How to find intervals of monotonicity of a function. b - final number

Theorem on the limit of a monotone function. The proof of the theorem is given using two methods. Definitions of strictly increasing, non-decreasing, strictly decreasing and non-increasing functions are also given. Definition of a monotonic function.

Definitions

Definitions of Increasing and Decreasing Functions
Let the function f (x) defined on some set real numbers x.
The function is called strictly increasing (strictly decreasing), if for all x′, x′′ ∈ X such that x′< x′′ выполняется неравенство:
f (x′)< f(x′′) ( f (x′) > f(x′′) ) .
The function is called non-decreasing (non-increasing), if for all x′, x′′ ∈ X such that x′< x′′ выполняется неравенство:
f (x′) ≤ f(x′′)( f (x′) ≥ f(x′′) ) .

This implies that a strictly increasing function is also nondecreasing. A strictly decreasing function is also nonincreasing.

Definition of a monotonic function
The function is called monotonous if it is non-decreasing or non-increasing.

To study the monotonicity of a function on some set X , you need to find the difference of its values ​​in two arbitrary points belonging to this set. If , then the function is strictly increasing; if , then the function does not decrease; if , then strictly decreases; if , then does not increase.

If on some set the function is positive: , then to determine monotonicity, one can examine the quotient of dividing its values ​​at two arbitrary points of this set. If , then the function is strictly increasing; if , then the function does not decrease; if , then strictly decreases; if , then does not increase.

Theorem
Let the function f (x) does not decrease over the interval (a,b), where .
If it is bounded from above by the number M : , then there is a finite left limit at the point b : . If f (x) not bounded above, then .
If f (x) is bounded from below by the number m : , then there is a finite right limit at the point a : . If f (x) not bounded below, then .

If the points a and b are at infinity, then in the expressions the limit signs mean that .
This theorem can be formulated more compactly.

Let the function f (x) does not decrease over the interval (a,b), where . Then there are one-sided limits at points a and b:
;
.

A similar theorem for a non-increasing function.

Let the function not increase on the interval , where . Then there are one-sided limits:
;
.

Consequence
Let the function be monotonic on the interval . Then at any point from this interval, there are one-sided finite limits of the function :
and .

Proof of the theorem

The function does not decrease

b - final number

Function limited from above

1.1.1. Let the function be bounded from above by the number M : for .

.
;
.

Since the function does not decrease, then for . Then
at .
Let's transform the last inequality:
;
;
.
Because , then . Then
at .

at .
"Definitions of one-sided limits of a function at a finite point").

The function is not limited from above

1. Let the function not decrease on the interval .
1.1. Let the number b be finite: .
1.1.2. Let the function be unbounded from above.
Let us prove that in this case there is a limit .

.

at .

Let's denote . Then for any exists , so that
at .
This means that the limit on the left at point b is (see "Definitions of one-sided infinite limits of a function at the end point").

b early plus infinity

Function limited from above

1. Let the function not decrease on the interval .
1.2.1. Let the function be bounded from above by the number M : for .
Let us prove that in this case there is a limit .

Since the function is bounded from above, there is a finite upper bound
.
According to the definition of exact upper face, are performed following conditions:
;
for any positive there is an argument for which
.

Since the function does not decrease, then for . Then at . Or
at .

So we have found that for any there exists a number , so that
at .
"Definitions of one-sided limits at infinity").

The function is not limited from above

1. Let the function not decrease on the interval .
1.2. Let the number b be plus infinity: .
1.2.2. Let the function be unbounded from above.
Let us prove that in this case there is a limit .

Since the function is not bounded from above, then for any number M there is an argument , for which
.

Since the function does not decrease, then for . Then at .

So, for any there is a number , so that
at .
This means that the limit at is (see "Definitions of One-Sided Infinite Limits at Infinity").

The function does not increase

Now consider the case when the function is not increasing. You can, as above, consider each option separately. But we will cover them right away. For this we use . Let us prove that in this case there is a limit .

Consider the finite lower bound of the set of function values:
.
Here B can be either a finite number or a point at infinity. According to the definition of the exact infimum, the following conditions are satisfied:
;
for any neighborhood of point B there is an argument for which
.
By the condition of the theorem, . So .

