Functional graphical method for solving exponential equations. Functional-graphical method for solving equations
In a standard school mathematics course, the properties of functions are used mainly to construct their graphs. The functional method for solving equations is used if and only if the equation F(x) = G(x) as a result of transformations or replacement of variables cannot be reduced to one or another standard equation that has a specific solution algorithm.
Unlike the graphical method, knowledge of the properties of functions allows you to find the exact roots of the equation, without the need to construct graphs of functions. Using the properties of functions helps to rationalize the solution of equations.
The following properties of the function are considered in the work: domain of definition of the function; function range; properties of monotonicity of a function; properties of convexity of a function; properties of even and odd functions.
Purpose of the work: to carry out some classification of non-standard equations using the general properties of functions, describe the essence of each property, give recommendations for its use, instructions for use.
All work is accompanied by the solution of specific problems proposed at the Unified State Examination in various years.
Chapter 1. Using the concept of domain of definition of a function.
Let's introduce a few key definitions.
The domain of definition of the function y = f(x) is the set of values of the variable x for which the function makes sense.
Let the equation f(x) = g(x) be given, where f(x) and g(x) are elementary functions defined on the sets D1, D2. Then the region D of admissible values of the equation will be a set consisting of those values of x that belong to both sets, that is, D = D1∩ D2. It is clear that when the set D is empty (D= ∅), then the equation has no solutions. (Appendix No. 1).
1. arcsin (x+2) +2x- x2 = x-2.
ODZ:-1 =0⇔-3
Answer: there are no solutions.
2. (x2-4x+3 +1)log5x5 + 1x(8x-2x2-6 + 1) = 0.
ODZ: x2-4x+3>=0,x>0.8x-2x2-6>=0⇔x∈(-infinity;1∪ 3;infinity),x>01
Check: x = 1.
(1-4+3 +1)log515 + (8-2-6 + 1) = 0,
0 = 0 - true.
x = 3. (9-12+3+1)log535 +13(24-18-6+1) = 0, log535 +13 = 0 - incorrect.
It often turns out to be sufficient to consider not the entire domain of definition of a function, but only its subset, on which the function takes values that satisfy certain conditions (for example, only non-negative values).
1. x+27-x(x-9 +1) = 1.
ODZ: x-9>=0, x>=9.
For x>=9 x+2>0, 7-x 0, thus, the product of the three factors on the left side of the equation is negative, and the right side of the equation is positive, which means the equation has no solutions.
Answer: ∅.
2. 3-x2+ x+2 = x-2.
ODZ: 3-x2>=0,x+2>=0,⇔ 3-x(3+x)>=0,x>=-2,⇔ -3=-2,⇔
On the set of admissible values, the left side of the equation is positive, and the right side is negative, which means that the equation has no solutions.
Answer: there are no solutions.
Chapter 2. Using the concept of function range.
The range of values of the function y = f(x) is the set of values of the variable y for acceptable values of the variable x.
A function y = f(x) is said to be bounded below (resp. above) on the set X if there exists a number M such that the inequality fx>=M holds on X (resp. fx
A function y = f(x) is called bounded on a given interval (contained in its domain of definition) if there is a number M >0 such that for all values of the argument belonging to this interval the inequality f(x) holds
Let the equation f(x) = g(x) be given, where g(x) are elementary functions defined on the sets D1, D2. Let us denote the range of variation of these functions as E1 and E2, respectively. If x1 is a solution to the equation, then the numerical equality f(x1) = g(x1) will hold, where f(x1) is the value of the function f(x) at x = x1, and g(x1) is the value of the function g(x) at x = x1. This means that if the equation has a solution, then the ranges of functions f(x) and g(x) have common elements (E1∩E2 !=∅). If the sets E1 and E2 do not contain such common elements, then the equation has no solutions.
Basic inequalities are used to evaluate expressions. (Appendix No. 2).
Let the equation f(x) = g(x) be given. If f(x)>=0 and g(x)
1. x2+2xsinxy+1=0.
Solution. There is a unit on the left side, which means we can use the basic trigonometric identity: x2+ 2xsinxy+ sin2xy+cos2xy=0.
The sum of the first three terms is a perfect square:
(x+sinxy)2+cos2xy =0.
Consequently, on the left side is the sum of squares; it is equal to zero when the expressions in the squares are simultaneously equal to zero. Let's write the system: cosxy=0,x+sinxy=0.
