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.
Decision:
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.
Decision:
Since the tangent to the function graph is parallel (or coincides) with a straight line (or, which is the same, ) having slope , 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.
Decision:
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.
Decision:
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?
Decision:
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 .
Decision:
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.
Decision:
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.
Decision:
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.
Decision:
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, friends! In this article, we will consider tasks for the primitive. These tasks are included in the exam in mathematics. Despite the fact that the sections themselves - differentiation and integration are quite capacious in the course of algebra and require a responsible approach to understanding, but the tasks themselves, which are included in open bank assignments in mathematics and will be extremely simple on the exam and are solved in one or two steps.
It is important to understand the essence of the antiderivative and, in particular, the geometric meaning of the integral. Consider briefly the theoretical foundations.
The geometric meaning of the integral
Briefly about the integral, we can say this: the integral is the area.
Definition: Let on coordinate plane given a graph of a positive function f given on the segment . subplot (or curvilinear trapezoid) is a figure bounded by the graph of the function f, the straight lines x \u003d a and x \u003d b and the x-axis.
Definition: Let Dana positive function f defined on the finite segment . The integral of a function f on a segment is the area of its subgraph.
As already mentioned, F (x) = f (x).What can we conclude?
He is simple. We need to determine how many points there are on this graph at which F′(x) = 0. We know that at those points where the tangent to the graph of the function is parallel to the x-axis. Let's show these points on the interval [–2;4]:
These are the extremum points of the given function F(x). There are ten of them.
Answer: 10
323078. The figure shows a graph of some function y = f (x) (two rays with a common starting point). Using the figure, calculate F(8) – F(2), where F(x) is one of the antiderivatives of f(x).
Let's rewrite the Newton-Leibniz theorem:Let f given function, F is its arbitrary antiderivative. Then
And this, as already mentioned, is the area of the subgraph of the function.
Thus, the task is reduced to finding the area of the trapezoid (interval from 2 to 8):
It is not difficult to calculate it by cells. We get 7. The sign is positive, since the figure is located above the x-axis (or in the positive half-plane of the y-axis).
Also in this case one could say this: the difference in the values of the antiderivatives at the points is the area of the figure.
Answer: 7
323079. The figure shows a graph of some function y = f (x). The function F (x) \u003d x 3 +30x 2 +302x–1.875 is one of the antiderivatives of the function y \u003d f (x). Find the area of the shaded figure.
As already mentioned about geometric sense integral, this is the area of \u200b\u200bthe figure bounded by the graph of the function f (x), the straight lines x \u003d a and x \u003d b and the axis ox.
Theorem (Newton–Leibniz):
Thus, the problem is reduced to calculating definite integral of this function on the interval from -11 to -9, or in other words, we need to find the difference between the values of the antiderivatives calculated at the indicated points:
Answer: 6
323080. The figure shows a graph of some function y = f (x).
The function F (x) = –x 3 –27x 2 –240x– 8 is one of the antiderivatives of the function f (x). Find the area of the shaded figure.
Theorem (Newton–Leibniz):
The task is reduced to calculating the definite integral of this function over the interval from –10 to –8:
Answer: 4
Another solution to this problem, from the site.
Derivatives and differentiation rules are still in. It is necessary to know them, not only for solving such tasks.
You can also see background information on the website and
Watch a short video, this is an excerpt from the movie "The Blind Side". We can say that this is a film about studies, about mercy, about the importance of supposedly “random” meetings in our lives ... But these words will not be enough, I recommend watching the film itself, I highly recommend it.
I wish you success!
Sincerely, Alexander Krutitskikh
P.S: I would be grateful if you tell about the site in social networks.
Job type: 7
Subject: An antiderivative of a function
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight line segments). Using the figure, compute F(9)-F(5), where F(x) is one of the antiderivatives of f(x).
Show SolutionDecision
According to the Newton-Leibniz formula, the difference F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid bounded by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. According to the graph, we determine that the specified curvilinear trapezoid is a trapezoid with bases equal to 4 and 3 and a height of 3.
