Parity of the function

The parity and oddness of a function are one of its main properties, and the study of the function on parity takes an impressive part of the school course in mathematics. It determines in many ways the behavior of the function and greatly facilitates the construction of the corresponding schedule.

Let us determine the parity of the function. Generally speaking, the function being examined is considered even if the corresponding values of y (functions) are equal for opposite values of the independent variable (x) in its domain of definition.

We give a more rigorous definition. Consider some function f (x) that is defined in D. It will be even if for any point x in the domain of definition:

• -x (the opposite point) also lies in this domain of definition,

• F (-x) = f (x).

From the above definition follows the condition necessary for the domain of definition of such a function, namely, the 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. From the above, thus, the conclusion follows: the even function has the form symmetric with respect to the axis of ordinates (Oy).

How to determine in practice the parity of a function?

Let the functional dependence be given by 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 the addition satisfies the commutative (moving) law, it is obvious that h (-x) = h (x) and the given functional dependence is even.

We 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. Carrying a minus, in the end, we have
H (-x) = - (11 ^ x-11 ^ (-x)) = -h (x). Consequently, h (x) is odd.

By the way, it should be recalled that there are functions that can not be classified according to these characteristics, they are called neither even nor odd.

Even functions have a number of interesting properties:

• As a result of the addition of such functions, an even number is obtained;
• As a result of subtraction of such functions, an even result is obtained;
• The inverse of the even function is also even;
• As a result of the multiplication of two such functions, an even number is obtained;
• As a result of multiplying the odd and even functions get odd;
• As a result of division of odd and even functions get odd;
• The derivative of such a function is odd;
• If we raise the odd function to a square, we get an even function.

The parity of a function can be used to solve equations.

To solve an equation of the type g (x) = 0, where the left side of the equation is an even function, it will be sufficient to find its solutions for non-negative values of the variable. The roots of the equation must be combined with the opposite numbers. One of them is subject to verification.

The same property of the function is successfully used to solve non-standard tasks 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 powers, then it is clear that replacing x by -x the given equation does not change. Hence it follows that if some number is its root, then it is the opposite number. The conclusion is obvious: the roots of the equation, other than zero, enter into the set of its solutions "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, naturally, for any value of the parameter it can not 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 value of the parameter. Indeed, it is easy to verify that the set of roots of the given equation contains solutions "in pairs". We verify that 0 is a root. When we substitute it into the equation, we get 2 = 2. Thus, in addition to "paired" 0 also is a root, which proves their odd number.