increasing on the interval \(X\) if for any \(x_1, x_2\in X\) such that \(x_1 The function is called non-decreasing \(\blacktriangleright\) The function \(f(x)\) is called decreasing on the interval \(X\) if for any \(x_1, x_2\in X\) such that \(x_1 The function is called non-increasing on the interval \(X\) if for any \(x_1, x_2\in X\) such that \(x_1 \(\blacktriangleright\) Increasing and decreasing functions are called strictly monotonous, and non-increasing and non-decreasing are simply monotonous. \(\blacktriangleright\) Basic properties: I. If the function \(f(x)\) is strictly monotone on \(X\) , then from the equality \(x_1=x_2\) (\(x_1,x_2\in X\) ) it follows \(f(x_1)= f(x_2)\) , and vice versa. Example: the function \(f(x)=\sqrt x\) is strictly increasing for all \(x\in \) , therefore the equation \(x^2=9\) has at most one solution on this interval, or rather one: \(x=-3\) . the function \(f(x)=-\dfrac 1(x+1)\) is strictly increasing for all \(x\in (-1;+\infty)\), so the equation \(-\dfrac 1(x +1)=0\) has no more than one solution on this interval, or rather none, because the numerator of the left-hand side can never be equal to zero. III. If the function \(f(x)\) is non-decreasing (non-increasing) and continuous on the segment \(\), and at the ends of the segment it takes the values \(f(a)=A, f(b)=B\) , then for \(C\in \) (\(C\in \) ) the equation \(f(x)=C\) always has at least one solution. Example: the function \(f(x)=x^3\) is strictly increasing (that is, strictly monotone) and continuous for all \(x\in\mathbb(R)\) , therefore for any \(C\in ( -\infty;+\infty)\) the equation \(x^3=C\) has exactly one solution: \(x=\sqrt(C)\) . Task 1 #3153 Task level: Easier than the Unified State Exam has exactly two roots. Let's rewrite the equation as: \[(3x^2)^3+3x^2=(x-a)^3+(x-a)\] Consider the function \(f(t)=t^3+t\) . Then the equation will be rewritten in the form: \ Let's study the function \(f(t)\) . \ Consequently, the function \(f(t)\) increases 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, it is necessary: \
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 Exam 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 \(w\). Since, in addition, \(f(w)\) is continuous, we can conclude that \(f (w)\) increases on the whole \(\mathbb(R)\) . \
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 Exam Find all positive values of the parameter \(a\) for which the equation has at least \(2\) solutions. Let's move all 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 \(t\in \mathbb(R)\) . This means that the equation \(f(ax)=f(x^2)\) is equivalent to the equation \(ax=x^2\) . The equation \(x^2-ax=0\) for \(a=0\) has one root \(x=0\), and for \(a\ne 0\) it has two various roots\(x_1=0\) and \(x_2=a\) . Answer: \((0;+\infty)\) . Task 4 #1232 Task level: Equal to the Unified State Exam Find all values of the parameter \(a\) , for each of which the equation \
has a unique solution. Let's multiply the right and left sides of the equation by \(2^(\sqrt(x+1))\) (since \(2^(\sqrt(x+1))>0\) ) and rewrite the equation in the form : \
