How to determine where the value of the derivative is greatest. Function derivative. The geometric meaning of the derivative. Tasks for determining the characteristics of the derivative from the graph of a function
In the interim ( but,b), but X- is a randomly chosen point of the given interval. Let's give an argument X incrementΔx (positive or negative).
The function y \u003d f (x) will receive an increment Δy equal to:
Δy = f(x + Δx)-f(x).
For infinitely small Δх incrementΔy is also infinitely small.
For example:
Consider the solution of the derivative of a function using the example of a free fall of a body.
Since t 2 \u003d t l + Δt, then
.
Calculating the limit, we find:
The notation t 1 is introduced to emphasize the constancy of t when calculating the limit of a function. Since t 1 is an arbitrary value of time, index 1 can be dropped; then we get:
It can be seen that the speed v, like the way s, eat function time. Function type v depends entirely on the type of function s, so the function s sort of "produces" a function v. Hence the name " derivative function».
Consider another example.
Find the value of the derivative of a function:
y = x 2 at x = 7.
Solution. At x = 7 we have y=7 2=49. Let's give an argument X increment Δ X. The argument becomes 7 + Δ X, and the function will get the value (7 + Δ x) 2.
There are new tasks. Let's take a look at their solution.
Job prototype B8 (#317543)
The figure shows a graph of the function y \u003d f (x) and points -2, -1, 1, 2 are marked. At which of these points is the value of the derivative the largest? Please indicate this point in your answer.
As we know it's called
limit of the ratio of function increment to argument increment when the argument increment goes to zero:
The derivative at a point shows function change rate at this point. The faster the function changes, that is, the greater the increment of the function, the greater the slope of the tangent. Since the task requires determining the point at which the value of the derivative is greatest, we exclude from consideration the points with abscissas -1 and 1 - at these points the function decreases, and the derivative is negative at them.
The function increases at points -2 and 2. However, it increases at them in different ways - at point -2, the graph of the function rises steeper than at point 2, and therefore, the increment of the function at this point, and hence the derivative, is greater.
Answer: -2
And a similar task:
Job prototype B8 (#317544)
The figure shows a function graph and points -2, -1, 1, 4 are marked. At which of these points is the value of the derivative the smallest? Please indicate this point in your answer.
The solution of this problem is similar to the solution of the previous one "exactly the opposite"
We are interested in the point at which the derivative takes the smallest value, that is, we are looking for the point at which the function decreases most rapidly - on the graph, this is the point at which the steepest "descent". This is the point with abscissa 4.
Showing the relationship of the sign of the derivative with the nature of the monotonicity of the function.
Please be extremely careful in the following. Look, the schedule of WHAT is given to you! Function or its derivative
Given a graph of the derivative, then we are only interested in function signs and zeros. No "knolls" and "hollows" are of interest to us in principle!
Task 1.
The figure shows a graph of a function defined on an interval. Determine the number of integer points where the derivative of the function is negative.
Solution:
In the figure, the areas of decreasing function are highlighted in color:
4 integer values fall into these areas of decreasing function.
Task 2.
The figure shows a graph of a function defined on an interval. Find the number of points where the tangent to the graph of the function is parallel or coincident with the line.
Solution:
Since the tangent to the function graph is parallel (or coincides) with a straight line (or, which is the same, ) having slope, equal to zero, then the tangent has a slope .
This in turn means that the tangent is parallel to the axis, since the slope is the tangent of the angle of inclination of the tangent to the axis.
Therefore, we find extremum points on the graph (maximum and minimum points), - it is in them that the functions tangent to the graph will be parallel to the axis.
There are 4 such points.
Task 3.
The figure shows a graph of the derivative of a function defined on the interval . Find the number of points where the tangent to the graph of the function is parallel or coincident with the line.
Solution:
Since the tangent to the graph of the function is parallel (or coincides) with a straight line, which has a slope, then the tangent has a slope.
This in turn means that at the points of contact.
Therefore, we look at how many points on the graph have an ordinate equal to .
As you can see, there are four such points.
Task 4.
The figure shows a graph of a function defined on an interval. Find the number of points where the derivative of the function is 0.
