Permissible range (ODZ), theory, examples, solutions. Whether to find ODZ? Find set of function values online with solution
Let's start by finding domain of definition of the sum of functions. It is clear that such a function makes sense for all such values of the variable for which all the functions that make up the sum make sense. Therefore, there is no doubt about the validity of the following statement:
If the function f is the sum of n functions f 1 , f 2 , …, f n , that is, the function f is given by the formula y=f 1 (x)+f 2 (x)+…+f n (x) , then the domain of the function f is the intersection of the domains of the functions f 1 , f 2 , …, f n . Let's write it as .
Let's agree to continue using records like the last one, by which we mean written inside a curly bracket, or the simultaneous fulfillment of any conditions. This is convenient and quite naturally resonates with the meaning of systems.
Example.
Given a function y=x 7 +x+5+tgx , and we need to find its domain.
Solution.
The function f is represented by the sum of four functions: f 1 is a power function with an exponent of 7 , f 2 is a power function with an exponent of 1 , f 3 is a constant function and f 4 is a tangent function.
Looking at the table of domains of definition of the basic elementary functions, we find that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) , D(f 3)=(−∞, +∞) , and the domain of the tangent is the set of all real numbers, except for the numbers .
The domain of the function f is the intersection of the domains of the functions f 1 , f 2 , f 3 and f 4 . It is quite obvious that this is the set of all real numbers, with the exception of the numbers .
Answer:
set of all real numbers except .
Let's move on to finding domains of the product of functions. For this case, a similar rule holds:
If the function f is the product of n functions f 1 , f 2 , …, f n , that is, the function f is given by the formula y=f 1 (x) f 2 (x) ... f n (x), then the domain of the function f is the intersection of the domains of the functions f 1 , f 2 , …, f n . So, .
It is understandable, in the indicated area all the functions of the product are defined, and hence the function f itself.
Example.
Y=3 arctgx lnx .
Solution.
The structure of the right side of the formula that defines the function can be considered as f 1 (x) f 2 (x) f 3 (x) , where f 1 is a constant function, f 2 is the arc tangent function, and f 3 is the logarithmic function with base e.
We know that D(f 1)=(−∞, +∞) , D(f 2)=(−∞, +∞) and D(f 3)=(0, +∞) . Then .
Answer:
the domain of the function y=3 arctgx lnx is the set of all real positive numbers.
Let us dwell separately on finding the domain of definition of the function given by the formula y=C·f(x) , where C is some real number. It is easy to show that the domain of this function and the domain of the function f coincide. Indeed, the function y=C f(x) is the product of a constant function and a function f . The domain of a constant function is the set of all real numbers, and the domain of the function f is D(f) . Then the domain of the function y=C f(x) is , which was to be shown.
So, the domains of the functions y=f(x) and y=C·f(x) , where С is some real number, coincide. For example, if the domain of the root is , it becomes clear that D(f) is the set of all x from the domain of the function f 2 for which f 2 (x) is included in the domain of the function f 1 .
In this way, domain of a complex function y=f 1 (f 2 (x)) is the intersection of two sets: the set of all x such that x∈D(f 2) and the set of all x such that f 2 (x)∈D(f 1) . That is, in our notation (this is essentially a system of inequalities).
Let's take a look at a few examples. In the process, we will not describe in detail, as this is beyond the scope of this article.
Example.
Find the domain of the function y=lnx 2 .
Solution.
The original function can be represented as y=f 1 (f 2 (x)) , where f 1 is a logarithm with base e, and f 2 is a power function with exponent 2.
Turning to the known domains of definition of the basic elementary functions, we have D(f 1)=(0, +∞) and D(f 2)=(−∞, +∞) .
Then
So we found the domain of definition of the function we needed, it is the set of all real numbers except zero.
Answer:
(−∞, 0)∪(0, +∞) .
Example.
What is the scope of the function ?
Solution.
This function is complex, it can be considered as y \u003d f 1 (f 2 (x)) , where f 1 is a power function with exponent, and f 2 is the arcsine function, and we need to find its domain.
Let's see what we know: D(f 1)=(0, +∞) and D(f 2)=[−1, 1] . It remains to find the intersection of sets of values x such that x∈D(f 2) and f 2 (x)∈D(f 1) :
For arcsinx>0, let's recall the properties of the arcsine function. The arcsine increases over the entire domain of definition [−1, 1] and vanishes at x=0 , therefore, arcsinx>0 for any x from the interval (0, 1] .