Since the function does not increase, then for . Because , then
at .
Or
at .
Further, we note that the inequality defines the left punctured neighborhood of the point b .

So, we have found that for any neighborhood of the point , there is such a punctured left neighborhood of the point b that
at .
This means that the limit on the left at point b is :

(see the universal definition of the limit of a function according to Cauchy).

Limit at point a

Now let's show that there is a limit at the point a and find its value.

Let's consider a function. By the condition of the theorem, the function is monotonic for . Let's replace the variable x with - x (or do the substitution and then replace the variable t with x ). Then the function is monotone for . Multiplying the inequalities by -1 and changing their order, we conclude that the function is monotonic for .

In a similar way, it is easy to show that if it does not decrease, then it does not increase. Then, according to what was proved above, there is a limit
.
If it doesn't increase, then it doesn't decrease. In this case, there is a limit
.

Now it remains to show that if there is a limit of the function at , then there is a limit of the function at , and these limits are equal:
.

Let's introduce the notation:
(1) .
Let's express f in terms of g :
.
Take an arbitrary positive number . Let there be an epsilon neighborhood of point A . Epsilon neighborhood is defined for both finite and infinite values ​​of A (see "Neighborhood of a point"). Since there is a limit (1), then, according to the definition of a limit, for any there exists such that
at .

Let a be a finite number. Let us express the left punctured neighborhood of the point -a using the inequalities:
at .
Let's replace x with -x and take into account that:
at .
The last two inequalities define a punctured right neighborhood of the point a . Then
at .

Let a be an infinite number, . We repeat the discussion.
at ;
at ;
at ;
at .

So, we have found that for any there exists such that
at .
It means that
.

The theorem has been proven.

We first met in the 7th grade algebra course. Looking at the graph of the function, we removed the relevant information: if moving along the graph from left to right, we are at the same time moving from bottom to top (as if climbing a hill), then we declared the function increasing (Fig. 124); if we move from top to bottom (go down the hill), then we declared the function to be decreasing (Fig. 125).

However, mathematicians are not very fond of this way of studying the properties of a function. They believe that definitions of concepts should not be based on a drawing - a drawing should only illustrate one or another property of a function on its chart. Let us give rigorous definitions of the concepts of increasing and decreasing functions.

Definition 1. The function y \u003d f (x) is called increasing on the interval X, if from the inequality x 1< х 2 - где хг и х2 - любые две точки промежутка X, следует неравенство f(x 1) < f(x 2).

Definition 2. The function y \u003d f (x) is called decreasing on the interval X, if from the inequality x 1< х 2 , где х 1 и х 2 - любые две точки промежутка X, следует inequality f(x1) > f(x2).

In practice, it is more convenient to use the following formulations:

the function increases if the larger value of the argument corresponds to the larger value of the function;
the function is decreasing if the larger value of the argument corresponds to the smaller value of the function.

Using these definitions and the properties established in § 33 numerical inequalities, we will be able to substantiate the conclusions about the increase or decrease of previously studied functions.

1. Linear function y = kx + m

If k > 0, then the function increases on the whole (Fig. 126); if k< 0, то функция убывает на всей числовой прямой (рис. 127).

Proof. Let f(x) = kx + m. If x 1< х 2 и k >Oh, then, according to property 3 of numerical inequalities (see § 33), kx 1< kx 2 . Далее, согласно свойству 2, из kx 1 < kx 2 следует, что kx 1 + m < kx 2 + m, т. е. f(х 1) < f(х 2).

So, from the inequality x 1< х 2 следует, что f(х 1) < f(x 2). Это и означает возрастание функции у = f(х), т.е. linear functions y = kx + m.

If x 1< х 2 и k < 0, то, согласно свойству 3 числовых неравенств, kx 1 >kx 2 , and according to property 2, from kx 1 > kx 2 it follows that kx 1 + m > kx 2 + t.

So, from the inequality x 1< х 2 следует, что f(х 1) >f(x 2). This means that the function y \u003d f (x) decreases, i.e. linear function y = kx + m.

