Let two utility functions be given
U(x) and U* (x) = h + y U(x) with d > 0.
The decision maker comes to the result A i h A2 on the basis of the second utility function when studying two alternatives. What would change if it instead focused on the first utility function?
What would your answer look like if the second utility function had the form U*(x) = h - y and (i) with y > 0?
How are the alternatives ordered when U*(x) = h?
* *
"To
1. Two utility functions lead to acceptance identical solutions when they can be mutually “translated” into each other through a positive linear transformation (see also p. 74 on this subject). If we can show that U(x) is a positive linear transformation of the function U*(x), then the choice of utility function will have no effect on the ordering of alternatives. We are looking for two numbers a and b for b > 0, so that it is true
a + bU*(x) = U(x).
If we substitute the second utility function, then we have
a + b (h + gU(x)) = U(x).
At the first stage, we define 6 in such a way that the factor by which U(x) is multiplied takes on the value of one. Obviously, we must denote b = 1 /d. Thus it turns out
a + - + U (x) = U (.r). 9
After this, we must choose a so that only U(x) remains on both sides of the equation. This will happen when a = -h/g.
Now we are looking for shape transformation
a + b(h-gU(x)) = U(x).
To obtain desired result, we must denote b = - - l/h. This would be a negative linear transformation and would reverse the rank order.
A decision maker given this utility function evaluates all alternatives with the same value. Therefore, when making a choice between alternatives A\ and A.2, it should come to the result A i ~
More on topic 2.1.5. Uniqueness of the utility function:
- 1. Consumer preferences and marginal utility. Utility function.
- 2.3.2. Quadratic utility function and expected utility
- Utility and the rational consumer. Total and marginal utility. Law of Diminishing Marginal Utility. The principle of utility maximization
- Quantitative utility theory. Concepts of utility, consumer choice, total and marginal utility.
If a rule is specified according to which a certain number u is associated with each point M of the plane (or some part of the plane), then it is said that on the plane (or on part of the plane, “a point function is given”; the definition of the function is symbolically expressed by an equality of the form u - The number u associated with a point M is called the value of this function at point M. For example, if A is a fixed point in the plane, M is arbitrary point, then the distance from A to M is a function of point M. B in this case f(M) = AM.
Let some function u = f(M) be given and at the same time a coordinate system be introduced. Then an arbitrary point M is determined by the coordinates x, y. Accordingly, the value of this function at point M is determined by the coordinates x, y, or, as they also say, u = f(M) is a function of two variables x and y. A function of two variables x, y is denoted by the symbol f(x, y); if f(M) = f(x, y) then the formula u = f(x, y) is called the expression of this function in the chosen coordinate system. So, in the previous example f(M)=AM; if you introduce Cartesian rectangular system coordinates with the origin at point A, we obtain the expression for this function:
u = √(x 2 + y 2)
146. Given two points P and Q, the distance between them is a, and the function f(M) = d 2 1 - d 2 2, where d 1 - MP and d 2 - MQ. Determine the expression of this function if point P is taken as the origin of coordinates and the Ox axis is directed along the segment PQ.
147. Under the conditions of problem 146, determine the expression of the function f(M) (directly and using coordinate transformation, using the result of problem 146), if:
1) the origin of coordinates is chosen in the middle of the segment PQ, the Ox axis is directed along the segment PQ.
2) the origin of coordinates is chosen at point P, and the Ox axis is directed along the segment QP.
148. Given: square ABCD with side a and function f(M) = d 2 1 - d 2 2 - d 2 3 + d 2 4, where d 1 = MA, d 2 = MB, d 3 = MC and d 4 = MD. Determine the expression of this function if the diagonals of the square are taken as the coordinate axes (and the Ox axis is directed along the segment AC, the Oy axis is directed along the segment BD).
149. Under the conditions of problem 148, determine the expression for f(M) (directly and using coordinate transformation, using the result of problem 148), if the origin of coordinates is chosen at point A, and the coordinate axes are directed along its sides (the Ox axis is along the segment AB, Oy axis - along segment AD).
150. Given the function f(x, y) = x 2 + y 2 - 6x + 8y. Determine the expression of this function in a new coordinate system if the origin of coordinates is moved (without changing the direction of the axes) to point O"(3; -4).
151. Given the function f(x, y) = x 2 - y 2 - 16. Determine the expression of this function in the new coordinate system if the coordinate axes are rotated at an angle of -45°.
152. Given a function f(x, y) = x 2 + y 2 . Determine the expression of this function in the new coordinate system if the coordinate axes are rotated by a certain angle α.
153. Find a point such that when the origin of coordinates is transferred to it, the expression of the function f(x,y) = x 2 - 4y 2 - 6x + 8y + 3 after the transformation does not contain terms of the first degree with respect to new variables.
