extract the root, that is, we perform the third action (we put a chocolate bar in a wrapper and put it in a briefcase with a ribbon). But there is no reason to be afraid: anyway, we will "unpack" this function in the same order as usual: from the end.

That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply all this.

In such cases, it is convenient to number the steps. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's take an example:

The later the action is performed, the more “external” the corresponding function will be. The sequence of actions - as before:

Here nesting is generally 4-level. Let's define a course of action.

1. A radical expression. ...

2. Root. ...

3. Sinus. ...

4. Square. ...

5. Putting everything together:

DERIVATIVE. BRIEFLY ABOUT THE MAIN

Derivative of a function- the ratio of the increment of the function to the increment of the argument with an infinitely small increment of the argument:

Basic derivatives:

Differentiation rules:

The constant is moved outside the derivative sign:

Derivative of the amount:

Derivative of the work:

Derivative of the quotient:

Derivative of a complex function:

Algorithm for finding the derivative of a complex function:

We define the "internal" function, we find its derivative.
We define the "external" function, we find its derivative.
We multiply the results of the first and second points.

\ (\ DeclareMathOperator (\ tg) (tg) \) \ (\ DeclareMathOperator (\ ctg) (ctg) \) \ (\ DeclareMathOperator (\ arctg) (arctg) \) \ (\ DeclareMathOperator (\ arcctg) (arcctg) \)

Content

Content elements

Derivative, tangent, antiderivative, graphs of functions and derivatives.

Derivative Let the function \ (f (x) \) be defined in some neighborhood of the point \ (x_0 \).

Derivative of the function \ (f \) at the point \ (x_0 \) called the limit

\ (f "(x_0) = \ lim_ (x \ rightarrow x_0) \ dfrac (f (x) -f (x_0)) (x-x_0), \)

if this limit exists.

The derivative of a function at a point characterizes the rate of change of this function at a given point.

Derivatives table

Function	Derivative
\ (const \)	$0$
\ (x \)	$1$
\ (x ^ n \)	\ (n \ cdot x ^ (n-1) \)
\ (\ dfrac (1) (x) \)	\ (- \ dfrac (1) (x ^ 2) \)
\ (\ sqrt (x) \)	\ (\ dfrac (1) (2 \ sqrt (x)) \)
\ (e ^ x \)	\ (e ^ x \)
\ (a ^ x \)	\ (a ^ x \ cdot \ ln (a) \)
\ (\ ln (x) \)	\ (\ dfrac (1) (x) \)
\ (\ log_a (x) \)	\ (\ dfrac (1) (x \ ln (a)) \)
\ (\ sin x \)	\ (\ cos x \)
\ (\ cos x \)	\ (- \ sin x \)
\ (\ tg x \)	\ (\ dfrac (1) (\ cos ^ 2 x) \)
\ (\ ctg x \)	\ (- \ dfrac (1) (\ sin ^ 2x) \)

Differentiation rules\ (f \) and \ (g \) - functions depending on the variable \ (x \); \ (c \) is a number.

2) \ ((c \ cdot f) "= c \ cdot f" \)

3) \ ((f + g) "= f" + g "\)

4) \ ((f \ cdot g) "= f" g + g "f \)

5) \ (\ left (\ dfrac (f) (g) \ right) "= \ dfrac (f" g-g "f) (g ^ 2) \)

6) \ (\ left (f \ left (g (x) \ right) \ right) "= f" \ left (g (x) \ right) \ cdot g "(x) \) - derivative of a compound function

The geometric meaning of the derivative Equation of a straight line- not parallel to the axis \ (Oy \) can be written as \ (y = kx + b \). The coefficient \ (k \) in this equation is called slope of the straight line... It is equal to the tangent angle of inclination this straight line.

Angle of inclination of a straight line- the angle between the positive direction of the \ (Ox \) axis and the given straight line, measured in the direction of the positive angles (that is, in the direction of least rotation from the \ (Ox \) axis to the \ (Oy \) axis).

The derivative of the function \ (f (x) \) at the point \ (x_0 \) is equal to the slope of the tangent to the graph of the function at this point: \ (f "(x_0) = \ tg \ alpha. \)

If \ (f "(x_0) = 0 \), then the tangent to the graph of the function \ (f (x) \) at the point \ (x_0 \) is parallel to the \ (Ox \) axis.