Consider the function \(y=2^t\cdot \log_(\frac(1)(9))((t+2))\) for \(t\geqslant 0\) (since \(\sqrt(x+1)\geqslant 0\) ). 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, as \(t\geqslant 0\) the function \(y\) decreases monotonically. The equation can be considered in the form \(y(t)=y(z)\) , where \(z=ax, t=\sqrt(x+1)\) . From the monotonicity of the function it follows 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: \[\begin(cases) a^2x^2-x-1=0\\ ax \geqslant 0 \end(cases)\] When \(a=0\) the system has one solution \(x=-1\) that satisfies the condition \(ax\geqslant 0\) . Consider the case \(a\ne 0\) . Discriminant of the first equation of the system \(D=1+4a^2>0\) for all \(a\) . Consequently, the equation always has two roots \(x_1\) and \(x_2\), and they are of different signs (since according to Vieta’s 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 condition is satisfied by a positive root. 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 Exam Find all values of the parameter \(a\) , for each of which the equation \
has at least one root from the segment \([-1;0]\) . 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 \(a=1\) : This means that for all \(a\ne 1\) the function \(f(x)\) is strictly increasing, therefore, the equation \(f(x)=0\) can have no more than one root. Taking into account the properties of the cubic function, the graph of \(f(x)\) for some fixed \(a\) will look like this: This means that in order for the equation to have a root from the segment \([-1;0]\), it is necessary: \[\begin(cases) f(0)\geqslant 0\\ f(-1)\leqslant 0 \end(cases) \Rightarrow \begin(cases) a(a^2+3)\leqslant 0\\ ( a+2)(a^2+a+4)\geqslant 0 \end(cases) \Rightarrow \begin(cases) a\leqslant 0\\ a\geqslant -2 \end(cases) \Rightarrow -2\leqslant a\leqslant 0\] Thus, \(a\in [-2;0]\) . Answer: \(a\in [-2;0]\) . Task 6 #2949 Task level: Equal to the Unified State Exam 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 equations: \(2x-2x^2\geqslant 0 \quad\Leftrightarrow\quad x\in \). Therefore, in order for an 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 have no solutions. Therefore, when \(a\in \left[-1;-1+\sin 1\right]\) 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 \(f(x)\) . Therefore, in order for the equation to have solutions, it is necessary that the graph \(f(x)\) intersect with the straight line \(y=-a\) (the figure shows one of the suitable options). That is, it is necessary that \
. For 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, the vertex of which is at the point \(x=\dfrac(9)(10)\) . Consequently, 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 \(x\geqslant 1\) represents 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. When \(x\geqslant 1\) \(f_a(x)\geqslant 3a+4, \ \ \ 0 \\cup Answer: \(a\in (-\infty;-1]\cup ,
limited on this segment; · the sum of increasing (decreasing) functions is an increasing (decreasing) function; · if function f increases (decreases) and n– an odd number, it also increases (decreases); · If f"(x)>0 for all xО(a,b), then the function y=f(x) is increasing on the interval (a,b); · If f"(x)<0
for all xО(a,b), then the function y=f(x) is decreasing on the interval (a,b); · If f(x) – continuous and monotonic function on the set X, then the equation f(x)=C, Where WITH– this constant may have X no more than one solution; · if on the domain of definition of the equation f(x)=g(x) function f(x) increases, and the function g(x) decreases, then the equation cannot have more than one solution. Theorem. (a sufficient condition for the monotonicity of a function). If continuous on the segment [ a, b] function y = f(X) at each point of the interval ( a, b) has a positive (negative) derivative, then this function increases (decreases) on the segment [ a, b]. Proof. Let >0 for everyone xО(a,b).
Consider two arbitrary values x 2 > x 1 , belonging to [ a, b]. According to Lagrange's formula x 1<с < х 2
.
