Home Salon Prove that a function is even. Examples. Even and odd functions. Sufficient conditions for the existence of an extremum

Prove that a function is even. Examples. Even and odd functions. Sufficient conditions for the existence of an extremum

Which were familiar to you to one degree or another. It was also noted there that the stock of function properties will be gradually replenished. Two new properties will be discussed in this section.

Definition 1.

The function y = f(x), x є X, is called even if for any value x from the set X the equality f (-x) = f (x) holds.

Definition 2.

The function y = f(x), x є X, is called odd if for any value x from the set X the equality f (-x) = -f (x) holds.

Prove that y = x 4 is an even function.

Solution. We have: f(x) = x 4, f(-x) = (-x) 4. But(-x) 4 = x 4. This means that for any x the equality f(-x) = f(x) holds, i.e. the function is even.

Similarly, it can be proven that the functions y - x 2, y = x 6, y - x 8 are even.

Prove that y = x 3 ~ an odd function.

Solution. We have: f(x) = x 3, f(-x) = (-x) 3. But (-x) 3 = -x 3. This means that for any x the equality f (-x) = -f (x) holds, i.e. the function is odd.

Similarly, it can be proven that the functions y = x, y = x 5, y = x 7 are odd.

You and I have already been convinced more than once that new terms in mathematics most often have an “earthly” origin, i.e. they can be explained somehow. This is the case with both even and odd functions. See: y - x 3, y = x 5, y = x 7 are odd functions, while y = x 2, y = x 4, y = x 6 are even functions. And in general, for any function of the form y = x" (below we will specifically study these functions), where n is a natural number, we can conclude: if n is an odd number, then the function y = x" is odd; if n is an even number, then the function y = xn is even.

There are also functions that are neither even nor odd. Such, for example, is the function y = 2x + 3. Indeed, f(1) = 5, and f (-1) = 1. As you can see, here, therefore, neither the identity f(-x) = f ( x), nor the identity f(-x) = -f(x).

So, a function can be even, odd, or neither.

The study of whether a given function is even or odd is usually called the study of parity.

Definitions 1 and 2 refer to the values ​​of the function at points x and -x. This assumes that the function is defined at both point x and point -x. This means that point -x belongs to the domain of definition of the function simultaneously with point x. If a numerical set X, together with each of its elements x, also contains the opposite element -x, then X is called a symmetric set. Let's say, (-2, 2), [-5, 5], (-oo, +oo) are symmetric sets, while \) .

Since \(x^2\geqslant 0\) , then the left side of the equation (*) is greater than or equal to \(0+ \mathrm(tg)^2\,1\) .

Thus, equality (*) can only be true when both sides of the equation are equal to \(\mathrm(tg)^2\,1\) . This means that \[\begin(cases) 2x^2+\mathrm(tg)^2\,1=\mathrm(tg)^2\,1 \\ \mathrm(tg)\,1\cdot \ mathrm(tg)\,(\cos x)=\mathrm(tg)^2\,1 \end(cases) \quad\Leftrightarrow\quad \begin(cases) x=0\\ \mathrm(tg)\, (\cos x)=\mathrm(tg)\,1 \end(cases)\quad\Leftrightarrow\quad x=0\] Therefore, the value \(a=-\mathrm(tg)\,1\) suits us .

Answer:

\(a\in \(-\mathrm(tg)\,1;0\)\)

Task 2 #3923

Task level: Equal to the Unified State Exam

Find all values ​​of the parameter \(a\) , for each of which the graph of the function \

symmetrical about the origin.

If the graph of a function is symmetrical with respect to the origin, then such a function is odd, that is, \(f(-x)=-f(x)\) holds for any \(x\) from the domain of definition of the function. Thus, it is required to find those parameter values ​​for which \(f(-x)=-f(x).\)

\[\begin(aligned) &3\mathrm(tg)\,\left(-\dfrac(ax)5\right)+2\sin \dfrac(8\pi a+3x)4= -\left(3\ mathrm(tg)\,\left(\dfrac(ax)5\right)+2\sin \dfrac(8\pi a-3x)4\right)\quad \Rightarrow\quad -3\mathrm(tg)\ ,\dfrac(ax)5+2\sin \dfrac(8\pi a+3x)4= -\left(3\mathrm(tg)\,\left(\dfrac(ax)5\right)+2\ sin \dfrac(8\pi a-3x)4\right) \quad \Rightarrow\\ \Rightarrow\quad &\sin \dfrac(8\pi a+3x)4+\sin \dfrac(8\pi a- 3x)4=0 \quad \Rightarrow \quad2\sin \dfrac12\left(\dfrac(8\pi a+3x)4+\dfrac(8\pi a-3x)4\right)\cdot \cos \dfrac12 \left(\dfrac(8\pi a+3x)4-\dfrac(8\pi a-3x)4\right)=0 \quad \Rightarrow\quad \sin (2\pi a)\cdot \cos \ frac34 x=0 \end(aligned)\]

The last equation must be satisfied for all \(x\) from the domain of definition \(f(x)\) , therefore, \(\sin(2\pi a)=0 \Rightarrow a=\dfrac n2, n\in\ mathbb(Z)\) .

