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Graphs of antiderivatives in the Unified State Exam. Antiderivative and integrals

Job type: 7
Topic: Antiderivative of function

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).

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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 curved 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

Job type: 7
Topic: Antiderivative of 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 segment [-3; 4].

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Solution

According to the definition of an antiderivative, the equality holds: F"(x)=f(x). Therefore, the equation f(x)=0 can be written as F"(x)=0. Since the figure shows the graph of the function y=F(x), we need to find those points in the interval [-3; 4], in which the derivative of the function F(x) is equal to zero. It is clear 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 in the indicated interval (four minimum points and three maximum points).

Answer

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

Job type: 7
Topic: Antiderivative of function

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(5)-F(0), where F(x) is one of the antiderivatives of the function f(x).

Show solution

Solution

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

Its area is equal \frac(5+3)(2)\cdot 3=12.

Answer

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

Job type: 7
Topic: Antiderivative of 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].

Show solution

Solution

According to the definition of an antiderivative, the equality holds: F"(x)=f(x). Therefore, the equation f(x)=0 can be written as F"(x)=0. Since the figure shows the graph of the function y=F(x), we need to find those points in the interval [-3; 3], in which the derivative of the function F(x) is equal to zero.

It is clear 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 in the indicated interval (two minimum points and three maximum points).

Answer

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

Job type: 7
Topic: Antiderivative of 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.

Show solution

Solution

The shaded figure is a curvilinear trapezoid bounded from above by the graph of the function y=f(x), 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. That's why 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 Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: Antiderivative of 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.

Antiderivative function f(x) in between (a; b) this function is called F(x), that equality holds for any X from a given interval.

If we take into account the fact that the derivative of a constant WITH is equal to zero, then the equality is true. So the function f(x) has many primitives F(x)+C, for an arbitrary constant WITH, and these antiderivatives differ from each other by an arbitrary constant value.

Definition of an indefinite integral.

The whole set of antiderivative functions f(x) is called the indefinite integral of this function and is denoted .

The expression is called integrand, A f(x) – integrand function. The integrand represents the differential of the function f(x).

The action of finding an unknown function given its differential is called uncertain integration, because the result of integration is more than one function F(x), and the set of its primitives F(x)+C.

Geometric meaning of the indefinite integral. The graph of the antiderivative D(x) is called the integral curve. In the x0y coordinate system, the graphs of all antiderivatives of a given function represent a family of curves that depend on the value of the constant C and are obtained from each other by a parallel shift along the 0y axis. For the example discussed above, we have:

J 2 x^x = x2 + C.

The family of antiderivatives (x + C) is geometrically interpreted by a set of parabolas.

If you need to find one from a family of antiderivatives, then additional conditions are set that allow you to determine the constant C. Usually, for this purpose, initial conditions are set: when the argument x = x0, the function has the value D(x0) = y0.

Example. It is required to find that antiderivative of the function y = 2 x that takes the value 3 at x0 = 1.

The required antiderivative: D(x) = x2 + 2.

Solution. ^2x^x = x2 + C; 12 + C = 3; C = 2.

2. Basic properties of the indefinite integral

1. The derivative of the indefinite integral is equal to the integrand function:

2. The differential of the indefinite integral is equal to the integrand expression:

3. Indefinite integral of the differential of some function equal to the sum this function itself and an arbitrary constant:

4. The constant factor can be taken out of the integral sign:

5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:

6. Property is a combination of properties 4 and 5:

7. Invariance property of the indefinite integral:

If , That

8. Property:

If , That

In fact, this property is a special case of integration using the variable change method, which is discussed in more detail in the next section.

Let's look at an example:

3. Integration method in which a given integral is reduced to one or more table integrals by means of identical transformations of the integrand (or expression) and the application of the properties of the indefinite integral, is called direct integration. When reducing this integral to a tabular one, the following differential transformations are often used (operation " subscribing to the differential sign»):

At all, f’(u)du = d(f(u)). This (formula is very often used when calculating integrals.