If a function is increasing (decreasing) in its entire domain of definition, then it can be called increasing (decreasing) without specifying the interval. For example, about the function y \u003d 2x - 3, we can say that it increases on the entire number line, but we can also say in short: y \u003d 2x - 3 - increasing
function.

2. Function y = x2

1. Consider the function y \u003d x 2 on the beam. Take two non-positive numbers x 1 and x 2 such that x 1< х 2 . Тогда, согласно свойству 3 числовых неравенств, выполняется неравенство - х 1 >- x 2 . Since the numbers - x 1 and - x 2 are non-negative, then, squaring both parts of the last inequality, we obtain an inequality of the same meaning (-x 1) 2 > (-x 2) 2, i.e. This means that f (x 1) > f (x 2).

So, from the inequality x 1< х 2 следует, что f(х 1) >f(x 2).

Therefore, the function y \u003d x 2 decreases on the beam (- 00, 0] (Fig. 128).

1. Consider a function on the interval (0, + 00).
Let x1< х 2 . Так как х 1 и х 2 - , то из х 1 < x 2 следует (см. пример 1 из § 33), т. е. f(x 1) >f(x2).

So, from the inequality x 1< х 2 следует, что f(x 1) >f(x2). This means that the function decreases on the open ray (0, + 00) (Fig. 129).

2. Consider a function on the interval (-oo, 0). Let x 1< х 2 , х 1 и х 2 - negative numbers. Then - x 1 > - x 2, and both parts of the last inequality - positive numbers, and therefore (we again used the inequality proved in Example 1 of § 33). Then we have , whence we get .

So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2) i.e. the function decreases on the open beam (- 00 , 0)

Usually the terms "increasing function", "decreasing function" combine common name monotonic function, and the study of a function for increasing and decreasing is called the study of a function for monotonicity.

Decision.

1) Let's plot the function y \u003d 2x 2 and take the branch of this parabola at x< 0 (рис. 130).

2) Let's build and select its part on the segment (Fig. 131).

3) We construct a hyperbola and select its part on the open ray (4, + 00) (Fig. 132).
4) All three "pieces" will be depicted in the same coordinate system - this is the graph of the function y \u003d f (x) (Fig. 133).

Let's read the graph of the function y \u003d f (x).

1. The scope of the function is the entire number line.

2. y \u003d 0 for x \u003d 0; y > 0 for x > 0.

3. The function decreases on the ray (-oo, 0], increases on the segment , decreases on the ray, convex upwards on the segment , convex downwards on the ray Consider the function \(f(t)=t^3+t\) . Then the equation will be rewritten in the form: \ We investigate the function \(f(t)\) . \ Therefore, the function \(f(t)\) is increasing for all \(t\) . This means that each value of the function \(f(t)\) corresponds to exactly one value of the argument \(t\) . Therefore, in order for the equation to have roots, you need: \ For the resulting equation to have two roots, its discriminant must be positive: \

Answer:

\(\left(-\infty;\dfrac1(12)\right)\)

Task 2 #2653

Task level: Equal to the Unified State Examination

Find all values ​​of the parameter \(a\) for which the equation \

has two roots.

(Task from subscribers.)

Let's make a replacement: \(ax^2-2x=t\) , \(x^2-1=u\) . Then the equation will take the form: \ Consider the function \(f(w)=7^w+\sqrtw\) . Then our equation will take the form:

Let's find the derivative \ Note that for all \(w\ne 0\) the derivative is \(f"(w)>0\) , since \(7^w>0\) , \(w^6>0\) . Note also that the function \(f(w)\) itself is defined for all \(w\) .Because, moreover, \(f(w)\) is continuous, we can conclude that \(f (w)\) is increasing on all \(\mathbb(R)\) .
Hence, the equality \(f(t)=f(u)\) is possible if and only if \(t=u\) . Let's go back to the original variables and solve the resulting equation:

\ In order for this equation to have two roots, it must be square and its discriminant must be positive:

\[\begin(cases) a-1\ne 0\\ 4-4(a-1)>0\end(cases) \quad\Leftrightarrow\quad \begin(cases)a\ne1\\a<2\end{cases}\]

Answer:

\((-\infty;1)\cup(1;2)\)

Task 3 #3921

Task level: Equal to the Unified State Examination

Find all positive values ​​of the parameter \(a\) for which the equation

has at least \(2\) solutions.