If cosxy=0, then sinxy= +-1, therefore this system is equivalent to a combination of two systems: x+1=0,cosxy=0 or x-1=0,cosxy=0.
Their solutions are pairs of numbers x=1, y = PI 2 + PIm, m∈Z, and x=-1, y = PI 2 + PIm, m∈Z.
Answer: x=1, y = PI 2 + PIm, m∈Z, and x=-1, y = PI 2 + PIm, m∈Z.
If on the interval X the largest value of one of the functions y = f(x), y = g(x) is equal to A and the smallest value of the other function is also equal to A, then the equation f(x) = g(x) is equivalent on the interval X to the system of equations fx=A,gx=A.
1. Find all values of a for which the equation has a solution
2cos222x-x2=a+3sin(22x-x2+1).
After replacing t= 22x-x2 we arrive at the equation cos(2t+PI3)=a-12.
The function t=2m increases, which means it reaches its greatest value at the highest value of m. But m=2х - x has the greatest value equal to 1. Then tmax = 22·1-1=2. Thus, the set of values of the function t = 22x-x2 is the interval (0;2, and the function cos(2t+PI3) is the interval -1;0.5). Consequently, the original equation has a solution for those and only those values of a that satisfy the inequalities -1Answer: -12. Solve the equation (log23)x+a+2 = (log94)x2+a2-6a-5.
Using the obvious inequalities
Answer: x= - 5+32 if a=1+32 and x=-5+32 if a= 1-32.
You can consider other equations in more detail. (Appendix No. 3).
Chapter 3. Using the property of monotonicity of a function.
A function y = f(x) is said to be increasing (respectively, decreasing) on a set X if on this set, as the argument increases, the values of the function increase (respectively, decrease).
In other words, the function y = f(x) increases on the set X if from x1∈X, x2∈X and x1 It decreases on this set if from x1∈X, x2∈X and x1 f(x2).
A function y = f(x) is said to be non-strictly increasing (respectively, non-strictly decreasing) on X if x1∈X, x2∈X and x1=f(x2)).
Functions that increase and decrease on X are called monotone on X, and functions that are not strictly increasing or decreasing on X are called non-strictly monotone on X.
To prove the monotonicity of functions, the following statements are used:
1. If a function f increases on a set X, then for any number C the function f + C also increases on X.
2. If the function f increases on the set X and C > 0, then the function Cf also increases on X.
3. If a function f increases on a set X, then the function - f decreases on this set.
4. If a function f increases on the set X and maintains sign on the set X, then the function 1f decreases on this set.
5. If functions f and g increase on a set X, then their sum f + g also increases on this set.
6. If the functions f and g are increasing and non-negative on the set X, then their product fg is also increasing on X.
7. If the function f is increasing and non-negative on the set X and n is a natural number, then the function fn is also increasing on X.
8. If both functions f(x) and g(x) are increasing or both are decreasing, then the function h(x) = f(g(x)) is an increasing function. If one of the functions is increasing. And the other is decreasing, then h(x) = f(g(x)) is a decreasing function.
Let us formulate theorems about equations.
Theorem 1.
If the function f(x) is monotonic on the interval X, then the equation f(x) = C has at most one root on the interval X.
Theorem 2.
If the function f(x) is monotonic on the interval X, then the equation f(g(x)) = f(h(x)) is equivalent on the interval X to the equation g(x) = h(x).
Theorem 3.
If the function f(x) increases on the interval X, and g(x) decreases on the interval X, then the equation g(x) = f(x) has at most one root on the interval X.
Theorem 4.
If the function f(x) increases on the interval X, then the equation f(f(x)) = x is equivalent on the interval X to the equation f(x) = x.
1. Find all values of a for which the equation has exactly three roots
4-x-alog3(x2-2x+3)+2-x2+2xlog13(2x-a+2)=0.
Solution. Let's transform this equation to the form
2x2-2xlog3(x2-2x+3)= 22x-a-1log3(2x-a+2).
If we put u = x2-2x, v=2x-a-1, then we arrive at the equation
2ulog3(u+3)= 2vlog3(v+3).
The function f (t) = 2tlog3(t+3) increases monotonically for t >-2, so from the last equation we can go to the equivalent u = v, x2-2x = 2x-a-1⇔(x-1)2=2x -a.