Solution:
The derivative is zero at the extremum points. We have 4 of them:
Task 5.
The figure shows a function graph and eleven points on the x-axis:. At how many of these points is the derivative of the function negative?
Solution:
On intervals of decreasing function, its derivative takes negative values. And the function decreases at points. There are 4 such points.
Task 6.
The figure shows a graph of a function defined on an interval. Find the sum of the extremum points of the function .
Solution:
extremum points are the maximum points (-3, -1, 1) and the minimum points (-2, 0, 3).
The sum of extreme points: -3-1+1-2+0+3=-2.
Task 7.
The figure shows a graph of the derivative of a function defined on the interval . Find the intervals of increasing function . In your answer, indicate the sum of integer points included in these intervals.
Solution:
The figure highlights the intervals on which the derivative of the function is non-negative.
There are no integer points on the small interval of increase, on the interval of increase there are four integer values: , , and .
Their sum:
Task 8.
The figure shows a graph of the derivative of a function defined on the interval . Find the intervals of increasing function . In your answer, write the length of the largest of them.
Solution:
In the figure, all the intervals on which the derivative is positive are highlighted, which means that the function itself increases on these intervals.
The length of the largest of them is 6.
Task 9.
The figure shows a graph of the derivative of a function defined on the interval . At what point on the segment does it take the greatest value.
Solution:
We look at how the graph behaves on the segment, namely, we are interested in derivative sign only .
The sign of the derivative on is minus, since the graph on this segment is below the axis.
Hello! Let's hit the approaching USE with high-quality systematic training, and perseverance in grinding the granite of science !!! INAt the end of the post there is a competitive task, be the first! In one of the articles in this section, we are with you, in which the graph of the function was given, and various questions were raised regarding extrema, intervals of increase (decrease) and others.
In this article, we will consider the tasks included in the USE in mathematics, in which the graph of the derivative of a function is given, and the following questions are posed:
1. At what point of a given segment does the function take on the largest (or smallest) value.
2. Find the number of maximum (or minimum) points of the function that belong to a given segment.
3. Find the number of extremum points of the function that belong to a given segment.
4. Find the extremum point of the function that belongs to the given segment.
5. Find intervals of increase (or decrease) of the function and in the answer indicate the sum of integer points included in these intervals.
6. Find intervals of increase (or decrease) of the function. In your answer, indicate the length of the largest of these intervals.
7. Find the number of points where the tangent to the graph of the function is parallel to the straight line y = kx + b or coincides with it.
8. Find the abscissa of the point at which the tangent to the graph of the function is parallel to the abscissa axis or coincides with it.
There may be other questions, but they will not cause you any difficulties if you understand and (links are provided to articles that provide the information necessary for solving, I recommend repeating).
Basic information (briefly):
1. The derivative on increasing intervals has a positive sign.
If the derivative at a certain point from some interval has a positive value, then the graph of the function on this interval increases.
2. On the intervals of decreasing, the derivative has a negative sign.
If the derivative at a certain point from some interval has a negative value, then the graph of the function on this interval decreases.
3. The derivative at the point x is equal to the slope of the tangent drawn to the graph of the function at the same point.
4. At the points of extremum (maximum-minimum) of the function, the derivative is equal to zero. The tangent to the graph of the function at this point is parallel to the x-axis.
This needs to be clearly understood and remembered!!!
The graph of the derivative "confuses" many people. Some inadvertently take it for the graph of the function itself. Therefore, in such buildings, where you see that a graph is given, immediately focus your attention in the condition on what is given: a graph of a function or a graph of a derivative of a function?
If it is a graph of the derivative of a function, then treat it like a "reflection" of the function itself, which simply gives you information about this function.
Consider the task:
The figure shows a graph y=f'(X)- derivative function f(X), defined on the interval (–2;21).
We will answer the following questions:
1. At what point of the segment is the function f(X) takes on the largest value.
On a given segment, the derivative of the function is negative, which means that the function decreases on this segment (it decreases from the left boundary of the interval to the right). Thus, the maximum value of the function is reached on the left boundary of the segment, i.e., at point 7.