Let's go back to the system:
Thus, the desired domain of definition of the function is a half-interval (0, 1] .
Answer:
(0, 1] .
Now let's move on to complex general functions y=f 1 (f 2 (…f n (x)))) . The domain of the function f in this case is found as .
Example.
Find the scope of a function .
Solution.
The given complex function can be written as y \u003d f 1 (f 2 (f 3 (x))), where f 1 - sin, f 2 - function of the root of the fourth degree, f 3 - lg.
We know that D(f 1)=(−∞, +∞) , D(f 2)=∪ .
All of this speaks to the importance of having a DHS.
Example 3
Find the ODZ expression x 3 + 2 x y − 4 .
Solution
Any number can be cubed. This expression does not have a fraction, so x and y can be anything. That is, ODZ is any number.
Answer: x and y are any values.
Example 4
Find the ODZ expression 1 3 - x + 1 0 .
Solution
It can be seen that there is one fraction, where the denominator is zero. This means that for any value of x, we will get a division by zero. This means that we can conclude that this expression is considered to be indefinite, that is, it does not have ODZ.
Answer: ∅ .
Example 5
Find the ODZ of the given expression x + 2 · y + 3 - 5 · x .
Solution
The presence of a square root indicates that this expression must be greater than or equal to zero. If it is negative, it has no meaning. Hence, it is necessary to write down an inequality of the form x + 2 · y + 3 ≥ 0 . That is, this is the desired range of acceptable values.
Answer: set of x and y , where x + 2 y + 3 ≥ 0 .
Example 6
Determine the ODZ expression of the form 1 x + 1 - 1 + log x + 8 (x 2 + 3) .
Solution
By condition, we have a fraction, so its denominator should not be equal to zero. We get that x + 1 - 1 ≠ 0 . The radical expression always makes sense when greater than or equal to zero, i.e. x + 1 ≥ 0 . Since it has a logarithm, its expression must be strictly positive, that is, x 2 + 3 > 0. The base of the logarithm must also be positive and different from 1 , then we add the conditions x + 8 > 0 and x + 8 ≠ 1 . From this it follows that the desired ODZ will take the form:
x + 1 - 1 ≠ 0 , x + 1 ≥ 0 , x 2 + 3 > 0 , x + 8 > 0 , x + 8 ≠ 1
In other words, it is called a system of inequalities with one variable. The solution will lead to such a record of the ODZ [ − 1 , 0) ∪ (0 , + ∞) .
Answer: [ − 1 , 0) ∪ (0 , + ∞)
Why is it important to take LHS into account when making changes?
For identical transformations, it is important to find the ODZ. There are cases when the existence of ODZ does not take place. To understand whether the solution has a given expression, you need to compare the ODZ of the variables of the original expression and the ODZ of the received expression.
Identity transformations:
- may not affect ODZ;
- may lead to an extension or addition to the DHS;
- can narrow the ODZ.
Let's look at an example.
Example 7
If we have an expression of the form x 2 + x + 3 · x , then its ODZ is defined on the entire domain of definition. Even with the reduction of similar terms and simplification of the expression, the ODZ does not change.
Example 8
If we take the example of the expression x + 3 x − 3 x , then things are different. We have a fractional expression. And we know that division by zero is not allowed. Then the ODZ has the form (− ∞ , 0) ∪ (0 , + ∞) . It can be seen that zero is not a solution, so we add it with a parenthesis.
Consider an example with the presence of a radical expression.
Example 9
If there is x - 1 · x - 3 , then you should pay attention to the ODZ, since it must be written as an inequality (x − 1) · (x − 3) ≥ 0 . It is possible to solve by the interval method, then we get that the ODZ will take the form (− ∞ , 1 ] ∪ [ 3 , + ∞) . After transforming x - 1 · x - 3 and applying the properties of the roots, we have that the ODZ can be supplemented and written down as a system of inequalities of the form x - 1 ≥ 0 , x - 3 ≥ 0 . When solving it, we obtain that [ 3 , + ∞) . Hence, the ODZ is written in full as follows: (− ∞ , 1 ] ∪ [ 3 , + ∞) .
Changes that narrow the DHS should be avoided.