This equation, as can be seen from the figure, has exactly three roots in the following cases:
1. The vertex of the graph of the function y = 2x-a is located at the vertex of the parabola y = (x-1)2, which corresponds to a = 1;
2. The left ray of the graph y = 2x-a touches the parabola, and the right one intersects it at two points; this is possible with a=12;
3. The right ray touches and the left ray intersects the parabola, which occurs when a=32.
Let us explain the second case. The equation of the left ray is y = 2a-2x, its slope is -2. Therefore, the angular coefficient of the tangent to the parabola is equal to
2(x -1) = -2 ⇒ x = 0 and the tangent point has coordinates (0; 1). From the condition that this point belongs to the ray, we find a=12.
The third case can be considered similarly or using symmetry considerations.
Answer: 0.5; 1;1.5.
We can consider other equations in more detail. (Appendix No. 4).
Chapter 4. Using the properties of convexity.
Let a function f(x) be defined on an interval X, it is called strictly convex downward (upward) on X if for any u and v from X, u!=v and 0
Geometrically, this means that any point of the chord BC (that is, a segment with ends at points B(u;f(u)) and C(v;f(v)), different from points B and C, lies above (below) the point And the graph of the function f(x), corresponding to the same argument value (Appendix No. 5).
Functions that are strictly convex up and down are called strictly convex.
The following statements are true.
Theorem 1.
Let the function f(x) be strictly downward convex on the interval X, u ,v ∈X, u
The following statement follows from Theorem 1.
Theorem 2.
If the function f(x) is strictly convex on the interval X, the functions u = u(x), v = v(x), u1=u1(x), v1 = v1(x) are such that for all x from the ODZ equations f(u)+f(v) = f(u1) + f(v1) (1) their values u(x), v(x), u1(x), v1(x) are contained in X and the condition u is satisfied +v = u1 +v1, then the equation f(u)+f(v) = f(u1) + f(v1) (2) on the ODZ is equivalent to the set of equations u (x) = u1(x), u(x) = v1(x) (3).
1. 41-sin4x+41-cos4x=412.
Solution. If we set fx= 41-x2, u=cos2x, v=sin2x, u1=v1=12, then this equation will be written in the form (1). Since f"x= -x24(1-x2)3, f""x=-2+x244(1-x2)7, then the function fx is strictly convex upward on the segment -1;1. Obviously, the remaining conditions are satisfied Theorem 2 and, therefore, the equation is equivalent to the equation cos2x = 0.5, x = PI4 +PIk2, where k∈Z.
Answer: x = PI4 +PIk2, where k∈Z.
Theorem 3.
Let the function fx be strictly convex on the interval X and u,v, λv+(1-λ)u∈X. Then the equality f (λv+(1-λ)u) = λf(v)+(1-λ)f(u) (4) is valid if and only if either u=v or λ=0, or λ=1.
Examples: sin2xcos3x+cos2xsin3x∙1+sin2xcos3x+cos2xsin3x= sin2xcos3x1+cos3x+cos2xsin3x1+sin3x.
The equation has the form (4) if fx=x1+x= x+x2, u=sin3x, v= cos3x, λ=sin2x.
It is obvious that the function fx is strictly convex downward on R. Therefore, by Theorem 3, the original equation is equivalent to the set of equations sinx=0, sin2x=1, cos3x=sin3x.
From here we get that its solutions will be PIk2, PI12+PIn3, where k,n∈Z.
Answer: PIk2, PI12+PIn3, where k,n∈Z.
The use of convexity properties is also used to solve more complex equations. (Appendix No. 6).
Chapter 5. Using the even or odd properties of functions.
A function fx is called even if for any value x taken from the domain of definition of the function, the value - x also belongs to the domain of definition and the equality f-x = fx holds. A function fx is called odd if for any value x taken from the domain of definition of the function, the value - x also belongs to the domain of definition and the equality f-x = - fx holds.
From the definition it follows that the domains of even and odd functions are symmetrical about zero (a necessary condition).
For any two symmetric values of the argument from the domain of definition, the even function takes equal numerical values, and the odd one - equal in absolute value, but of opposite sign.
Theorem 1.
The sum, difference, product and quotient of two even functions are even functions.
Theorem 2.
The product and quotient of two odd functions are even functions.
Let us have the equation F(x)=0, where F(x) is an even or odd function.