Tangent equation

The equation of the tangent to the graph of the function \ (f (x) \) at the point \ (x_0 \):

\ (y = f (x_0) + f "(x_0) (x-x_0) \)

Monotonicity of a function If the derivative of a function is positive at all points of the interval, then the function increases in this interval.

If the derivative of a function is negative at all points of the interval, then the function decreases in this interval.

Minimum, maximum and inflection points positive on negative at this point, then \ (x_0 \) is the maximum point of the function \ (f \).

If the function \ (f \) is continuous at the point \ (x_0 \), and the value of the derivative of this function \ (f "\) changes from negative on positive at this point, then \ (x_0 \) is the minimum point of the function \ (f \).

The points at which the derivative \ (f "\) is zero or does not exist are called critical points function \ (f \).

Interior points of the domain of definition of the function \ (f (x) \), in which \ (f "(x) = 0 \) can be points of minimum, maximum or inflection.

The physical meaning of the derivative If a material point moves rectilinearly and its coordinate changes depending on time according to the law \ (x = x (t) \), then the speed of this point is equal to the derivative of the coordinate with respect to time:

The acceleration of a material point is equal to the derivative of the speed of this point with respect to time:

\ (a (t) = v "(t). \)

Sergey Nikiforov

If the derivative of a function is of constant sign on an interval, and the function itself is continuous on its boundaries, then the boundary points are added to both increasing and decreasing intervals, which fully corresponds to the definition of increasing and decreasing functions.

Farit Yamaev 26.10.2016 18:50

Hello. How (on what basis) can one assert that at the point where the derivative is zero, the function increases. Give reasons. Otherwise, it's just someone's whim. By what theorem? And also the proof. Thanks.

Support

The value of the derivative at a point is not directly related to the increase in the function on the interval. Consider, for example, functions - they all increase on the segment

Vladlen Pisarev 02.11.2016 22:21

If a function increases on the interval (a; b) and is defined and continuous at points a and b, then it increases on the interval. Those. point x = 2 is included in this interval.

Although, as a rule, increasing and decreasing is considered not on a segment, but on an interval.

But at the very point x = 2, the function has a local minimum. And how to explain to children that when they are looking for points of increase (decrease), then the points of local extremum are not counted, but they enter the intervals of increase (decrease).

Considering that the first part of the exam is for the "middle group of kindergarten", then such nuances are probably too much.

Separately, many thanks for the "Solve the Unified State Exam" to all employees - an excellent guide.

Sergey Nikiforov

A simple explanation can be obtained by starting from the definition of an increasing / decreasing function. Let me remind you that it sounds like this: a function is called increasing / decreasing in the interval, if a larger function argument corresponds to a larger / smaller function value. This definition does not use the concept of a derivative in any way, so questions about the points where the derivative vanishes cannot arise.

Irina Ishmakova 20.11.2017 11:46

Good afternoon. Here in the comments I see the belief that borders should be included. Let's say I agree with this. But please look at your solution to problem 7089. There, when specifying ascending intervals, boundaries are not included. And this affects the answer. Those. solutions of tasks 6429 and 7089 contradict each other. Please clarify this situation.

Alexander Ivanov

Items 6429 and 7089 have completely different questions.

In one about the intervals of increasing, and in the other about the intervals with a positive derivative.

There is no contradiction.

The extrema are included in the intervals of increasing and decreasing, but the points at which the derivative is equal to zero are not included in the intervals at which the derivative is positive.

A Z 28.01.2019 19:09

Colleagues, there is a concept of increasing at a point

(see Fichtengolts for example)

and your understanding of increasing at x = 2 is contrary to the classical definition.

Increasing and decreasing is a process and I would like to adhere to this principle.

In any interval that contains the point x = 2, the function is not increasing. Therefore, the inclusion of a given point x = 2 is a special process.

Usually, to avoid confusion, the inclusion of the ends of intervals is spoken of separately.

Alexander Ivanov

The function y = f (x) is called increasing on a certain interval if the larger value of the argument from this interval corresponds to the larger value of the function.

At the point x = 2, the function is differentiable, and on the interval (2; 6) the derivative is positive, hence, on the interval)