1) Basic concept of inequality
2) Basic properties numerical inequalities. Inequalities containing a variable.
3) Graphic solution inequalities of the second degree
4) Systems of inequalities. Inequalities and systems of inequalities with two variables.
5) Solving rational inequalities using the interval method
6) Solving inequalities containing a variable under the modulus sign
1. Basic concept of inequality
Inequality is a relationship between numbers (or any mathematical expression that can take numerical value), indicating which one is larger or smaller than the other. The following actions can be performed on these expressions according to certain rules: addition, subtraction, multiplication and division (and when multiplying or dividing N. by a negative number its meaning is reversed). One of the main concepts linear programming — linear inequalities kind
a 1 x 1 + a 2 x 2 +... + a n x n * b,
Where a 1 ,..., a n, b- constants and the * sign is one of the inequality signs, for example. ≥,
algebraic
· transcendental
Algebraic inequalities are divided into inequalities of the first, second, etc. degrees.
The inequality is algebraic, of the second degree.
Inequality is transcendental.
2. Basic properties of numerical inequalities. Inequalities involving a variable
1) Schedule quadratic function y = ax 2 + bx + c is a parabola with branches pointing upward if a > 0, and down if a (sometimes they say that a parabola is convexly directed downward if a > 0 and convex upward if A). In this case, three cases are possible:
2) The parabola intersects the 0x axis (i.e. the equation ax 2 + bx + c = 0 has two various roots). That is, if a
y = ax 2 + bx + ca>0 D>0 y = ax 2 + bx + ca D>0,
A parabola has a vertex on the 0x axis (i.e., equation ax 2 + x + c = 0 has one root, the so-called double root) That is, if d = 0, then for a>0 the solution to the inequality is the entire number line, and for a ax 2 + x + c
y = ax 2 + bx + ca>0 D= 0 y = ax 2 + bx + ca D=0,
3) If d0 and below it at a
y = ax 2 + bx + ca>0 D0 y = ax 2 + bx + ca D 0,
4) Solve the inequality graphically
1. Let f(x) = 3x 2 -4x - 7 then find those x for which f(x) ;
2. Let's find the zeros of the function.
f(x) at x.
The answer is f(x) at x.
Let f(x)=x 2 +4x +5 then Let us find such x for which f(x)>0,
D=-4 No zeros.
4. Systems of inequalities. Inequalities and systems of inequalities with two variables
1) The set of solutions to a system of inequalities is the intersection of the sets of solutions to the inequalities included in it.
2) The set of solutions to the inequality f(x;y)>0 can be graphically represented on coordinate plane. Typically, the line defined by the equation f(x;y) = 0 divides the plane into 2 parts, one of which is the solution to the inequality. To determine which part, you need to substitute the coordinates of an arbitrary point M(x0;y0) that does not lie on the line f(x;y)=0 into the inequality. If f(x0;y0) > 0, then the solution to the inequality is the part of the plane containing the point M0. if f(x0;y0)
3) The set of solutions to a system of inequalities is the intersection of the sets of solutions to the inequalities included in it. Let, for example, be given a system of inequalities:
For the first inequality, the set of solutions is a circle of radius 2 and centered at the origin, and for the second, it is a half-plane located above the straight line 2x+3y=0. The set of solutions of this system is the intersection of these sets, i.e. semicircle.
4) Example. Solve the system of inequalities:
The solution to the 1st inequality is the set , the 2nd is the set (2;7) and the third is the set .
The intersection of these sets is the interval (2;3], which is the set of solutions to the system of inequalities.
5. Solving rational inequalities using the interval method
The interval method is based on the following property of the binomial ( Ha): dot x=α divides the number line into two parts - to the right of the point α binomial (x‑α)>0, and to the left of the point α (x-α) .
Suppose we need to solve the inequality (x-α 1)(x-α 2)...(x-α n)>0, where α 1, α 2 ...α n-1, α n are fixed numbers, among which there are no equals, and such that α 1 (x-α 1)(x-α 2)...(x‑ α n)>0 using the interval method, proceed as follows: the numbers α 1, α 2 ...α n-1, α n are plotted on the numerical axis; in the interval to the right of the largest of them, i.e. numbers αn, put a “plus” sign, in the interval following it from right to left put a “minus” sign, then a “plus” sign, then a “minus” sign, etc. Then the set of all solutions to the inequality (x-α 1)(x‑α 2)...(x-α n)>0 will be the union of all intervals in which the plus sign is placed, and the set of solutions to the inequality (x-α 1)(x-α 2)...(x‑α n) will be the union of all intervals in which the minus sign is placed.
1) The solution of rational inequalities (i.e. inequalities of the form P(x) Q(x) where are polynomials) is based on the following property of a continuous function: if a continuous function vanishes at points x1 and x2 (x1;x2) and between these points has no other roots, then in the intervals (x1; x2) the function retains its sign.
Therefore, to find intervals of constant sign of the function y=f(x) on the number line, mark all the points at which the function f(x) vanishes or suffers a discontinuity. These points divide the number line into several intervals, inside each of which the function f(x) is continuous and does not vanish, i.e. saves the sign. To determine this sign, it is enough to find the sign of the function at any point of the considered interval of the number line.
2) To determine intervals of constant sign of a rational function, i.e. To solve a rational inequality, we mark on the number line the roots of the numerator and the roots of the denominator, which are also the roots and breakpoints of the rational function.
Solving inequalities using the interval method
Solution. The range of acceptable values is determined by the system of inequalities:
For function f(x)= - 20. Find f(x):
where x= 29 and x = 13.
f(30) = - 20 = 0,3 > 0,
f(5) = - 1 - 20 = - 10
Answer: }
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