To solve the equation F(x) = 0, where F(x) is an even or odd function, it is enough to find positive (or negative) roots symmetrical to those obtained, and for an odd function the root will be x = 0 if this value is within the domain of definition F(x). For an even function, the value x = 0 is checked by direct substitution into the equation.
We have even functions on both sides of the equation. Therefore, it is enough to find solutions for x>=0. Since x=0 is not a root of the equation, consider two intervals: (0;2, 2;infinity.
a) On the interval (0;2 we have:
8x= 2x+2-x+2, 23x=24, x= 43.
b) On the interval 2;infinity we have:
8x= 2x+2+x-2.23x=22x, x=0.
But since x = 0 is not a root of the equation, then for x>0 this equation has a root x = 43. Then x = - 43 is also a root of the equation.
Answer: 43; - 43.
The author believes that the work can be used by teachers and students of general education in elective classes, in preparation for mathematical Olympiads, passing the Unified State Exam, and entrance exams to technical educational institutions.
The accuracy of such a solution is low, but with the help of a graph you can intelligently choose the first approximation from which to begin further solving the equation. There are two ways to solve equations graphically.
First way . All terms of the equation are transferred to the left side, i.e. the equation is presented in the form f(x) = 0. After this, a graph of the function y = f(x) is constructed, where f(x) is the left side of the equation. Abscissas of the points of intersection of the graph of the function y = f(x) with the axis Ox and are the roots of the equation, because at these points y = 0.
Second way . All terms of the equation are divided into two groups, one of them is written on the left side of the equation, and the other on the right, i.e. represent it in the form j(x) = g(x). After this, graphs of two functions y = j(x) and y = g(x) are plotted. The abscissas of the intersection points of the graphs of these two functions serve as the roots of this equation. Let the point of intersection of the graphs have an abscissa x o, the ordinates of both graphs at this point are equal to each other, i.e. j(x o) = g(x o). From this equality it follows that x 0 is the root of the equation.
Root separation
The process of finding approximate values of the roots of the equation is divided into two stages:
1) separation of roots;
2) refinement of the roots to a given accuracy.
The x root of the equation f(x) = 0 is considered separated on the interval if the equation f(x) = 0 has no other roots on this interval.
Separating roots means dividing the entire range of acceptable values into segments, each of which contains one root.
Graphic method of root separation - in this case, proceed in the same way as with the graphical method of solving equations.
If the curve touches the x-axis, then at this point the equation has a double root (for example, the equation x 3 - 3x + 2 = 0 has three roots: x 1 = -2; x 2 = x 3 = 1).
If the equation has a threefold real root, then at the point of contact with the axis X the curve y = f(x) has an inflection point (for example, the equation x 3 - 3x 2 + 3x - 1 = 0 has a root x 1 = x 2 = x 3 = 1).
Analytical root separation method . To do this, use some properties of functions.
Theorem 1 . If the function f(x) is continuous on a segment and takes values of different signs at the ends of this segment, then inside the segment there is at least one root of the equation f(x) = 0.
Theorem 2. If the function f(x) is continuous and monotonic on a segment and takes values of different signs at the ends of the segment, then the segment contains the root of the equation f(x) = 0, and this root is unique.
Theorem 3 . If the function f(x) is continuous on a segment and takes values of different signs at the ends of this segment, and the derivative f "(x) maintains a constant sign inside the segment, then inside the segment there is a root of the equation f(x) = 0 and, moreover, a unique one.
If the function f(x) is given analytically, then domain of existence (domain of definition) of the function is the set of all those real values of the argument for which the analytical expression defining the function does not lose its numerical meaning and takes only real values.
The function y = f(x) is called increasing , if as the argument increases, the value of the function increases, and decreasing , if as the argument increases, the value of the function decreases.
The function is called monotonous , if in a given interval it either only increases or only decreases.
Let the function f(x) be continuous on the segment and take values of different signs at the ends of the segment, and the derivative f "(x) maintains a constant sign on the interval. Then if at all points of the interval the first derivative is positive, i.e. f "(x) >0, then the function f(x) in this interval increases . If at all points of the interval the first derivative is negative, i.e. f "(x)<0, то функция в этом интервале decreases .
Let the function f(x) on an interval have a second-order derivative that maintains a constant sign throughout the entire interval. Then if f ""(x)>0, then the graph of the function is convex down ; if f ""(x)<0, то график функции является convex up .