154. Find a point such that when the origin of coordinates is transferred to it, the expression of the function f(x, y) = x 2 - 4xy + 4y 2 + 2x + y - 7 does not contain terms of the first degree with respect to new variables.
155. By what angle should the coordinate axes be rotated so that the expression of the function f (x, y) = x 2 - 2xy + y 2 - 6x + 3 after the transformation does not contain a term with the product of new variables?
156. By what angle should the coordinate axes be rotated so that the expression of the function f(x, y) = 3x 2 + 2√3xy + y 2 after the transformation does not contain a term with the product of new variables?
2. Functions. The simplest properties of functions 21 2.11. Prove that if f (x) is a periodic function with period T, then the function f (ax) is also periodic with period T /a. Solution. Indeed, f = f (ax + T) = f (ax), i.e. T /a is one of the periods of the function f (ax). 2.12. Find the period of the function f (x) = cos2 x. 1 + cos 2x Solution. We can write: cos2 x = . We see that period 2 cos functions 2 x is the same as the period of the cos 2x function. Since the period of the function cos x is equal to 2π, then according to Problem 2.11 the period of the function cos 2x is equal to π. 2.13. Find the period of the functions: a) f (x) = sin 2πx; b) f (x) = | cos x|. Answer: a) T = 1; b) T = π. Tasks for independent decision 2.14. Let f (x) = x2 and ϕ(x) = 2x. Find: a) f [ϕ(x)], b) ϕ. 2.15. Find f (x + 1) if f (x − 1) = x2. 1 2.16. The function f (x) = is given. 1−x Find ϕ(x) = f (f ). 2.17. Given a function f (x) = 3x2 − 4x − 2. Prove that the function f (2x + 1) can be represented as f (2x+1) = = Ax2 + Bx + C. Find the values of the constants A, B, C 2.18. Given two linear functions f1 (x) = 5x + 4 and f2 (x) = 3x − 1. Prove that the function f (x) = f2 is also linear, that is, it has the form f (x) = Ax + B. Find the values of the con- stant A and B. 3x + 7 5x + 4 2.19. Given two functions f1 (x) = and f2 (x) = , 5x + 6 2x − 8 called fractional linear. Prove that the function f (x) = f1 is also fractional linear, that is, it has the form Ax + B f (x) = . Give the values of the constants A, B, C, D. Cx + D 22 Introduction to mathematical analysis 2.20. For some function f: X ⊂ R → Y ⊂ R, it is known that f (3x + 5) = 45x2 − 12x + 3. Prove that the function f (x) can be represented as f (x) = Ax2 + Bx + C. Find the values of the constants A, B, C. 2.21. Find the domain of definition following functions: √ 2+x a) f (x) = x + 1; b) f (x) = lg ; √ 2−x c) f (x) = 2 + x − x2 ; d) f (x) = arcsin(log2 x); 1 + x2 d) f (x) = cos(sin x) + arcsin. 2x 2.22. Find the domain of definition of the following functions: √ 1 a) f (x) = x2 + 33x + 270; b) f (x) = 2; x + 26x + 168 x+2 c) f (x) = log[(1 + x)(12 − x)]; d) f (x) = arcsin; x−6 d) f (x) = (x + 9)(x + 8)(x − 14); 15 f) f (x) = arcsin; x − 11 −x f) f (x) = x2 + 13x + 42 + arcsin . 13 2.23. Construct the domain of definition of the following functions: a) f (x, y) = log2 (x + y); √ b) f (x, y) = x2 − 4 + 4 − y 2 ; x2 + y 2 c) f (x, y) = arcsin; 4 √ g) f (x, y) = xy. 2.24. Find the domain of definition of the following functions: 1 − log x 3 − 2x arcsin a) f (x) = 1 ; b) f (x) = √ 5 . √ x2 − 4x 3−x 2. Functions. The simplest properties of functions 23 2.25. Find and construct the domain of definition of the following functions: 4x − y 2 a) f (x, y) = ; log(1 − x2 − y 2) x2 + 2x + y 2 b) f (x, y) = . x2 − 2x + y 2 2.26. Prove that the functions 2 2x + 2−x a) f1 (x) = 2−x , f2 (x) = , 2 f3 (x) = |x + 1| + |x − 1| even; 2x − 2−x 3x + 1 b) ϕ1 (x) = , ϕ2 (x) = x , 2 3 −1 1+x ϕ3 (x) = lg odd; 1−x 2 c) ψ1 (x) = sin x − cos x, ψ2 (x) = 2x−x, ψ3 (x) = x3 + x2 − 2 of general form. 2.27. The functions are given: 1 a) y = sin2 x; b) y = sin x2 ; c) y = 1 + tan x; d) y = sin. x Which of them are periodic? 2x 2.28. Prove that the function y = has an inverse, 1 + 2x, and find it. 2.29. Prove that the function y = x2 − 2x has two inverses: y1 = 1 + x + 1 and y2 = 1 − x + 1. 