(With) > 0
And x 2 – x 1 > 0,
therefore > 0,
whence > , that is, the function f(x) increases on the interval [ a, b]. The second part of the theorem is proved in a similar way. Theorem 3. (a necessary sign of the existence of an extremum of a function). If the function differentiable at point c at=f(X) has an extremum at this point, then . Proof. Let, for example, the function at= f(X) has a maximum at point c. This means that there is a punctured neighborhood of the point c such that for all points x this neighborhood is satisfied f(x) < f
(c),
that is f(c) is the largest value of the function in this neighborhood. Then by Fermat's theorem. The case of a minimum at point c is proved in a similar way. Comment. A function may have an extremum at a point at which its derivative does not exist. For example, a function has a minimum at point x =
0, although it doesn't exist. The points at which the derivative of a function is zero or does not exist are called critical points of the function. However, the function does not have an extremum at all critical points. For example, the function y = x 3 has no extrema, although its derivative
=0. Theorem 4. (a sufficient sign of the existence of an extremum). If continuous function y = f(x) has a derivative at all points of a certain interval containing the critical point C (except, perhaps, for this point itself), and if the derivative, when the argument passes from left to right through the critical point C, changes sign from plus to minus, then the function at point C has the maximum, and when the sign changes from minus to plus, the minimum. Proof. Let c be a critical point and let, for example, when the argument passes through the point c changes sign from plus to minus. This means that on some interval (c–e; c) the function increases, and on the interval (c; c+e)– decreases (at e>0). Therefore, at point c the function has a maximum. The case of a minimum is proved in a similar way. Comment. If the derivative does not change sign when the argument passes through the critical point, then the function at this point does not have an extremum. Since the definitions of limit and continuity for a function of several variables practically coincide with the corresponding definitions for a function of one variable, then for functions of several variables all the properties of limits and continuous functions are preserved ©2015-2019 site Theorem on the limit of a monotone function. A 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 of increasing and decreasing functions It follows that a strictly increasing function is also non-decreasing. A strictly decreasing function is also non-increasing. Definition of a monotonic function To study the monotonicity of a function on a certain set X, you need to find the difference of its values at 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 it does not increase. If on a certain set the function is positive: , then to determine monotonicity, you can study 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 it does not increase. Theorem If points a and b are at infinity, then in the expressions the limit signs mean that . Let the function f (x) does not decrease on 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 Since the function does not decrease, then when . Then 1. Let the function not decrease on the interval. Let's denote . Then for anyone there is, so 1. Let the function not decrease on the interval. Since the function is bounded above, there is a finite supremum Since the function does not decrease, then when . Then at . Or So, we found that for anyone there is a number, so 1. Let the function not decrease on the interval. Since the function is not bounded above, then for any number M there is an argument for which Since the function does not decrease, then when . Then at . So for any there is a number , so Now consider the case when the function does not increase. You can, as above, consider each option separately. But we'll cover them right away. For this we use . Let us prove that in this case there is a limit. Consider the finite infimum of the set of function values: Since the function does not increase, then when . Since then So, we found that for any neighborhood of the point, there is a punctured left neighborhood of the point b such that Now we will show that there is a limit at point a and find its value. Let's consider the function. According to the conditions of the theorem, the function is monotonic for . Let's replace the variable x with - x (or do a substitution and then replace the variable t with x ). Then the function is monotonic for . Multiplying inequalities by -1
and changing their order we come to the conclusion 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 proven above, there is a limit Now it remains to show that if there is a limit of a function at , then there is a limit of the function at , and these limits are equal: Let us introduce the notation: Let a be a finite number. Let us express the left punctured neighborhood of the point -a using the inequalities: Let a be an infinite number, . We repeat the reasoning. So, we found that for anyone there is such that The theorem has been proven. Additional materials Manuals and simulators in the Integral online store for grade 10 from 1C
