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How to prove that a function is odd. Even and odd functions. Algorithm for studying the function y = f(x) for parity |
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To do this, use graph paper or a graphing calculator. Select any number of independent variable values x (\displaystyle x) and plug them into the function to calculate the values of the dependent variable y (\displaystyle y). Plot the found coordinates of the points on the coordinate plane, and then connect these points to build a graph of the function.
Check whether the graph of the function is symmetrical about the Y axis. Symmetry means a mirror image of the graph relative to the ordinate. If the part of the graph to the right of the Y-axis (positive values of the independent variable) is the same as the part of the graph to the left of the Y-axis (negative values of the independent variable), the graph is symmetrical about the Y-axis. If the function is symmetrical about the y-axis, the function is even. Check whether the graph of the function is symmetrical about the origin. The origin is the point with coordinates (0,0). Symmetry about the origin means that a positive value y (\displaystyle y)(with a positive value x (\displaystyle x)) corresponds to a negative value y (\displaystyle y)(with a negative value x (\displaystyle x)), and vice versa. Odd functions have symmetry about the origin. Check if the graph of the function has any symmetry. The last type of function is a function whose graph has no symmetry, that is, there is no mirror image both relative to the ordinate axis and relative to the origin. For example, given the function .
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. We have already seen 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 satisfied when both sides of the equation are equal to \(\mathrm(tg)^2\,1\) . And 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 about 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 \(f(x)\) , 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, when \(-\dfrac83\leqslant x\leqslant 0\)\(f(x)=ax^2\) . Thus, when \(-\dfrac83\leqslant x\leqslant \dfrac83\), and this is a segment of length \(\dfrac(16)3\) , function \(f(x)=ax^2\) . 1) Let \(a>0\) . Then the graph of the function \(f(x)\) will look like this: 2) Let \(a<0\)
. Тогда картинка окажется симметричной относительно начала координат: 3) The case when \(a=0\) is not suitable, since then \(f(x)=0\) for all \(x\) , \(g(x)=2\sqrtx\) and the equation will have only 1 root. Answer: \(a\in \left\(-\dfrac(18)(41);\dfrac(18)(23)\right\)\) Task 4 #3072 Task level: Equal to the Unified State Exam Find all values of \(a\) , for each of which the equation \ has at least one root. (Task from subscribers) Let's rewrite the equation in the form \
and consider two functions: \(g(x)=7\sqrt(2x^2+49)\) and \(f(x)=3|x-7a|-6|x|-a^2+7a\ ) . In order for the equation to have at least one solution, it is necessary that the graphs of the functions \(f\) and \(g\) have at least one intersection point. Therefore, you need: \ \\] Answer: \(a\in \(-7\)\cup\) Task 5 #3912 Task level: Equal to the Unified State Exam Find all values of the parameter \(a\) , for each of which the equation \ has six different solutions. Let's make the replacement \((\sqrt2)^(x^3-3x^2+4)=t\) , \(t>0\) . Then the equation will take the form \
We will gradually write out the conditions under which the original equation will have six solutions. Thus, the solution plan becomes clear. Let's write down the conditions that must be met point by point. 1) For the equation \((*)\) to have two different solutions, its discriminant must be positive: \ 2) It is also necessary that both roots be positive (since \(t>0\) ). If the product of two roots is positive and their sum is positive, then the roots themselves will be positive. Therefore, you need: \[\begin(cases) 12-a>0\\-(a-10)>0\end(cases)\quad\Leftrightarrow\quad a<10\] Thus, we have already provided ourselves with two different positive roots \(t_1\) and \(t_2\) . 3)
Let's look at this equation \
For what \(t\) will it have three different solutions? Thus, we have determined that both roots of the equation \((*)\) must lie in the interval \((1;4)\) . How to write this condition? had four different roots, different from zero, representing, together with \(x=0\), an arithmetic progression. Note that the function \(y=25x^4+25(a-1)x^2-4(a-7)\) is even, which means that if \(x_0\) is the root of the equation \((*)\ ) , then \(-x_0\) will also be its root. Then it is necessary that the roots of this equation be numbers ordered in ascending order: \(-2d, -d, d, 2d\) (then \(d>0\)). It is then that these five numbers will form an arithmetic progression (with the difference \(d\)). For these roots to be the numbers \(-2d, -d, d, 2d\) , it is necessary that the numbers \(d^(\,2), 4d^(\,2)\) be the roots of the equation \(25t^2 +25(a-1)t-4(a-7)=0\) . Then, according to Vieta’s theorem: Let's rewrite the equation in the form \
and consider two functions: \(g(x)=20a-a^2-2^(x^2+2)\) and \(f(x)=13|x|-2|5x+12a|\) . In order for the equation to have at least one solution, it is necessary that the graphs of the functions \(f\) and \(g\) have at least one intersection point. Therefore, you need: \
Solving this set of systems, we get the answer: \\]
Answer: \(a\in \(-2\)\cup\) The dependence of a variable y on a variable x, in which each value of x corresponds to a single value of y is called a function. For designation use the notation y=f(x). Each function has a number of basic properties, such as monotonicity, parity, periodicity and others. Take a closer look at the parity property. A function y=f(x) is called even if it satisfies the following two conditions: 2. The value of the function at point x, belonging to the domain of definition of the function, must be equal to the value of the function at point -x. That is, for any point x, from the domain of definition of the function the following equality must be satisfied: f(x) = f(-x). If you plot a graph of an even function, it will be symmetrical about the Oy axis. For example, the function y=x^2 is even. Let's check it out. The domain of definition is the entire numerical axis, which means it is symmetrical about point O. Let's take an arbitrary x=3. f(x)=3^2=9. f(-x)=(-3)^2=9. Therefore f(x) = f(-x). Thus, both conditions are met, which means the function is even. Below is a graph of the function y=x^2. The figure shows that the graph is symmetrical about the Oy axis. A function y=f(x) is called odd if it satisfies the following two conditions: 1. The domain of definition of a given function must be symmetrical with respect to point O. That is, if some point a belongs to the domain of definition of the function, then the corresponding point -a must also belong to the domain of definition of the given function. 2. For any point x, the following equality must be satisfied from the domain of definition of the function: f(x) = -f(x). The graph of an odd function is symmetrical with respect to point O - the origin of coordinates. For example, the function y=x^3 is odd. Let's check it out. The domain of definition is the entire numerical axis, which means it is symmetrical about point O. Let's take an arbitrary x=2. f(x)=2^3=8. f(-x)=(-2)^3=-8. Therefore f(x) = -f(x). Thus, both conditions are met, which means the function is odd. Below is a graph of the function y=x^3. The figure clearly shows that the odd function y=x^3 is symmetrical about the origin. |
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