2.30. Prove that the following functions are bounded from below: a) f1 (x) = x6 − 6x4 + 11x2 ; b) f2 (x) = x4 − 8x3 + 22x2. 2.31. Prove that the following functions are bounded from above: 1 5 a) f1 (x) = √ ; b) f1 (x) = √ . 4x2 − 16x + 36 5x 2 − 10x + 55 24 Introduction to Calculus 2.32. Find the smallest and highest value the following functions: a) f1 (x) = 3 sin x + 4 cos x; b) f2 (x) = 5 sin x + 12 cos x. 2.33. Describe the form of the graph of the following functions: a) z = 1 − x2 − y 2 ; b) z = x2 + y 2 ; c) z = x2 + y 2 ; d) z = x2 − y 2 . 2.34. Draw level lines for these functions, giving z values from −3 to +3 through 1: a) z = xy; b) z = y(x2 + 1). 2.35. Graph the function y = 2 −3(x + 1) − 0.5 s √ by transforming the graph of the function y = x. 2.36. Plot the graph of the function y = 3 sin(2x − 4) by transforming the graph of the function y = sin x. 2.37. Using basic function research (without using derivatives), plot graphs of the following functions: 1 x a) y = 2 ; b) y = 2; x +1 x +1 1 c) y = x4 − 2x2 + 5; d) y = 2; x + 4x + 5 2x − 5 d) y = ; e) y = x2 + 6x + 9 + 10. x−3 2.38. Plot graphs of the following functions: x, if − ∞< x < 1;
1 1
а) f (x) = x + , если 1 ≤ x ≤ 3;
2
2
4, если 3 < x < +∞;
б) f (x) = |x − 1| + |x + 3|;
в) f (x) = |x2 − 2x + 1|;
г) f (x) = sin x + | sin x|, если 0 ≤ x ≤ 3π;
д) f (x) = arccos(cos x);
t+5 t+1
е) f (t) = ; ж) f (t) = .
t−7 t2 + 2t + 2
3. Предел функции 25
3. Предел функции
Рекомендуется по textbook study subsections 1.4 and 1.5. Should be paid Special attention to subsection 1.4 and know all types of neighborhoods, their designations and forms of writing in the form of inequalities. The statement lim f (x) = A means: for any neighborhood x→x0 of the element A (in particular, arbitrarily small) of the element A there is a punctured neighborhood V (x0) of the element x0 such that from the condition x ∈ V˙ (x0) ∩ X follows, f (x) ∈ U (A), where X is the domain of definition of the function f (x), and x0 is the limit point of the set X. Often, instead of an arbitrary neighborhood U (A), one considers symmetric neighborhood Uε (A). In this case, the neighborhood ˙ V (x0) can turn out to be either symmetric or asymmetric, but from any asymmetric neighborhood it is possible to select a symmetric neighborhood Vδ (x0). Since the neighborhood V (x0) is punctured, i.e. does not contain the point x0, then x = x0, and at the point x0 the function f (x) may not be defined. To prove that lim f (x) = A, it is enough to find x→x0 the set (x) of those values of x for which the inclusion f (x) ⊂ U (A) holds for any neighborhood U (A). If the found set (x) is a neighborhood of x0, then the statement lim f (x) = A is true, in otherwise it x→x0 is false. In particular, if the function f (x) at the point x0 is defined and lim f (x) = f (x0), then the set (x) will also contain x→x0 the point x0. The given definition of a limit is applicable for any class of functions. In this section we will mainly deal with numerical functions one numeric argument. 3.1. Based on the definition of the limit, prove: 1 1 a) lim x = x0 ; b) lim = ; x→x0 x→2 x 2 1 1 1 c) lim = lim = lim = 0; x→+∞ x x→−∞ x x→∞ x 26 Introduction to mathematical analysis 1 1 d) lim = +∞; e) lim = −∞; x→0+0 x x→0−0 x 1 f) lim = 2; g) lim x2 = 4. x→1 x x→2 Solution: a) the statement lim x = x0 directly x→x0 follows from the definition of the limit. If the neighborhood Uε (x0) ˙ (|x − x0 |< ε) дана, то в качестве окрестности Vδ (x0) можно
принять |x − x0 | < δ = ε, т.е. положить δ = ε;
1 1
б) докажем, что lim = . По определению предела
x→2 x 2
мы должны доказать, что для любой заданной окрестности
1 ˙
Uε , ε >0 there is a neighborhood V (2) such that if 2 ˙ 1 1 1 1 x ∈ V (2), then −< ε, т.е. ∈ Uε , что равносильно сле-
x 2 x 2
дующим двум неравенствам:
1 1
−ε < − < +ε
или x 2
1 1 1
− ε < < + ε.
2 x 2
Так как при достаточ-
но малом ε все части
этого неравенства по-
ложительны, то
2 2