What we will study: Guys, earlier we looked at a lot of various functions and built their graphs. Now let's introduce new rules that work for all the functions that we have considered and will continue to consider. A function is a correspondence y= f(x), in which each value of x is associated with a single value of y. Let's look at the graph of some function: Our graph shows: the larger x, the smaller y. So let's define a decreasing function. A function is called decreasing if a larger value of the argument corresponds to a smaller value of the function. If x2 > x1, then f(x2) Now let's look at the graph of this function: If a function increases or decreases over a certain interval, then it is said that it is monotonic on this interval. If you look at our tangents or visually draw any other tangent, you will notice that the angle between the tangent and the positive direction of the x-axis will be acute. This means that the tangent has a positive slope. Tangent slope equal to the value derivative in the abscissa of the point of tangency. Thus, the value of the derivative is positive at all points in our graph. For an increasing function, the following inequality holds: f"(x) ≥ 0, for any point x. Guys, now let's look at the graph of some decreasing function and construct tangents to the graph of the function. Let's look at the tangents and visually draw any other tangent. We will notice that the angle between the tangent and the positive direction of the x-axis is obtuse, which means the tangent has a negative slope. Thus, the value of the derivative is negative at all points in our graph. For a decreasing function, the following inequality holds: f"(x) ≤ 0, for any point x. So, the monotonicity of a function depends on the sign of the derivative: If a function increases on an interval and has a derivative on this interval, then this derivative will not be negative. If a function decreases on an interval and has a derivative on this interval, then this derivative will not be positive. Important, so that the intervals on which we consider the function are open! Theorem 1. If the inequality f'(x) ≥ 0 holds at all points of an open interval X (and the equality of the derivative to zero either does not hold or holds, but only at a finite set of points), then the function y= f(x) increases on the interval X. Theorem 2. If the inequality f'(x) ≤ 0 holds at all points of an open interval X (and the equality of the derivative to zero either does not hold or holds, but only at a finite set of points), then the function y= f(x) decreases on the interval X. Theorem 3. If at all points of the open interval X the equality 1) Prove that the function y= x 7 + 3x 5 + 2x - 1 is increasing on the entire number line. Solution: Let's find the derivative of our function: y"= 7 6 + 15x 4 + 2. Since the degree at x is even, then power function takes only positive values. Then y" > 0 for any x, which means by Theorem 1, our function increases on the entire number line. 2) Prove that the function is decreasing: y= sin(2x) - 3x. Let's find the derivative of our function: y"= 2cos(2x) - 3. 3) Examine the monotonicity of the function: y= x 2 + 3x - 1. Solution: Let's find the derivative of our function: y"= 2x + 3. 4) Examine the monotonicity of the function: y= $\sqrt(3x - 1)$. Solution: Let's find the derivative of our function: y"= $\frac(3)(2\sqrt(3x - 1))$. Our inequality is greater than or equal to zero: $\sqrt(3x-1)$ ≤ 0, Finding the intervals of increase, decrease and extrema of a function is both an independent task and an essential part of other tasks, in particular, full function study. Initial information about the increase, decrease and extrema of the function is given in theoretical chapter on derivative, which I highly recommend for preliminary study (or repetition)– also for the reason that the following material is based on the very essentially derivative, being a harmonious continuation of this article. Although, if time is short, then a purely formal practice of examples from today’s lesson is also possible. And today there is a spirit of rare unanimity in the air, and I can directly feel that everyone present is burning with desire learn to explore a function using its derivative. Therefore, reasonable, good, eternal terminology immediately appears on your monitor screens. For what? One of the reasons is the most practical: so that it is clear what is generally required of you in a particular task! Let's consider some function. To put it simply, we assume that she continuous on the entire number line: Just in case, let’s immediately get rid of possible illusions, especially for those readers who have recently become acquainted with intervals of constant sign of the function. Now we NOT INTERESTED, how the graph of the function is located relative to the axis (above, below, where the axis intersects). To be convincing, mentally erase the axes and leave one graph. Because that’s where the interest lies. Function increases on an interval if for any two points of this interval connected by the relation , the inequality is true. That is, a larger value of the argument corresponds to a larger value of the function, and its graph goes “from bottom to top”. The demonstration function grows over the interval. Likewise, the function decreases on an interval if for any two points of a given interval such that , the inequality is true. That is, a larger value of the argument corresponds to a smaller value of the function, and its graph goes “from top to bottom”. Our function decreases on intervals If a function increases or decreases over an interval, then it is called strictly monotonous at this interval. What is monotony? Take it literally – monotony. You can also define non-decreasing function (relaxed condition in the first definition) and non-increasing function (softened condition in the 2nd definition). A non-decreasing or non-increasing function on an interval is called a monotonic function on a given interval (strict monotony - special case“just” monotony). The theory also considers other approaches to determining the increase/decrease of a function, including on half-intervals, segments, but in order not to pour oil-oil-oil on your head, we will agree to operate with open intervals with categorical definitions - this is clearer, and for solving many practical problems quite enough. Thus, in my articles the wording “monotonicity of a function” will almost always be hidden intervals strict monotony