Its area is equal to \frac(4+3)(2)\cdot 3=10.5.
Answer
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=F(x) - one of the antiderivatives of some function f(x) defined on the interval (-5; 5). Using the figure, determine the number of solutions to the equation f(x)=0 on the interval [-3; 4].
Decision
According to the definition of the antiderivative, the equality holds: F "(x) \u003d f (x). Therefore, the equation f (x) \u003d 0 can be written as F "(x) \u003d 0. Since the figure shows the graph of the function y=F(x), we need to find those interval points [-3; 4], in which the derivative of the function F(x) is equal to zero. It can be seen from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 7 of them on the indicated interval (four minimum points and three maximum points).
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight line segments). Using the figure, compute F(5)-F(0), where F(x) is one of the antiderivatives of f(x).
Decision
According to the Newton-Leibniz formula, the difference F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid bounded by the graph of the function y=f(x), straight lines y=0 , x=5 and x=0. According to the graph, we determine that the specified curvilinear trapezoid is a trapezoid with bases equal to 5 and 3 and a height of 3.
Its area is equal to \frac(5+3)(2)\cdot 3=12.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=F(x) — one of the antiderivatives of some function f(x), defined on the interval (-5; 4). Using the figure, determine the number of solutions to the equation f (x) = 0 on the segment (-3; 3].
Decision
According to the definition of the antiderivative, the equality holds: F "(x) \u003d f (x). Therefore, the equation f (x) \u003d 0 can be written as F "(x) \u003d 0. Since the figure shows the graph of the function y=F(x), we need to find those interval points [-3; 3], in which the derivative of the function F(x) is equal to zero.
It can be seen from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 5 of them on the specified interval (two minimum points and three maximum points).
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of some function y=f(x). The function F(x)=-x^3+4.5x^2-7 is one of the antiderivatives of the function f(x).
Find the area of the shaded figure.
Decision
The shaded figure is a curvilinear trapezoid bounded from above by the graph of the function y=f(x), the straight lines y=0, x=1 and x=3. According to the Newton-Leibniz formula, its area S is equal to the difference F(3)-F(1), where F(x) is the antiderivative of the function f(x) specified in the condition. So S= F(3)-F(1)= -3^3 +(4,5)\cdot 3^2 -7-(-1^3 +(4,5)\cdot 1^2 -7)= 6,5-(-3,5)= 10.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of some function y=f(x). The function F(x)=x^3+6x^2+13x-5 is one of the antiderivatives of the function f(x). Find the area of the shaded figure.
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=\frac(2)(3)x^3-20x^2+201x-\frac(5)(9)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323383. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(4)(9)x^3-\frac(34)(3)x^2-\frac(280)(3)x-\frac(18)(5 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323385. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(1)(6)x^3-\frac(17)(4)x^2-35x-\frac(5)(11)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323387. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(1)(5)x^3-\frac(9)(2)x^2-30x-\frac(11)(8)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323389. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(11)(30)x^3-\frac(33)(4)x^2-\frac(297)(5)x-\frac(1)(2 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323391. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(7)(27)x^3-\frac(35)(6)x^2-42x-\frac(7)(4)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323393. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(1)(4)x^3-\frac(21)(4)x^2-\frac(135)(4)x-\frac(13)(2 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323395. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-x^3-21x^2-144x-\frac(11)(4)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323397. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). The function \(F(x)=-\frac(5)(8)x^3-\frac(105)(8)x^2-90x-\frac(1)(2)\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
Task #:
323399. Prototype No.:
The figure shows a graph of some function \(y=f(x)\). Function \(F(x)=-\frac(1)(10)x^3-\frac(21)(10)x^2-\frac(72)(5)x-\frac(4)(3 )\) is one of the antiderivatives of the function \(f(x)\). Find the area of the shaded figure.
Answer:
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