Find the integral

Solution. Let's use the properties of the integral and reduce this integral to several tabular ones.

4. Integration by substitution method.

The essence of the method is that we introduce a new variable, express the integrand through this variable, and as a result we arrive at a tabular (or simpler) form of the integral.

Very often, the substitution method comes to the rescue when integrating trigonometric functions and functions with radicals.

Example.

Find the indefinite integral .

Solution.

Let's introduce a new variable. Let's express X through z:

We substitute the resulting expressions into the original integral:

From the table of antiderivatives we have .

It remains to return to the original variable X:

Answer:

Function F(x ) called antiderivative for function f(x) on a given interval, if for all x from this interval the equality holds

F"(x ) = f(x ) .

For example, the function F(x) = x 2 f(x ) = 2X , because

F"(x) = (x 2 )" = 2x = f(x). ◄

The main property of the antiderivative

If F(x) - antiderivative of a function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written in the form F(x) + C, Where WITH is an arbitrary constant.

For example.

Function F(x) = x 2 + 1 is an antiderivative of the function

f(x ) = 2X , because F"(x) = (x 2 + 1 )" = 2 x = f(x);

function F(x) = x 2 - 1 is an antiderivative of the function

f(x ) = 2X , because F"(x) = (x 2 - 1)" = 2x = f(x) ;

function F(x) = x 2 - 3 is an antiderivative of the function

f(x) = 2X , because F"(x) = (x 2 - 3)" = 2 x = f(x);

any function F(x) = x 2 + WITH , Where WITH - an arbitrary constant, and only such a function is an antiderivative of the function f(x) = 2X . ◄

Rules for calculating antiderivatives

  1. If F(x) - antiderivative for f(x) , A G(x) - antiderivative for g(x) , That F(x) + G(x) - antiderivative for f(x) + g(x) . In other words, the antiderivative of the sum is equal to the sum of the antiderivatives .
  2. If F(x) - antiderivative for f(x) , And k - constant, then k · F(x) - antiderivative for k · f(x) . In other words, the constant factor can be taken out of the sign of the derivative .
  3. If F(x) - antiderivative for f(x) , And k,b- constant, and k ≠ 0 , That 1 / k F( k x+ b ) - antiderivative for f(k x+ b) .

Indefinite integral

Not definite integral from function f(x) called expression F(x) + C, that is, the set of all antiderivatives of a given function f(x) . The indefinite integral is denoted as follows:

∫ f(x) dx = F(x) + C ,

f(x)- they call integrand function ;

f(x) dx- they call integrand ;

x - they call integration variable ;

F(x) - one of the primitive functions f(x) ;

WITH is an arbitrary constant.

For example, ∫ 2 x dx =X 2 + WITH , ∫ cosx dx = sin X + WITH and so on. ◄

The word "integral" comes from the Latin word integer , which means "restored". Considering the indefinite integral of 2 x, we seem to restore the function X 2 , whose derivative is equal to 2 x. Restoring a function from its derivative, or, what is the same, finding an indefinite integral over a given integrand is called integration this function. Integration is the inverse operation of differentiation. In order to check whether the integration was performed correctly, it is enough to differentiate the result and obtain the integrand.

Basic properties of the indefinite integral

  1. The derivative of the indefinite integral is equal to the integrand:

    2. (∫ f(x) dx )" = f(x) .

  2. The constant factor of the integrand can be taken out of the integral sign:

    3. ∫ k · f(x) dx = k · ∫ f(x) dx .

  3. The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions:

    4. ∫ ( f(x) ± g(x ) ) dx = ∫ f(x) dx ± ∫ g(x ) dx .

  4. If k,b- constant, and k ≠ 0 , That

    5. ∫ f ( k x+ b) dx = 1 / k F( k x+ b ) + C .