Answer:

\(\dfrac n2, n\in\mathbb(Z)\)

Task 3 #3069

Task level: Equal to the Unified State Exam

Find all values ​​of the parameter \(a\) , for each of which the equation \ has 4 solutions, where \(f\) is an even periodic function with period \(T=\dfrac(16)3\) defined on the entire number line , and \(f(x)=ax^2\) for \(0\leqslant x\leqslant \dfrac83.\)

(Task from subscribers)

Since \(f(x)\) is an even function, its graph is symmetrical with respect to the ordinate axis, therefore, for \(-\dfrac83\leqslant x\leqslant 0\) \(f(x)=ax^2\) . Thus, for \(-\dfrac83\leqslant x\leqslant \dfrac83\) , and this is a segment of length \(\dfrac(16)3\), the function is \(f(x)=ax^2\) .

1) Let \(a>0\) . Then the graph of the function \(f(x)\) will look like this:


Then, in order for the equation to have 4 solutions, it is necessary that the graph \(g(x)=|a+2|\cdot \sqrtx\) pass through the point \(A\) :


Therefore, \[\dfrac(64)9a=|a+2|\cdot \sqrt8 \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &9(a+2)=32a\\ &9 (a+2)=-32a\end(aligned)\end(gathered)\right. \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &a=\dfrac(18)(23)\\ &a=-\dfrac(18)(41) \end(aligned) \end( gathered)\right.\] Since \(a>0\) , then \(a=\dfrac(18)(23)\) is suitable.

2) Let \(a0\) ). If the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Therefore, you need: \[\begin(cases) 12-a>0\\-(a-10)>0\end(cases)\quad\Leftrightarrow\quad a 0 on (x_(1); x_(2) ) \cup (x_(3); +\infty)

Intervals where the function is negative, that is, f(x)< 0 - отрезки оси абсцисс, которые отвечают точкам графика функции, лежащих ниже оси абсцисс.

f(x)< 0 на (-\infty; x_{1}) \cup (x_{2}; x_{3})

Limited function

A function y=f(x), x \in X is usually called bounded below when there is a number A for which the inequality f(x) \geq A holds for any x \in X .

An example of a function bounded from below: y=\sqrt(1+x^(2)) since y=\sqrt(1+x^(2)) \geq 1 for any x .

A function y=f(x), x \in X is called bounded above if there is a number B for which the inequality f(x) \neq B holds for any x \in X .

An example of a function bounded from below: y=\sqrt(1-x^(2)), x \in [-1;1] since y=\sqrt(1+x^(2)) \neq 1 for any x \ in [-1;1] .

A function y=f(x), x \in X is usually called bounded when there is a number K > 0 for which the inequality \left | f(x)\right | \neq K for any x \in X .

An example of a bounded function: y=\sin x is bounded on the entire number line, since \left | \sin x \right | \neq 1 .

Increasing and decreasing function

It is customary to speak of a function that increases over the interval under consideration as an increasing function when a larger value of x corresponds to a larger value of the function y=f(x) . It follows that taking two arbitrary values ​​of the argument x_(1) and x_(2) from the interval under consideration, with x_(1) > x_(2) , the result will be y(x_(1)) > y(x_(2)).

A function that decreases on the interval under consideration is called a decreasing function when a larger value of x corresponds to a smaller value of the function y(x) . It follows that, taking from the interval under consideration two arbitrary values ​​of the argument x_(1) and x_(2) , and x_(1) > x_(2) , the result will be y(x_(1))< y(x_{2}) .

The roots of a function are usually called the points at which the function F=y(x) intersects the abscissa axis (they are obtained by solving the equation y(x)=0).

a) If for x > 0 an even function increases, then it decreases for x< 0

b) When an even function decreases at x > 0, then it increases at x< 0

c) When an odd function increases at x > 0, then it also increases at x< 0

d) When an odd function decreases for x > 0, then it will also decrease for x< 0

Extrema of the function

The minimum point of the function y=f(x) is usually called a point x=x_(0) whose neighborhood will have other points (except for the point x=x_(0)), and for them then the inequality f(x ) > f(x_(0)) . y_(min) - designation of the function at the min point.

The maximum point of the function y=f(x) is usually called a point x=x_(0) whose neighborhood will have other points (except for the point x=x_(0)), and for them then the inequality f(x )< f(x^{0}) . y_{max} - обозначение функции в точке max.

Prerequisite

According to Fermat's theorem: f"(x)=0 when the function f(x) that is differentiable at the point x_(0) will have an extremum at this point.

Sufficient condition
  • When the derivative changes sign from plus to minus, then x_(0) will be the minimum point;
  • x_(0) - will be a maximum point only when the derivative changes sign from minus to plus when passing through the stationary point x_(0) .
  • The largest and smallest value of a function on an interval

    Calculation steps:

  • The derivative f"(x) is sought;
  • Stationary and critical points of the function are found and those belonging to the segment are selected;
  • The values ​​of the function f(x) are found at stationary and critical points and ends of the segment. The smaller of the results obtained will be the smallest value of the function, and the larger one will be the largest.
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