Points at which the first derivative of a function is equal to zero, as well as those at which it does not exist (for example, it turns to infinity), but the function maintains continuity, are called critical .
Procedure for separating roots using the analytical method:
1) Find f "(x) - the first derivative.
2) Make a table of signs of the function f(x), assuming X equal to:
a) critical values (roots) of the derivative or those closest to them;
b) boundary values (based on the range of permissible values of the unknown).
Example. Separate the roots of the equation 2 x - 5x - 3 = 0.
We have f(x) = 2 x - 5x - 3 . The domain of definition of the function f(x) is the entire numerical axis.
Let's calculate the first derivative f "(x) = 2 x ln(2) - 5.
We equate this derivative to zero:
2 x log(2) - 5 = 0 ; 2 x log(2) = 5 ; 2 x = 5/ln(2) ; xlg(2) = lg(5) - lg(ln(2)) .
We compile a table of signs of the function f(x), assuming X equal to: a) critical values (roots of the derivative) or closest to them; b) boundary values (based on the range of permissible values of the unknown):
The roots of the equation lie in the intervals (-1.0) and (4.5).
Functional-graphical method for solving the inequality f(x)< g(x). 1. Подбором найдем корень уравнения f(x)=g(x), используя свойства монотонных функций; 2. Построим схематически графики обеих функций, проходящие через точку с найденной абсциссой; 3. Выберем решение неравенства, соответствующее знаку неравенства; 4. Запишем ответ. :
Slide 9 from the presentation "Exponential Equations and Inequalities". The size of the archive with the presentation is 174 KB.Algebra 11th grade
summary of other presentations“Equations of the third degree” - (1). Tartaglia refuses. On February 12, Cardano repeats his request. "Great Art" X3 + px + q = 0. Example: x3 – 5 x2 + 8 x – 4 = 0 x3 – 2 x2 –3 x2 + 8x – 4 = 0 x2 (x – 2) – (3 x2 – 8x + 4) = 0 3 x2 – 8x + 4 = 0 x = 2 x = 2/3 x2 (x – 2) – (3 (x –2) (x – 2/3)) = 0 x2 (x – 2) – (( x – 2) (3x – 2)) = 0 (x – 2)(x2 – 3x + 2) = 0 x – 2 = 0 x2 – 3x + 2 = 0 x = 2 x = 2 x = 1 Answer: x = 2; x = 1. Our formula gives: Municipal educational institution “Secondary school No. 24”. X3 + ax = b (1). Here p = 6 and q = -2. First example:
“Application of the definite integral” - Ch. 4. Development of an elective on the topic “Definite Integral”. Definite integral. §4. Properties of a definite integral. Ch. 2. Various approaches to integral theory in textbooks for schoolchildren. §1. Volume of a body of revolution. §6. Introduction. Darboux sums. §3. Mechanical work. Goal: Approaches to the construction of integral theory: Introductory remarks. §2. Integration methods. §3. Conclusion. Chapter 3. Application of a definite integral. §1.
“Exponential equations and inequalities” - 2) Equivalent to the inequality f(x)< g(x), 0<а<1. "Что значит решить задачу? Обоснование: 12). Сравните основание а с единицей: Если 0 Sections:
Mathematics Class:
11
To conduct this lesson, groups of children were organized in the class and were asked to remember a certain method for solving equations, select 5-8 equations, solve them and prepare a presentation. Equipment: Computer, projector. Presentation . The teacher's presentation included presentations from the children, but they had different backgrounds. During the classes Today in the lesson we will recall the functional-graphical method of solving equations, consider when it is used, what difficulties may arise when solving it, and we will choose methods for solving equations. Let us recall the basic methods for solving equations.(slide number 2) The first group examines the graphical method. The second group talks about the majorant method. The majorant method is a method for finding the boundedness of a function. Majorization - finding the limit points of a function. M - majorante. If we have f(x) = g(x) and the ODZ is known, and if .№1 Solve the equation: x = 4 - solution to the equation. #2 Solve the equation Solution: Let's estimate the right and left sides of the equation: A) b) An evaluation of the parts of the equation shows that the left side is not less than, and the right side is not more than two for any admissible values of the variable x. Therefore, this equation is equivalent to the system The first equation of the system has only one root x=-2. Substituting this value into the second equation, we obtain the correct numerical equality: Answer: x=-2. The third group explains the use of the root uniqueness theorem. If one of the functions (F(x)) decreases and the other (G(x)) increases on some domain of definition, then the equation F(x)=G(x) has at most one solution. #1 Solve the equation Solution: domain of definition of this equation x>0. We examine the monotonicity of the function. The first of them is decreasing (since it is a logarithmic function with a base greater than zero but less than one), and the second is increasing (it is a linear function with a positive coefficient at x). By selection one can easily find the root of the equation x=3, which is the only solution to this equation. Answer: x=3. The teacher reminds. where else the monotonicity of a function is used when solving equations. A) - From an equation of the form h(f(x))=h(g(x)) we pass to an equation of the form f(x)=g(x) If the function is monotonic №5 sin (4x+?