(strictly increasing or strictly decreasing function). Neighborhood of a point. Words after which students run away wherever they can and hide in horror in the corners. ...Although after the post Cauchy limits They’re probably not hiding anymore, but just shuddering slightly =) Don’t worry, there won’t be any proofs of theorems now mathematical analysis– I needed the surroundings to formulate definitions more strictly extremum points. Let's remember: Neighborhood of a point called the interval that contains this point, while for convenience the interval is often assumed to be symmetrical. For example, a point and its standard neighborhood: The point is called strict maximum point, If exists her neighborhood, for all values of which, except for the point itself, the inequality . In our specific example, this is a dot. The point is called strict minimum point, If exists her neighborhood, for all values of which, except for the point itself, the inequality . In the drawing there is point “a”. Note
: the requirement of neighborhood symmetry is not at all necessary. In addition, it is important the very fact of existence neighborhood (whether tiny or microscopic) that satisfies the specified conditions The points are called strictly extremum points or simply extremum points functions. That is, it is a generalized term for maximum points and minimum points. How do we understand the word “extreme”? Yes, just as directly as monotony. Extreme points of roller coasters. As in the case of monotonicity, loose postulates exist and are even more common in theory (which, of course, the strict cases considered fall under!): The point is called maximum point, If exists its surroundings are such that for all Note that according to the last two definitions, any point of a constant function (or a “flat section” of a function) is considered both a maximum and a minimum point! The function, by the way, is both non-increasing and non-decreasing, that is, monotonic. However, we will leave these considerations to theorists, since in practice we almost always contemplate traditional “hills” and “hollows” (see drawing) with a unique “king of the hill” or “princess of the swamp”. As a variety, it occurs tip, directed up or down, for example, the minimum of the function at the point. Oh, and speaking of royalty: Common name – extremes functions. Please be careful with your words! Extremum points– these are “X” values. ! Note
: sometimes the listed terms refer to the “X-Y” points that lie directly on the GRAPH OF the function ITSELF. How many extrema can a function have? None, 1, 2, 3, ... etc. to infinity. For example, sine has infinitely many minima and maxima. IMPORTANT! The term "maximum of function" not identical the term “maximum value of a function”. It is easy to notice that the value is maximum only in a local neighborhood, and at the top left there are “cooler comrades”. Likewise, “minimum of a function” is not the same as “minimum value of a function,” and in the drawing we see that the value is minimum only in a certain area. In this regard, extremum points are also called local extremum points, and the extrema – local extremes
. They walk and wander nearby and global brethren. So, any parabola has at its vertex global minimum or global maximum. Further, I will not distinguish between types of extremes, and the explanation is voiced more for general educational purposes - the additional adjectives “local”/“global” should not take you by surprise. Let’s summarize our short excursion into the theory with a test shot: what does the task “find the monotonicity intervals and extremum points of the function” mean? The wording encourages you to find: – intervals of increasing/decreasing function (non-decreasing, non-increasing appears much less often); – maximum and/or minimum points (if any exist). Well, to avoid failure, it’s better to find the minimums/maximums themselves ;-) How to determine all this? Using the derivative function! Many rules, in fact, are already known and understood from lesson about the meaning of a derivative. Tangent derivative With cotangent and its derivative The arcsine increases over the interval - the derivative here is positive: I think it won’t be too difficult for you to carry out similar reasoning for the arc cosine and its derivative. All of the above cases, many of which are tabular derivatives, I remind you, follow directly from derivative definitions. To better understand what the graph of this function looks like: where it goes “bottom up”, where “top down”, where it reaches minimums and maximums (if it reaches at all). Not all functions are so simple - in most cases we have no idea at all about the graph of a particular function. It's time to move on to more meaningful examples and consider algorithm for finding intervals of monotonicity and extrema of a function: Example 1 Find intervals of increase/decrease and extrema of the function Solution: 1) The first step is to find domain of a function, and also take note of the break points (if they exist). IN in this case the function is continuous on the entire number line, and this action is to a certain extent formal. But in a number of cases, serious passions flare up here, so let’s treat the paragraph without disdain. 2) The second point of the algorithm is due to If there is an extremum at a point, then either the value does not exist. Confused by the ending? Extremum of the “modulus x” function .