Example 10
Consider an example of the expression x - 1 · x - 3 when x = - 1 . When substituting, we get that - 1 - 1 · - 1 - 3 = 8 = 2 2 . If this expression is transformed and brought to the form x - 1 x - 3, then when calculating we get that 2 - 1 2 - 3 the expression does not make sense, since the radical expression should not be negative.
Identical transformations should be followed, which will not change the DHS.
If there are examples that extend it, then it should be added to the DPV.
Example 11
Consider the example of a fraction of the form x x 3 + x. If we reduce by x , then we get that 1 x 2 + 1 . Then the ODZ expands and becomes (− ∞ 0) ∪ (0 , + ∞) . Moreover, when calculating, we are already working with the second simplified fraction.
In the presence of logarithms, the situation is slightly different.
Example 12
If there is an expression of the form ln x + ln (x + 3) , it is replaced by ln (x (x + 3)) , based on the property of the logarithm. This shows that the ODZ from (0 , + ∞) to (− ∞ , − 3) ∪ (0 , + ∞) . Therefore, to determine the ODZ ln (x (x + 3)) it is necessary to perform calculations on the ODZ, that is, (0 , + ∞) sets.
When solving, it is always necessary to pay attention to the structure and form of the expression given by the condition. If the domain of definition is found correctly, the result will be positive.
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Job type: 13
Condition
a) Solve the equation 2(\sin x-\cos x)=tgx-1.
b) \left[ \frac(3\pi )2;\,3\pi \right].
Show SolutionSolution
a) Opening the brackets and moving all the terms to the left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tg x \cos x, we obtain the equation 1+2 tan x \cos x-2 \cos x-tg x=0, which, by grouping, can be reduced to the form (1-tg x)(1-2 \cos x)=0.
1) 1-tgx=0, tanx=1, x=\frac\pi 4+\pi n, n \in \mathbb Z;
2) 1-2 \cos x=0, \cosx=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.
b) With the help of a numerical circle, we select the roots belonging to the interval \left[ \frac(3\pi )2;\, 3\pi \right].
x_1=\frac\pi 4+2\pi =\frac(9\pi )4,
x_2=\frac\pi 3+2\pi =\frac(7\pi )3,
x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.
Answer
a) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;
b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi )4.
Job type: 13
Topic: Tolerance Range (ODV)
Condition
a) Solve the Equation (2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0.
b) Indicate the roots of this equation that belong to the interval \left(0;\,\frac(3\pi )2\right] ;
Show SolutionSolution
a) ODZ: \begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)
The original equation on the ODZ is equivalent to the set of equations
\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right.
Let's solve the first equation. To do this, we will replace \cos 4x=t, t \in [-1; one]. Then \sin^24x=1-t^2. We get:
2(1-t^2)-3t=0,
2t^2+3t-2=0,
t_1=\frac12, t_2=-2, t_2\notin [-1; one].
\cos4x=\frac12,
4x=\pm \frac\pi 3+2\pi n,
x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.
Let's solve the second equation.
tg x=0,\, x=\pi k, k \in \mathbb Z.
Using the unit circle, we find solutions that satisfy the ODZ.
The sign "+" marks the 1st and 3rd quarters, in which tg x>0.
We get: x=\pi k, k \in \mathbb Z; x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.
b) Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right].
x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ; x=\frac(13\pi )(12); x=\frac(17\pi )(12).
Answer
a) \pi k, k \in \mathbb Z; \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z.
b) \pi; \frac\pi(12); \frac(5\pi )(12); \frac(13\pi )(12); \frac(17\pi )(12).
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 13
Topic: Tolerance Range (ODV)
Condition
a) Solve the equation: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;
b) Specify all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].
Show SolutionSolution
a) Because \sin \frac\pi 3=\cos \frac\pi 6, then \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, hence, the given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.
But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) and
\cos 2x=2 \cos ^2 x-1, so the equation becomes
(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,
(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.
Then either 2 \cos ^2 x-\cos x-1=0 or 2 \cos ^2 x+\cos x-1=0.
Solving the first equation as a quadratic equation for \cos x, we get:
(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore, either \cos x=1 or \cosx=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cosx=-\frac12, then x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.
Similarly, solving the second equation, we get either \cos x=-1, or \cosx=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If a \cosx=\frac12, then x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.
Let's combine the obtained solutions:
x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.
b) We select the roots that fall within the given interval using a number circle.
We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.
Answer
a) m\pi, m \in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;
b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 13
Topic: Tolerance Range (ODV)
Condition
a) Solve the Equation 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).
b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).