/6) = sin 3x WRONG! (periodic function). And then we pronounce the correct answer. WRONG! (even degree) And then we pronounce the correct answer: B) Method of using functional equations. Theorem. If the function y = f(x) is an increasing (or decreasing) function on the domain of permissible values of the equation f(g(x)) = f(h(x)), then the equations f(g(x)) = f(h( x)) and g(x)=f(x) are equivalent. No. 1 Solve the equation: Consider the functional equation f(2x+1) = f(-x), where f(x) = f() Find the derivative Determine its sign. Because the derivative is always positive, then the function is increasing on the entire number line, then we move on to the equation Solve the equation. X 6 -|13 + 12x| 3 = 27cos x 2- 27cos(13 + 12x). 1) the equation is reduced to the form x6 - 27cos x2 = |13 + 12x|3 - 27cos(13 + 12x), f(x2) = f(13 + 12x), where f(t) = |t|3-27сost; 2)The function f is even and for t > 0 has the following derivative Consequently, the function f increases on the positive semi-axis, which means that it takes each of its values at exactly two points symmetrical with respect to zero. This equation is equivalent Answer: -1, 13, -6+?/23. Tasks to be solved in class. Answer Reflection. 1. What new did you learn? 2. Which method do you do better? House task: Select 2 equations for each method and solve them. Target
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consider ZNO problems using functional-graphical methods using the example of an exponential function y = a x, a>0, a1 Lesson objectives: repeat the property of monotonicity and limitedness of the exponential function; repeat the algorithm for constructing function graphs using transformations; find multiple values and multiple definitions of a function by type of formula and using a chart; solve exponential equations, inequalities, and systems using graphs and properties of functions. working with function graphs containing a module; consider the graphs of a complex function and their range of values; During the classes: 1. Introductory speech by the teacher. Motivation for studying this topic Slide 1
Exponential function. “Functional - graphical methods for solving equations and inequalities” The functional-graphical method is based on the use of graphic illustrations, the application of the properties of a function and allows you to solve many problems in mathematics. Slide 2-3 Goals and lesson objectives. Today we will look at ZNO problems of different levels of complexity using functional-graphical methods using the example of the exponential function y = a x, a>o, a1. Using a graphic program, we will create illustrations for the problems. Slide 4
Why is it so important to know the properties of the exponential function? According to the law of exponential function, all living things on Earth would reproduce if there were favorable conditions for this, i.e. there were no natural enemies and there was plenty of food. Proof of this is the spread of rabbits in Australia, which were not there before. It was enough to release a couple of individuals, and after some time their offspring became a national disaster. In nature, technology and economics there are numerous processes during which the value of a quantity changes the same number of times, i.e. according to the law of exponential function. These processes are called processes organic growth or organic attenuation. For example, bacterial growth under ideal conditions corresponds to the process of organic growth; radioactive decay of substances– the process of organic attenuation. Subject to the laws of organic growth growth of deposit at the Savings Bank, hemoglobin restoration in the blood of a donor or a wounded person who has lost a lot of blood. Give your examples Application in real life (dose of medication). Message about medication dosage: Everyone knows that the pills recommended by the doctor for treatment must be taken several times a day, otherwise they will be ineffective. The need to re-administer the drug to maintain a constant concentration in the blood is caused by the destruction of the drug occurring in the body. The figure shows how, in most cases, the concentration of drugs in the blood of a person or animal changes after a single administration. Slide 5.
