Parity and oddness of a function are one of its main properties, and the study of a function for parity occupies an impressive part of the school course in mathematics. It largely determines the nature of the behavior of the function and greatly facilitates the construction of the corresponding graph.
Define the parity of the function. Generally speaking, the function under investigation is considered even if, for opposite values of the independent variable (x) located in its domain of definition, the corresponding values of y (function) turn out to be equal.
We give a more rigorous definition. Consider some function f (x), which is given in the domain D. It will be even if, for any point x in the domain of definition:
- -x (opposite point) also lies in this area of definition,
The above definition implies the condition necessary for the domain of definition of such a function, namely, symmetry with respect to the point O, which is the origin, since if some point b is contained in the domain of definition of an even function, then the corresponding point - b also lies in this region. Thus, the conclusion follows: the even function has a symmetrical form with respect to the ordinate axis (Oy).
How to determine the parity of a function in practice?
Let the functional dependence be defined using the formula h (x) = 11 ^ x + 11 ^ (- x). Following the algorithm that follows directly from the definition, we first of all examine its domain of definition. Obviously, it is defined for all values of the argument, that is, the first condition is satisfied.
The next step is to replace the argument (x) with its opposite value (-x).
We get:
h (-x) = 11 ^ (- x) + 11 ^ x.
Since addition satisfies the commutative (moving) law, it is obvious that h (-x) = h (x) and the given functional dependence is even.
Let us verify the parity of the function h (x) = 11 ^ x-11 ^ (- x). Following the same algorithm, we obtain that h (-x) = 11 ^ (- x) -11 ^ x. Taking out the minus, as a result, we have
h (-x) = - (11 ^ x-11 ^ (- x)) = - h (x). Therefore, h (x) is odd.
By the way, it should be recalled that there are functions that cannot be classified according to these criteria; they are called either odd or even.
Even functions have a number of interesting properties:
- as a result of the addition of such functions, they get even;
- as a result of subtracting such functions, they get even;
- the inverse of the even function is also even;
- as a result of multiplying two such functions, they get even;
- as a result of multiplying the odd and even functions, they get the odd;
- as a result of the division of the odd and even functions, they get the odd;
- the derivative of such a function is odd;
- if you square an odd function, we get an even one.
The parity of the function can be used in solving equations.
To solve an equation of type g (x) = 0, where the left side of the equation is an even function, it will be quite enough to find its solutions for non-negative values of the variable. The obtained roots of the equation must be combined with opposite numbers. One of them is subject to verification.
The same property of the function is successfully used to solve non-standard problems with a parameter.
For example, is there any value of the parameter a for which the equation 2x ^ 6-x ^ 4-ax ^ 2 = 1 will have three roots?
If we take into account that the variable enters the equation in even degrees, then it is clear that replacing x with - x the given equation will not change. It follows that if a certain number is its root, then it is the opposite number. The conclusion is obvious: the roots of the equation, non-zero, are included in the set of its solutions in pairs.
It is clear that the number 0 itself is not the root of the equation , that is, the number of roots of such an equation can only be even and, of course, for any value of the parameter it cannot have three roots.
But the number of roots of the equation 2 ^ x + 2 ^ (- x) = ax ^ 4 + 2x ^ 2 + 2 can be odd, and for any parameter value. Indeed, it is easy to verify that the set of roots of this equation contains solutions in pairs. Check if 0 is the root. When substituting it in the equation, we get 2 = 2. Thus, in addition to "paired" 0 is also the root, which proves their odd number.