The condition is necessary, but not enough, and the converse is not always true. So, it does not yet follow from the equality that the function reaches a maximum or minimum at point . A classic example has already been highlighted above - this is a cubic parabola and its critical point. But be that as it may, necessary condition extremum dictates the need to find suspicious points. To do this, find the derivative and solve the equation: At the beginning of the first article about function graphs I told you how to quickly build a parabola using an example Example 2 Find intervals of monotonicity and extrema of the function This is an example for independent decision. Complete solution and an approximate final sample of the task at the end of the lesson. The long-awaited moment of meeting with fractional-rational functions has arrived: Example 3 Explore a function using the first derivative Pay attention to how variably one and the same task can be reformulated. Solution: 1) The function suffers infinite discontinuities at points. 2) We detect critical points. Let's find the first derivative and equate it to zero: Let's solve the equation. A fraction is zero when its numerator is zero: Thus, we get three critical points: 3) We plot ALL detected points on the number line and interval method we define the signs of the DERIVATIVE: The action, as you understand, needs to be carried out for each of the six intervals. By the way, note that the numerator factor and denominator are strictly positive for any point in any interval, which greatly simplifies the task. So, the derivative told us that the FUNCTION ITSELF increases by At the point the function reaches its maximum: Think about why you don't have to recalculate the second value ;-) When passing through a point, the derivative does not change sign, so the function has NO EXTREMUM there - it both decreased and remained decreasing. ! Let's repeat important point
: points are not considered critical - they contain a function not determined. Accordingly, here In principle there can be no extremes(even if the derivative changes sign). Answer: function increases by Knowledge of monotonicity intervals and extrema, coupled with established asymptotes already gives a very good idea of appearance function graphics. A person of average training is able to verbally determine that the graph of a function has two vertical asymptotes and an oblique asymptote. Here is our hero: Example 4 Find the extrema of the function Example 5 Find monotonicity intervals, maxima and minima of the function …it’s almost like some kind of “X in a cube” holiday today.... Each task has its own substantive nuances and technical subtleties, which are commented on at the end of the lesson.
This means that the equality \(f(t)=f(u)\) is possible if and only if \(t=u\) . Let's return to the original variables and solve the resulting equation:
\
\
\
We need to find the values of \(a\) at which the equation will have at least two roots, also taking into account the fact that \(a>0\) .
Therefore, the answer is: \(a\in (0;+\infty)\) .
\(f"(x)=6(x-1)^2 \Rightarrow f(x)=2(x-1)^3 \Rightarrow\) 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 \(1\) .
Note that \(f(0)=f(1)=0\) . So, schematically the graph \(f(x)\) looks like this: 
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Page creation date: 2016-02-12Definitions
Let the function f (x) is defined on some set of 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′′) )
.
The function is called monotonous, if it is non-decreasing or non-increasing.
Let the function f (x) does not decrease on the interval (a, b), Where .
If it is bounded above by the number M:, then there is a finite left limit at the point b:. If f (x) is not limited from above, then .
If f (x) is bounded below by the number m : , then there is a finite right limit at the point a : . If f (x) is not bounded below, then .
This theorem can be formulated more compactly.
;
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;
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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 is not decreasing
b - final number
The function is limited from above
1.1.1. Let the function be bounded from above by the number M: for .
.
;
.
at .
Let's transform the last inequality:
;
;
.
Because , then . Then
at .
at .
"Definitions of one-sided limits of a function at an end point").The function is not limited from above
1.1. Let the number b be finite: .
1.1.2. Let the function not be bounded above.
Let us prove that in this case there is a limit.
.
at .
at .
This means that the limit on the left at point b is (see "Definitions of one-sided infinite limits of a function at an end point").b early plus infinity
The function is limited from above
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.
.
According to the definition of the exact upper bound, following conditions:
;
for any positive there is an argument for which
.
at .
at .
"Definitions of one-sided limits at infinity").The function is not limited from above
1.2. Let the number b be equal to plus infinity: .
1.2.2. Let the function not be bounded above.
Let us prove that in this case there is a limit.