Answer: 7
2. At what point of the segment is the function f(X)
From this graph of the derivative, we can say the following. On a given segment, the derivative of the function is positive, which means that the function increases on this segment (it increases from the left border of the interval to the right one). Thus, the smallest value of the function is reached on the left border of the segment, that is, at the point x = 3.
Answer: 3
3. Find the number of maximum points of the function f(X)
The maximum points correspond to the points where the sign of the derivative changes from positive to negative. Consider where the sign changes in this way.
On the segment (3;6) the derivative is positive, on the segment (6;16) it is negative.
On the segment (16;18) the derivative is positive, on the segment (18;20) it is negative.
Thus, on a given segment, the function has two maximum points x = 6 and x = 18.
Answer: 2
4. Find the number of minimum points of the function f(X) belonging to the segment .
The minimum points correspond to the points where the sign of the derivative changes from negative to positive. We have a negative derivative on the interval (0; 3), and positive on the interval (3; 4).
Thus, on the segment, the function has only one minimum point x = 3.
*Be careful when writing the answer - the number of points is recorded, not the x value, such a mistake can be made due to inattention.
Answer: 1
5. Find the number of extremum points of the function f(X) belonging to the segment .
Please note that you need to find number extremum points (these are both maximum and minimum points).
The extremum points correspond to the points where the sign of the derivative changes (from positive to negative or vice versa). On the graph given in the condition, these are the zeros of the function. The derivative vanishes at points 3, 6, 16, 18.
Thus, the function has 4 extremum points on the segment.
Answer: 4
6. Find the intervals of increasing function f(X)
Intervals of increase of this function f(X) correspond to the intervals on which its derivative is positive, that is, the intervals (3;6) and (16;18). Please note that the boundaries of the interval are not included in it (round brackets - boundaries are not included in the interval, square brackets are included). These intervals contain integer points 4, 5, 17. Their sum is: 4 + 5 + 17 = 26
Answer: 26
7. Find the intervals of decreasing function f(X) on the given interval. In your answer, indicate the sum of integer points included in these intervals.
Function Decreasing Intervals f(X) correspond to intervals on which the derivative of the function is negative. In this problem, these are the intervals (–2;3), (6;16), (18;21).
These intervals contain the following integer points: -1, 0, 1, 2, 7, 8, 9, 10, 11, 12, 13, 14, 15, 19, 20. Their sum is:
(–1) + 0 + 1 + 2 + 7 + 8 + 9 + 10 +
11 + 12 + 13 + 14 + 15 + 19 + 20 = 140
Answer: 140
*Pay attention in the condition: whether the boundaries are included in the interval or not. If the boundaries are included, then these boundaries must also be taken into account in the intervals considered in the solution process.
8. Find the intervals of increasing function f(X)
Function increase intervals f(X) correspond to the intervals on which the derivative of the function is positive. We have already indicated them: (3;6) and (16;18). The largest of them is the interval (3;6), its length is 3.
Answer: 3
9. Find the intervals of decreasing function f(X). In your answer, write the length of the largest of them.
Function Decreasing Intervals f(X) correspond to intervals on which the derivative of the function is negative. We have already indicated them, these are the intervals (–2; 3), (6; 16), (18; 21), their lengths are respectively equal to 5, 10, 3.
The length of the largest is 10.
Answer: 10
10. Find the number of points where the tangent to the graph of the function f(X) parallel to the line y \u003d 2x + 3 or coincides with it.
The value of the derivative at the point of contact is equal to the slope of the tangent. Since the tangent is parallel to the straight line y \u003d 2x + 3 or coincides with it, then their slopes are equal to 2. Therefore, it is necessary to find the number of points at which y (x 0) \u003d 2. Geometrically, this corresponds to the number of intersection points of the derivative graph with the straight line y = 2. There are 4 such points on this interval.
Answer: 4
11. Find the extremum point of the function f(X) belonging to the segment .
An extremum point of a function is a point at which its derivative is equal to zero, and in the vicinity of this point, the derivative changes sign (from positive to negative or vice versa). On the segment, the graph of the derivative crosses the x-axis, the derivative changes sign from negative to positive. Therefore, the point x = 3 is an extremum point.