Show SolutionSolution
a) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of the equation will be x values such that \cos x \neq 0 and tg x \neq -1. We transform the equation using the double angle cosine formula 2 \cos ^2 \frac x2=1+\cos x. We get the equation: 5(1+\cos x) =\frac(11+5tgx)(1+tgx).
notice, that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cosx=\frac(\dfrac65)(1+tgx), \cosx+\sinx=\frac65.
2. Transform \sin x+\cos x using the reduction formula and the formula for the sum of cosines: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.
From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,
or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.
That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,
or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.
The found values of x belong to the domain of definition.
b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be respectively the numbers a=\frac\pi 4+arccos \frac(3\sqrt 2)5 and b=\frac\pi 4-arccos \frac(3\sqrt 2)5.
1. Let us prove an auxiliary inequality:
\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.
Really, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.
Note also that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, means \frac(3\sqrt 2)5<1.
2. From inequalities (1) by the property of the arccosine we get:
arccos 1 0 From here \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,
0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,
Without ODZ WITH ODZ Answer: x=5 ODZ: => => Answer: no roots The range of valid values protects us from such serious errors. To be honest, it is because of the ODZ that many “drummers” turn into “triples”. Considering that the search and accounting for ODZ is an insignificant step in the solution, they skip it, and then they are surprised: “why did the teacher put 2?”. Yes, that's why I put it because the answer is wrong! These are not "nitpicks" of the teacher, but a very specific mistake, the same as an incorrect calculation or a lost sign. Additional equations: a) = ; b) -42=14x+; c) =0; d) |x-5|=2x-2 Chapter 2 ODZ. What for? When? How? Acceptable range - there is a solution Answer: no roots. Answer: no roots. 0, the equation has no roots Answer: no roots. Additional examples: a) + =5; b) + = 23x-18; c) =0. ODZ: x=2, x=3 Check: x=2, + , 0<1, верно Check: x=3, + , 0<1, верно. Answer: x=2, x=3. Check: x=0, > , 0>0, wrong Check: x=1, > , 1>0, true Answer: x=1. Check: +=3, 0=3, wrong. Answer: no roots. Additional examples: a) = ; b) + =0; c) + \u003d x -1 Danger of ODZ Note that identical transformations can: It is also known that as a result of some transformations that change the original ODZ, it can lead to incorrect decisions. Let's explain each case with an example. 1) Consider the expression x + 4x + 7x, the ODZ of the variable x for this is the set R. We present similar terms. As a result, it will take the form x 2 +11x. Obviously, the ODZ of the variable x of this expression is also the set R. Thus, the performed transformation did not change the ODZ. 2) Take the equation x+ - =0. In this case, ODZ: x≠0. This expression also contains similar terms, after reduction of which, we come to the expression x, for which the ODZ is R. What we see: as a result of the transformation, the ODZ has expanded (zero has been added to the ODZ of the x variable for the original expression). 3) Let's take an expression. The ODV of the variable x is determined by the inequality (x−5) (x−2)≥0, ODV: (−∞, 2]∪∪/ Access mode: Materials of the sites www.fipi.ru, www.eg
Attachment 1 Practical work "ODZ: when, why and how?" Option 1 Option 2 │х+14│= 2 - 2х │3-х│=1 - 3х Annex 2 Answers to the tasks of practical work "ODZ: when, why and how?" Option 1 Option 2 Answer: no roots Answer: x is any number except x=5 9x+ = +27 ODZ: x≠3 Answer: no roots ODZ: x=-3, x=5. Answer: -3;5. y= -decreases, y= -increases So the equation has at most one root. Answer: x=6. ODZ: → →х≥5 Answer: x≥5, x≤-6. │х+14│=2-2х ODZ:2-2х≥0, х≤1 х=-4, х=16, 16 does not belong to ODZ Decreases - increases The equation has at most one root. Answer: no roots. 0, ODZ: x≥3, x≤2 Answer: x≥3, x≤2 8x+ = -32, ODZ: x≠-4. Answer: no roots. x=7, x=1. Answer: no solution Increasing - decreasing Answer: x=2. 0 ODZ: x≠15 Answer: x is any number except x=15. │3-х│=1-3х, ODZ: 1-3х≥0, х≤ x=-1, x=1 does not belong to the ODZ. Answer: x=-1.