Let us move all the terms containing \(ax\) to the left, and those containing \(x^2\) to the right, and consider the function
\

Then the original equation will take the form:
\

Let's find the derivative:
\

Because \((t-2)^2 \geqslant 0, \ e^t>0, \ 1+\cos(2t) \geqslant 0\), then \(f"(t)\geqslant 0\) for any \(t\in \mathbb(R)\) .

Moreover, \(f"(t)=0\) if \((t-2)^2=0\) and \(1+\cos(2t)=0\) at the same time, which is not true for any \ (t\) Therefore, \(f"(t)> 0\) for any \(t\in \mathbb(R)\) .

Thus the function \(f(t)\) is strictly increasing for all \(t\in \mathbb(R)\) .

So the equation \(f(ax)=f(x^2)\) is equivalent to the equation \(ax=x^2\) .

The equation \(x^2-ax=0\) with \(a=0\) has one root \(x=0\) , and with \(a\ne 0\) it has two different root\(x_1=0\) and \(x_2=a\) .
We need to find the values ​​\(a\) for which the equation will have at least two roots, also taking into account the fact that \(a>0\) .
Therefore, the answer is: \(a\in (0;+\infty)\) .

Answer:

\((0;+\infty)\) .

Task 4 #1232

Task level: Equal to the Unified State Examination

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

has a unique solution.

Multiply the right and left sides of the equation by \(2^(\sqrt(x+1))\) (because \(2^(\sqrt(x+1))>0\) ) and rewrite the equation as : \

Consider the function \(y=2^t\cdot \log_(\frac(1)(9))((t+2))\) for \(t\geqslant 0\) (because \(\sqrt(x+1)\geqslant 0\) ).

Derivative \(y"=\left(-2^t\cdot \log_9((t+2))\right)"=-\dfrac(2^t)(\ln9)\cdot \left(\ln 2\cdot \ln((t+2))+\dfrac(1)(t+2)\right)\).

Because \(2^t>0, \ \dfrac(1)(t+2)>0, \ \ln((t+2))>0\) for all \(t\geqslant 0\) , then \(y"<0\) при всех \(t\geqslant 0\) .

Consequently, for \(t\geqslant 0\) the function \(y\) decreases monotonically.

The equation can be viewed as \(y(t)=y(z)\) , where \(z=ax, t=\sqrt(x+1)\) . It follows from the monotonicity of the function that equality is possible only if \(t=z\) .

This means that the equation is equivalent to the equation: \(ax=\sqrt(x+1)\) , which in turn is equivalent to the system: \[\begin(cases) a^2x^2-x-1=0\\ ax \geqslant 0 \end(cases)\]

For \(a=0\) the system has one solution \(x=-1\) , which satisfies the condition \(ax\geqslant 0\) .

Consider the case \(a\ne 0\) . The discriminant of the first equation of the system \(D=1+4a^2>0\) for all \(a\) . Therefore, the equation always has two roots \(x_1\) and \(x_2\) , and they have different signs (because by the Vieta theorem \(x_1\cdot x_2=-\dfrac(1)(a^2)<0\) ).

This means that for \(a<0\) условию \(ax\geqslant 0\) подходит отрицательный корень, при \(a>0\) the positive root fits the condition. Therefore, the system always has a unique solution.

So \(a\in \mathbb(R)\) .

Answer:

\(a\in \mathbb(R)\) .

Task 5 #1234

Task level: Equal to the Unified State Examination

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

has at least one root from the interval \([-1;0]\) .

Consider the function \(f(x)=2x^3-3x(ax+x-a^2-1)-3a-a^3\) for some fixed \(a\) . Let's find its derivative: \(f"(x)=6x^2-6ax-6x+3a^2+3=3(x^2-2ax+a^2+x^2-2x+1)=3((x-a)^2 +(x-1)^2)\).

Note that \(f"(x)\geqslant 0\) for all values ​​of \(x\) and \(a\) , and is equal to \(0\) only for \(x=a=1\) . But for \(a=1\) :
\(f"(x)=6(x-1)^2 \Rightarrow f(x)=2(x-1)^3 \Rightarrow\) the equation \(2(x-1)^3=0\) has a single root \(x=1\) that does not satisfy the condition. Therefore, \(a\) cannot be equal to \(1\) .