Table of antiderivatives and indefinite integrals


f(x)
F(x) + C
∫ f(x) dx = F(x) + C
I.
$$0$$
$$C$$
$$\int 0dx=C$$
II.
$$k$$
$$kx+C$$
$$\int kdx=kx+C$$
III.
$$x^n~(n\neq-1)$$
$$\frac(x^(n+1))(n+1)+C$$
$$\int x^ndx=\frac(x^(n+1))(n+1)+C$$
IV.
$$\frac(1)(x)$$
$$\ln |x|+C$$
$$\int\frac(dx)(x)=\ln |x|+C$$
V.
$$\sin x$$
$$-\cos x+C$$
$$\int\sin x~dx=-\cos x+C$$
VI.
$$\cos x$$
$$\sin x+C$$
$$\int\cos x~dx=\sin x+C$$
VII.
$$\frac(1)(\cos^2x)$$
$$\textrm(tg) ~x+C$$
$$\int\frac(dx)(\cos^2x)=\textrm(tg) ~x+C$$
VIII.
$$\frac(1)(\sin^2x)$$
$$-\textrm(ctg) ~x+C$$
$$\int\frac(dx)(\sin^2x)=-\textrm(ctg) ~x+C$$
IX.
$$e^x$$
$$e^x+C$$
$$\int e^xdx=e^x+C$$
X.
$$a^x$$
$$\frac(a^x)(\ln a)+C$$
$$\int a^xdx=\frac(a^x)(\ln a)+C$$
XI.
$$\frac(1)(\sqrt(1-x^2))$$
$$\arcsin x +C$$
$$\int\frac(dx)(\sqrt(1-x^2))=\arcsin x +C$$
XII.
$$\frac(1)(\sqrt(a^2-x^2))$$
$$\arcsin \frac(x)(a)+C$$
$$\int\frac(dx)(\sqrt(a^2-x^2))=\arcsin \frac(x)(a)+C$$
XIII.
$$\frac(1)(1+x^2)$$
$$\textrm(arctg) ~x+C$$
$$\int \frac(dx)(1+x^2)=\textrm(arctg) ~x+C$$
XIV.
$$\frac(1)(a^2+x^2)$$
$$\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$
$$\int \frac(dx)(a^2+x^2)=\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$
XV.
$$\frac(1)(\sqrt(a^2+x^2))$$
$$\ln|x+\sqrt(a^2+x^2)|+C$$
$$\int\frac(dx)(\sqrt(a^2+x^2))=\ln|x+\sqrt(a^2+x^2)|+C$$
XVI.
$$\frac(1)(x^2-a^2)~(a\neq0)$$
$$\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+C$$
$$\int\frac(dx)(x^2-a^2)=\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+ C$$
XVII.
$$\textrm(tg) ~x$$
$$-\ln |\cos x|+C$$
$$\int \textrm(tg) ~x ~dx=-\ln |\cos x|+C$$
XVIII.
$$\textrm(ctg) ~x$$
$$\ln |\sin x|+C$$
$$\int \textrm(ctg) ~x ~dx=\ln |\sin x|+C$$
XIX.
$$ \frac(1)(\sin x) $$
$$\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$
$$\int \frac(dx)(\sin x)=\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$
XX.
$$ \frac(1)(\cos x) $$
$$\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right) \end(vmatrix)+C $$
$$\int \frac(dx)(\cos x)=\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right ) \end(vmatrix)+C $$
The antiderivative and indefinite integrals given in this table are usually called tabular antiderivatives And table integrals .

Definite integral

Let in between [a; b] a continuous function is given y = f(x) , Then definite integral from a to b functions f(x) is called the increment of the antiderivative F(x) this function, that is

$$\int_(a)^(b)f(x)dx=F(x)|(_a^b) = ~~F(a)-F(b).$$

Numbers a And b are called accordingly lower And top limits of integration.

