Relationship between the graph of a function and the graph of an antiderivative.

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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 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

Rules of differentiation\(f\) and \(g\) are 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 complex function

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

Straight angle- the angle between the positive direction of the \(Ox\) axis and this straight line, measured in the direction of the positive angles (that is, in the direction of the smallest 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 axis \(Ox\).

Tangent equation

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

If the derivative of a function is negative at all points of the interval, then the function decreases on 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 with 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 equal to zero or does not exist are called critical points functions \(f\).

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

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:

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

51. The figure shows a graph y=f "(x)- derivative of a function f(x), defined on the interval (− 4; 6). Find the abscissa of the point at which the tangent to the graph of the function y=f(x) parallel to the line y=3x or coincides with it.

Answer: 5

52. The figure shows a graph y=F(x) f(x) f(x) positive?

Answer: 7

53. The figure shows a graph y=F(x) one of the antiderivatives of some function f(x) and eight points are marked on the x-axis: x1, x2, x3, x4, x5, x6, x7, x8. At how many of these points is the function f(x) negative?

Answer: 3

54. The figure shows a graph y=F(x) one of the antiderivatives of some function f(x) and ten points are marked on the x-axis: x1, x2, x3, x4, x5, x6, x7, x8, x9, x10. At how many of these points is the function f(x) positive?

Answer: 6

55. The figure shows a graph y=F(x f(x), defined on the interval (− 7; 5). Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [− 5; 2].

Answer: 3

56. The figure shows a graph y=F(x) one of the antiderivatives of some function f (x), defined on the interval (− 8; 7). Using the figure, determine the number of solutions to the equation f(x)= 0 on the interval [− 5; 5].

Answer: 4

57. The figure shows a graph y=F(x) one of the antiderivatives of some function f(x), defined on the interval (1;13). Using the figure, determine the number of solutions to the equation f (x)=0 on the segment .

Answer: 4

58. The figure shows a graph of a certain function y=f(x)(two rays with a common starting point). Using the figure, calculate F(−1)−F(−8), Where F(x) f(x).

Answer: 20

59. The figure shows a graph of a certain function y=f(x) (two rays with a common starting point). Using the figure, calculate F(−1)−F(−9), Where F(x)- one of antiderivative functions f(x).

Answer: 24

60. The figure shows a graph of a certain function y=f(x). Function

-one of the primitive functions f(x). Find the area of ​​the shaded figure.

Answer: 6

61. The figure shows a graph of a certain function y=f(x). Function

One of the primitive functions f(x). Find the area of ​​the shaded figure.

Answer: 14.5

parallel to the tangent to the graph of the function

Answer:0.5

Find the abscissa of the tangent point.

Answer: -1

is tangent to the graph of the function

Find c.

Answer: 20

is tangent to the graph of the function

Find a.

Answer:0.125

is tangent to the graph of the function

Find b, taking into account that the abscissa of the tangent point is greater than 0.

Answer: -33

67. Material point moves in a straight line according to the law

Where x t- time in seconds, measured from the moment the movement began. At what point in time (in seconds) was its speed equal to 96 m/s?

Answer: 18

68. A material point moves rectilinearly according to the law

Where x- distance from the reference point in meters, t- time in seconds, measured from the moment the movement began. At what point in time (in seconds) was its speed equal to 48 m/s?

Answer: 9

69. A material point moves rectilinearly according to the law

Where x t t=6 With.

Answer: 20

70. A material point moves rectilinearly according to the law

Where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. Find its speed (in m/s) at the moment of time t=3 With.

Answer: 59

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

Solution

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

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

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

Answer

Condition

The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight segments). Using the figure, calculate F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x).

Solution

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 limited by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and height 3.

Its area is equal \frac(4+3)(2)\cdot 3=10.5.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level" Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of y=f"(x) - the derivative of the function f(x), defined on the interval (-4; 10). Find the intervals of decreasing function f(x). In your answer, indicate the length of the largest of them.

Solution

As is known, the function f(x) decreases on those intervals at each point of which the derivative f"(x) is less than zero. Considering that it is necessary to find the length of the largest of them, three such intervals are naturally distinguished from the figure: (-4; -2) ; (0; 3); (5; 9).

The length of the largest of them - (5; 9) is 4.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of y=f"(x) - the derivative of the function f(x), defined on the interval (-8; 7). Find the number of maximum points of the function f(x) belonging to the interval [-6; -2].

Solution

The graph shows that the derivative f"(x) of the function f(x) changes sign from plus to minus (at such points there will be a maximum) at exactly one point (between -5 and -4) from the interval [-6; -2 ] Therefore, on the interval [-6; -2] there is exactly one maximum point.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of the function y=f(x), defined on the interval (-2; 8). Determine the number of points at which the derivative of the function f(x) is equal to 0.

Solution

The equality of the derivative at a point to zero means that the tangent to the graph of the function drawn at this point is parallel to the Ox axis. Therefore, we find points at which the tangent to the graph of the function is parallel to the Ox axis. On this chart such points are extremum points (maximum or minimum points). As you can see, there are 5 extremum points.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

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

Solution

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

We get: x_0 = 4.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The figure shows a graph of the function y=f(x) and points -6, -1, 1, 4 are marked on the abscissa. At which of these points is the derivative the smallest? Please indicate this point in your answer.