Answer: 3
12. Find the abscissas of the points where the tangents to the graph y \u003d f (x) are parallel to the abscissa axis or coincide with it. In your answer, indicate the largest of them.
The tangent to the graph y \u003d f (x) can be parallel to the x-axis or coincide with it, only at points where the derivative is zero (these can be extremum points or stationary points, in the vicinity of which the derivative does not change its sign). This graph shows that the derivative is zero at points 3, 6, 16,18. The largest is 18.
The argument can be structured like this:
The value of the derivative at the point of contact is equal to the slope of the tangent. Since the tangent is parallel or coincident with the x-axis, its slope is 0 (indeed, the tangent of an angle of zero degrees is zero). Therefore, we are looking for a point at which the slope is equal to zero, which means that the derivative is equal to zero. The derivative is equal to zero at the point where its graph crosses the x-axis, and these are points 3, 6, 16,18.
Answer: 18
The figure shows a graph y=f'(X)- derivative function f(X) defined on the interval (–8;4). At what point of the segment [–7;–3] is the function f(X) takes the smallest value.
The figure shows a graph y=f'(X)- derivative function f(X), defined on the interval (–7;14). Find the number of maximum points of a function f(X) belonging to the segment [–6;9].
The figure shows a graph y=f'(X)- derivative function f(X) defined on the interval (–18;6). Find the number of minimum points of a function f(X) belonging to the segment [–13;1].
The figure shows a graph y=f'(X)- derivative function f(X), defined on the interval (–11; –11). Find the number of extremum points of a function f(X), belonging to the segment [–10; -10].
The figure shows a graph y=f'(X)- derivative function f(X) defined on the interval (–7;4). Find the intervals of increasing function f(X). In your answer, indicate the sum of integer points included in these intervals.
The figure shows a graph y=f'(X)- derivative function f(X), defined on the interval (–5; 7). Find the intervals of decreasing function f(X). In your answer, indicate the sum of integer points included in these intervals.
The figure shows a graph y=f'(X)- derivative function f(X) defined on the interval (–11;3). Find the intervals of increasing function f(X). In your answer, write the length of the largest of them.
F The figure shows a graph
The condition of the problem is the same (which we considered). Find the sum of three numbers:
1. The sum of the squares of the extrema of the function f (x).
2. The difference of the squares of the sum of the maximum points and the sum of the minimum points of the function f (x).
3. The number of tangents to f (x) parallel to the straight line y \u003d -3x + 5.
The first one to give the correct answer will receive an incentive prize - 150 rubles. Write your answers in the comments. If this is your first comment on the blog, then it will not appear immediately, a little later (do not worry, the time of writing a comment is recorded).
Good luck to you!
Sincerely, Alexander Krutitsikh.
P.S: I would be grateful if you tell about the site in social networks.
Dear friends! The group of tasks related to the derivative includes tasks - in the condition, the graph of the function is given, several points on this graph and the question is:
At what point is the value of the derivative the largest (smallest)?
Let's briefly repeat:
The derivative at the point is equal to the slope of the tangent passing throughthis point on the graph.
Atthe global coefficient of the tangent in turn equal to tangent the slope of this tangent.
*This refers to the angle between the tangent and the x-axis.
1. On intervals of increasing function, the derivative has a positive value.
2. On the intervals of its decrease, the derivative has a negative value.
Consider the following sketch:
At points 1,2,4, the derivative of the function has a negative value, since these points belong to the decreasing intervals.
At points 3,5,6, the derivative of the function has a positive value, since these points belong to the intervals of increase.
As you can see, everything is clear with the value of the derivative, that is, it is not difficult to determine what sign it has (positive or negative) at a certain point on the graph.
Moreover, if we mentally construct tangents at these points, we will see that the lines passing through points 3, 5 and 6 form angles with the oX axis lying in the range from 0 to 90 °, and the lines passing through points 1, 2 and 4 form with the oX axis, angles ranging from 90 o to 180 o.
* The relationship is clear: the tangents passing through the points belonging to the intervals of increasing functions form with the oX axis sharp corners, the tangents passing through the points belonging to the intervals of decreasing functions form obtuse angles with the oX axis.