Hence, for all \(a\ne 1\) the function \(f(x)\) is strictly increasing, hence the equation \(f(x)=0\) can have at most one root. Given the properties of the cubic function, the graph \(f(x)\) for some fixed \(a\) will look like this:

So, in order for the equation to have a root from the segment \([-1;0]\) , it is necessary: \[\begin(cases) f(0)\geqslant 0\\ f(-1)\leqslant 0 \end(cases) \Rightarrow \begin(cases) a(a^2+3)\leqslant 0\\ ( a+2)(a^2+a+4)\geqslant 0 \end(cases) \Rightarrow \begin(cases) a\leqslant 0\\ a\geqslant -2 \end(cases) \Rightarrow -2\leqslant a\leqslant 0\]

So \(a\in [-2;0]\) .

Answer:

\(a\in [-2;0]\) .

Task 6 #2949

Task level: Equal to the Unified State Examination

Find all values ​​of the parameter \(a\) , for each of which the equation \[(\sin^2x-5\sin x-2a(\sin x-3)+6)\cdot (\sqrt2a+8x\sqrt(2x-2x^2))=0\]

has roots.

(Task from subscribers)

odz equation: \(2x-2x^2\geqslant 0 \quad\Leftrightarrow\quad x\in \). Therefore, in order for the equation to have roots, it is necessary that at least one of the equations \[\sin^2x-5\sin x-2a(\sin x-3)+6=0 \quad (\small(\text(or)))\quad \sqrt2a+8x\sqrt(2x-2x^ 2)=0\] had decisions on ODZ.

1) Consider the first equation \[\sin^2x-5\sin x-2a(\sin x-3)+6=0 \quad\Leftrightarrow\quad \left[\begin(gathered)\begin(aligned) &\sin x=2a+ 2\\ &\sin x=3\\ \end(aligned) \end(gathered)\right. \quad\Leftrightarrow\quad \sin x=2a+2\] This equation must have roots in \(\) . Consider a circle:

Thus, we see that for any \(2a+2\in [\sin 0;\sin 1]\) the equation will have one solution, and for all others it will not have solutions. Therefore, at \(a\in \left[-1;-1+\sin 1\right]\) the equation has solutions.

2) Consider the second equation \[\sqrt2a+8x\sqrt(2x-2x^2)=0 \quad\Leftrightarrow\quad 8x\sqrt(x-x^2)=-a\]

Consider the function \(f(x)=8x\sqrt(x-x^2)\) . Let's find its derivative: \ On the ODZ, the derivative has one zero: \(x=\frac34\) , which is also the maximum point of the function \(f(x)\) .
Note that \(f(0)=f(1)=0\) . So, schematically, the graph \(f(x)\) looks like this:

Therefore, in order for the equation to have solutions, it is necessary that the graph \ (f (x) \) intersect with the line \ (y \u003d -a \) (one of the suitable options is shown in the figure). That is, it is necessary that \ . With these \(x\) :

The function \(y_1=\sqrt(x-1)\) is strictly increasing. The graph of the function \(y_2=5x^2-9x\) is a parabola whose vertex is at the point \(x=\dfrac(9)(10)\) . Therefore, for all \(x\geqslant 1\) the function \(y_2\) is also strictly increasing (the right branch of the parabola). Because the sum of strictly increasing functions is strictly increasing, then \(f_a(x)\) is strictly increasing (the constant \(3a+8\) does not affect the monotonicity of the function).

The function \(g_a(x)=\dfrac(a^2)(x)\) for all \(x\geqslant 1\) is a part of the right branch of the hyperbola and is strictly decreasing.

Solving the equation \(f_a(x)=g_a(x)\) means finding the intersection points of the functions \(f\) and \(g\) . From their opposite monotonicity it follows that the equation can have at most one root.

For \(x\geqslant 1\) \(f_a(x)\geqslant 3a+4, \ \ \ 0 . Therefore, the equation will have a unique solution if:

\\cup

Answer:

\(a\in(-\infty;-1]\cup)