Basic rules for calculating a definite integral

1. \(\int_(a)^(a)f(x)dx=0\);

2. \(\int_(a)^(b)f(x)dx=- \int_(b)^(a)f(x)dx\);

3. \(\int_(a)^(b)kf(x)dx=k\int_(a)^(b)f(x)dx,\) where k - constant;

4. \(\int_(a)^(b)(f(x) ± g(x))dx=\int_(a)^(b)f(x) dx±\int_(a)^(b) g(x)dx\);

5. \(\int_(a)^(b)f(x)dx=\int_(a)^(c)f(x)dx+\int_(c)^(b)f(x)dx\);

6. \(\int_(-a)^(a)f(x)dx=2\int_(0)^(a)f(x)dx\), where f(x) — even function;

7. \(\int_(-a)^(a)f(x)dx=0\), where f(x) is an odd function.

Comment . In all cases, it is assumed that the integrands are integrable on numerical intervals, the boundaries of which are the limits of integration.

Geometric and physical meaning of the definite integral

Geometric meaning
definite integral

Physical meaning
definite integral



Square S curvilinear trapezoid (a figure limited by the graph of a continuous positive on the interval [a; b] functions f(x) , axis Ox and straight x=a , x=b ) is calculated by the formula

$$S=\int_(a)^(b)f(x)dx.$$

Path s, which the material point has overcome, moving rectilinearly with a speed varying according to the law v(t) , for a period of time a ; b] , then the area of ​​the figure limited by the graphs of these functions and straight lines x = a , x = b , calculated by the formula

$$S=\int_(a)^(b)(f(x)-g(x))dx.$$

For example. Let's calculate the area of ​​the figure bounded by lines

y = x 2 And y = 2-x .


Let us schematically depict the graphs of these functions and highlight in a different color the figure whose area needs to be found. To find the limits of integration, we solve the equation:

x 2 = 2-x ; x 2 + x- 2 = 0 ; x 1 = -2, x 2 = 1 .

$$S=\int_(-2)^(1)((2-x)-x^2)dx=$$

$$=\int_(-2)^(1)(2-x-x^2)dx=\left (2x-\frac(x^2)(2)-\frac(x^3)(2) \right )\bigm|(_(-2)^(~1))=4\frac(1)(2). $$

◄

Volume of a body of revolution


If a body is obtained as a result of rotation about an axis Ox curvilinear trapezoid bounded by a continuous and non-negative graph on the interval [a; b] functions y = f(x) and straight x = a And x = b , then it is called body of rotation .

The volume of a body of revolution is calculated by the formula

$$V=\pi\int_(a)^(b)f^2(x)dx.$$

If a body of revolution is obtained as a result of the rotation of a figure bounded above and below by graphs of functions y = f(x) And y = g(x) , accordingly, then

$$V=\pi\int_(a)^(b)(f^2(x)-g^2(x))dx.$$

For example. Let's calculate the volume of a cone with radius r and height h .

Let us position the cone in a rectangular coordinate system so that its axis coincides with the axis Ox , and the center of the base was located at the origin. Generator rotation AB defines a cone. Since the equation AB

$$\frac(x)(h)+\frac(y)(r)=1,$$

$$y=r-\frac(rx)(h)$$
and for the volume of the cone we have

$$V=\pi\int_(0)^(h)(r-\frac(rx)(h))^2dx=\pi r^2\int_(0)^(h)(1-\frac( x)(h))^2dx=-\pi r^2h\cdot \frac((1-\frac(x)(h))^3)(3)|(_0^h)=-\pi r^ 2h\left (0-\frac(1)(3) \right)=\frac(\pi r^2h)(3).$$

◄

One of the operations of differentiation is finding the derivative (differential) and applying it to the study of functions.

The inverse problem is no less important. If the behavior of a function in the vicinity of each point of its definition is known, then how can one reconstruct the function as a whole, i.e. throughout the entire scope of its definition. This problem is the subject of study of the so-called integral calculus.