Now the important question!
How does the value of the derivative change? After all, the tangent different points graphics continuous function forms different angles, depending on which point on the graph it passes through.
*Or, speaking plain language, the tangent is located, as it were, “more horizontally” or “more vertically”. Look:
Straight lines form angles with the oX axis ranging from 0 to 90 o
Straight lines form angles with the oX axis ranging from 90 o to 180 o
So if there are any questions:
- at which of the given points on the graph does the value of the derivative have the smallest value?
- at which of the given points on the graph does the value of the derivative have the greatest value?
then for the answer it is necessary to understand how the value of the tangent of the angle of the tangent changes in the range from 0 to 180 o.
*As already mentioned, the value of the derivative of the function at a point is equal to the tangent of the slope of the tangent to the x-axis.
The tangent value changes as follows:
When the slope of the straight line changes from 0 o to 90 o, the value of the tangent, and hence the derivative, changes from 0 to +∞, respectively;
When the slope of the straight line changes from 90 o to 180 o, the value of the tangent, and hence the derivative, changes accordingly –∞ to 0.
This can be clearly seen from the graph of the tangent function:
In simple terms:
When the angle of inclination of the tangent is from 0 o to 90 o
The closer it is to 0 o, the greater the value of the derivative will be close to zero (on the positive side).
The closer the angle is to 90°, the more the value of the derivative will increase towards +∞.
When the angle of inclination of the tangent is from 90 o to 180 o
The closer it is to 90 o, the more the value of the derivative will decrease towards –∞.
The closer the angle is to 180 o, the greater the value of the derivative will be close to zero (on the negative side).
317543. The figure shows a graph of the function y = f(x) and marked points–2, –1, 1, 2. At which of these points is the value of the derivative greatest? Please indicate this point in your answer.
We have four points: two of them belong to the intervals on which the function decreases (these are points –1 and 1) and two to the intervals on which the function increases (these are points –2 and 2).
We can immediately conclude that at points -1 and 1 the derivative has a negative value, at points -2 and 2 it has a positive value. Therefore, in this case it is necessary to analyze points -2 and 2 and determine which of them will have the largest value. Let's construct tangents passing through the indicated points:
The value of the tangent of the angle between line a and the abscissa axis will be greater than the value of the tangent of the angle between line b and this axis. This means that the value of the derivative at the point -2 will be the largest.
Let's answer the following question: at which of the points -2, -1, 1 or 2 is the value of the derivative the largest negative? Please indicate this point in your answer.
The derivative will have a negative value at the points belonging to the decreasing intervals, so consider the points -2 and 1. Let's construct the tangents passing through them:
We see that obtuse angle between the line b and the axis oX is "closer" to 180 about , so its tangent will be greater than the tangent of the angle formed by the straight line a and the x-axis.
Thus, at the point x = 1, the value of the derivative will be the largest negative.
317544. The figure shows a graph of the function y = f(x) and marked points–2, –1, 1, 4. At which of these points is the value of the derivative the smallest? Please indicate this point in your answer.
We have four points: two of them belong to the intervals on which the function decreases (these are points -1 and 4) and two to the intervals on which the function increases (these are points -2 and 1).
We can immediately conclude that at points -1 and 4 the derivative has a negative value, at points -2 and 1 it has a positive value. Therefore, in this case, it is necessary to analyze points –1 and 4 and determine which of them will have the smallest value. Let's construct tangents passing through the indicated points:
The value of the tangent of the angle between line a and the abscissa axis will be greater than the value of the tangent of the angle between line b and this axis. This means that the value of the derivative at the point x = 4 will be the smallest.
Answer: 4
I hope I didn't "overload" you with the amount of writing. In fact, everything is very simple, one has only to understand the properties of the derivative, its geometric meaning and how the value of the tangent of the angle changes from 0 to 180 o.
1. First, determine the signs of the derivative at these points (+ or -) and select the necessary points (depending on the question posed).
2. Construct tangents at these points.
3. Using the tangesoid plot, schematically mark the corners and displayAlexander.
P.S: I would be grateful if you tell about the site in social networks.