The decrease in drug concentration can be approximated by an exponential whose exponent contains time. Obviously, the rate of destruction of the drug in the body must be proportional to the intensity of metabolic processes. There is one known tragic case that occurred due to ignorance of this addiction. From a scientific point of view, the drug LSD, which causes peculiar hallucinations in normal people, is very interesting for psychiatrists and neurophysiologists. Some researchers decided to study the elephant's reaction to this drug. To do this, they took the amount of LSD that infuriates cats and multiplied it by the number of times the mass of an elephant is greater than the mass of a cat, believing that the dose of the drug administered should be directly proportional to the mass of the animal. The administration of such a dose of LSD to an elephant led to its death within 5 minutes, from which the authors concluded that elephants have increased sensitivity to this drug. A review of this work that appeared later in the press called it an “elephant-like mistake” by the authors of the experiment. 2. Updating students' knowledge. What does it mean to study a function? (formulate a definition, describe properties, draw a graph) What function is called exponential? Give an example. What basic properties of the exponential function do you know? Scope of significance (limitedness) domain monotonicity (condition of increasing and decreasing) Slide 6
. Specify a variety of function values (according to the finished drawing) Slide 7.
Name the condition for increasing and decreasing function and correlate the formula of the function with its graph Slide 8.
Based on the finished drawing, describe the algorithm for constructing function graphs Slide a) y=3 x + 2 b) y=3 x-2 – 2 3.Diagnostic independent work (using a PC). The class is divided into two groups. The main part of the class performs test tasks. Strong students perform more complex tasks. Independent work in the programPower
point(for the main part Independent work (for the strong part of the class) Slide9.
Write down the algorithm for constructing a graph of a function, name its domain of definition, range of value, intervals of increase and decrease. Slide 10.
Match the function formula with its graph Students check their answers without correcting mistakes; independent work is handed over to the teacher Slides 11-21.
Checking the test for the main part 4. Studying a new topic. Application of the functional-graphic method for solving equations, inequalities, systems, determining the range of values of a complex function Slides 22-23. Functionally graphical method for solving equations
To solve an equation of the form f(x)=g(x) using the functional-graphical method you need: Construct graphs of the functions y=f(x) and y=g(x) in the same coordinate system. Determine the coordinates of the intersection point of the graphs of these functions. Write down the answer. SOLVING EQUATIONS
Slide 24-25.
Does the equation have a root and if so, is it positive or negative? 5. Doing practical work. SOLVING EQUATIONS. SLIDES 27-30 This equation can be solved graphically. Students are asked to complete the task and then answer the question: “Is it necessary to construct graphs of functions to solve this equation?” Answer: “The function increases over the entire domain of definition, and the function decreases. Consequently, the graphs of such functions have at most one intersection point, which means that the equation has at most one root. By selection we find that “. Solve equation 3 x = (x-1) 2 + 3 Solution: We use the functional method for solving equations: because this system has a unique solution, then by selection method we find x = 1 SOLVING INEQUALITIES. Slides 31-33
G Solve inequality: a) cos x 1 + 3 x Solution: Answer: ( ; ) Solve the inequality graphically. (The graph of the exponential function lies above the function written on the right side of the equation.) Answer: x>2. ABOUT . The exponential function contains the modulus sign in the exponent. slide 34-35
Let's repeat the module definition. Make notes in your notebook: 1). 2). A graphical illustration is presented on the slide. Explain how the graphs are constructed. E(y)=(0;1] To solve this equation, you need to remember the property of boundedness of the exponential function. The function takes values >
1, a – 1 <
>
1, therefore equality is possible only if both sides of the equation are simultaneously equal to 1. This means that Solving this system, we find that X = 0. .Finding the range of values of a complex function. Slides 36-37.
Using the ability to construct a graph of a quadratic function, determine sequentially the coordinates of the vertex of the parabola and find the range of values. Question: determine the nature of the monotonicity of the function. The exponential function y = 16 t increases, since 16>1. At the lowest value of the function indicator The graph illustrates our conclusion.
, 
, because 
, A ;
, because 
.
f"(t)= therefore f"(t)> 0 for everyone
the following set:
Subject: "Exponential function. Functional-graphical methods for solving equations, inequalities, systems"
SLIDE 26
Graphical methods make it possible to solve inequalities containing different functions. To do this, after constructing graphs of the functions on the left and right sides of the inequality and determining the abscissa of the point of intersection of the graphs, it is necessary to determine the interval in which all the points of one of the graphs lie above (below 0 points of the second.
Answer: x>0.
(write on the board)
, is the vertex of the parabola.
.