Integration is the inverse action of differentiation. Or restoring the function f(x) from a given derivative f`(x). The Latin word “integro” means restoration.

Example No. 1.

Let (f(x))’ = 3x 2. Let's find f(x).

Solution:

Based on the rule of differentiation, it is not difficult to guess that f(x) = x 3, because

(x 3)’ = 3x 2 However, you can easily notice that f(x) is not found uniquely. As f(x), you can take f(x)= x 3 +1 f(x)= x 3 +2 f(x)= x 3 -3, etc.

Because the derivative of each of them is 3x 2. (The derivative of a constant is 0). All these functions differ from each other by a constant term. That's why general solution the problem can be written in the form f(x)= x 3 +C, where C is any constant real number.

Any of the found functions f(x) is called antiderivative for the function F`(x)= 3x 2

Definition.

A function F(x) is called antiderivative for a function f(x) on a given interval J if for all x from this interval F`(x)= f(x). So the function F(x)=x 3 is antiderivative for f(x)=3x 2 on (- ∞ ; ∞). Since, for all x ~R the equality is true: F`(x)=(x 3)`=3x 2

As we have already noticed, this function has an infinite number of antiderivatives.

Example No. 2.

The function is antiderivative for all on the interval (0; +∞), because for all h from this interval, equality holds.

The problem of integration is to given function find all its antiderivatives. When solving this problem, the following statement plays an important role:

A sign of constancy of function. If F"(x) = 0 on some interval I, then the function F is constant on this interval.

Proof.

Let us fix some x 0 from the interval I. Then for any number x from such an interval, by virtue of the Lagrange formula, we can indicate a number c contained between x and x 0 such that

F(x) - F(x 0) = F"(c)(x-x 0).

By condition, F’ (c) = 0, since c ∈1, therefore,

F(x) - F(x 0) = 0.

So, for all x from the interval I

that is, the function F maintains a constant value.

All antiderivative functions f can be written using one formula, which is called general form of antiderivatives for the function f. The following theorem is true ( main property of antiderivatives):

Theorem. Any antiderivative for a function f on the interval I can be written in the form

F(x) + C, (1) where F (x) is one of the antiderivatives for the function f (x) on the interval I, and C is an arbitrary constant.

Let us explain this statement, in which two properties of the antiderivative are briefly formulated:

  1. Whatever number we put in expression (1) instead of C, we obtain the antiderivative for f on the interval I;
  2. no matter what antiderivative Ф for f on the interval I is taken, it is possible to select a number C such that for all x from the interval I the equality

Proof.

  1. By condition, the function F is antiderivative for f on the interval I. Consequently, F"(x)= f (x) for any x∈1, therefore (F(x) + C)" = F"(x) + C"= f(x)+0=f(x), i.e. F(x) + C is the antiderivative for the function f.
  2. Let Ф (x) be one of the antiderivatives for the function f on the same interval I, i.e. Ф "(x) = f (х) for all x∈I.

Then (Ф(x) - F (x))" = Ф"(x)-F' (x) = f(x)-f(x)=0.

From here it follows c. the power of the sign of constancy of the function, that the difference Ф(х) - F(х) is a function that takes some constant value C on the interval I.

Thus, for all x from the interval I the equality Ф(x) - F(x)=С is true, which is what needed to be proved. The main property of the antiderivative can be given geometric meaning: graphs of any two antiderivatives for the function f are obtained from each other by parallel translation along the Oy axis

Questions for notes

The function F(x) is an antiderivative of the function f(x). Find F(1) if f(x)=9x2 - 6x + 1 and F(-1) = 2.

Find all antiderivatives for the function

For the function (x) = cos2 * sin2x, find the antiderivative of F(x) if F(0) = 0.

For a function, find an antiderivative whose graph passes through the point

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 the primitive 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 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