.
at .
This means that the limit at is equal to (see "Definitions of one-sided infinite limits at infinity").The function is not increasing
.
Here B can be either a finite number or a point at infinity. According to the definition of an exact lower bound, the following conditions are satisfied:
;
for any neighborhood of point B there is an argument for which
.
According to the conditions of the theorem, . That's why .
at .
Or
at .
Next, we note that the inequality defines the left punctured neighborhood of the point b.
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
.
If it does not increase, it does not decrease. In this case there is a limit
.
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(1)
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Let's express f in terms of g:
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Let's take an arbitrary positive number. Let there be an epsilon neighborhood of point A. The 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 .
at .
Let's replace x with -x and take into account that:
at .
The last two inequalities define the punctured right neighborhood of the point a. Then
at .
at ;
at ;
at ;
at .
at .
It means that
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Lesson and presentation in algebra in 10th grade on the topic: "Investigation of a function for monotonicity. Research algorithm"
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Algebraic problems with parameters, grades 9–11
Software environment "1C: Mathematical Constructor 6.1"
1. Decreasing and increasing functions.
2. Relationship between derivative and monotonicity of a function.
3. Two important theorems on monotonicity.
4. Examples.Decreasing and increasing functions
Let's look at the concept of increasing and decreasing functions. Guys, what is a function? 
This graph shows that the larger x, the larger y. So let's define an increasing function. A function is called increasing if a larger value of the argument corresponds to a larger value of the function.
If x2 > x1, then f(x2 > f(x1) or: the greater x, the greater y.Relationship between derivative and monotonicity of a function
Guys, now let's think about how you can apply the concept of derivative when studying function graphs. Let's draw a graph of an increasing differentiable function and draw a couple of tangents to our graph. 
Two important theorems on monotonicity
f’(x)= 0, then the function y= f(x) is constant on this interval.Examples of studying a function for monotonicity
Let's solve the inequality:
2cos(2x) - 3 ≤ 0,
2cos(2x) ≤ 3,
cos(2x) ≤ 3/2.
Because -1 ≤ cos(x) ≤ 1, which means our inequality is satisfied for any x, then by Theorem 2 the function y= sin(2x) - 3x decreases.
Let's solve the inequality:
2x + 3 ≥ 0,
x ≥ -3/2.
Then our function increases for x ≥ -3/2, and decreases for x ≤ -3/2.
Answer: For x ≥ -3/2, the function increases, for x ≤ -3/2, the function decreases.
Let's solve the inequality: $\frac(3)(2\sqrt(3x - 1))$ ≥ 0.
$\sqrt(3x - 1)$ ≥ 0,
3x - 1 ≥ 0,
x ≥ 1/3.
Let's solve the inequality:
$\frac(3)(2\sqrt(3x-1))$ ≤ 0,
3x - 1 ≤ 0.
But this is impossible, because Square root is defined only for positive expressions, which means our function has no decreasing intervals.
Answer: for x ≥ 1/3 the function increases.Problems to solve independently
a) Prove that the function y= x 9 + 4x 3 + 1x - 10 is increasing along the entire number line.
b) Prove that the function is decreasing: y= cos(5x) - 7x.
c) Examine the monotonicity of the function: y= 2x 3 + 3x 2 - x + 5.
d) Examine the monotonicity of the function: y = $\frac(3x-1)(3x+1)$. Increasing, decreasing and extrema of a function
Monotonicity of the function. Extremum points and extrema of a function
.
Actually, the definitions:
The point is called minimum point, If exists its surroundings are such that for all values of this neighborhood, the inequality holds.
– the meaning is called maximum functions;
– the meaning is called minimum functions.
Extremes– “game” meanings.How to find intervals of increasing, decreasing,
extremum points and extrema of the function?
brings the cheerful news that function is increasing throughout domain of definition.
the situation is exactly the opposite.
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When the function is defined, but not differentiable. However, at the critical point there is a right-handed derivative and a right-handed tangent, and at the other edge there are their left-handed counterparts.Why explore a function using its derivative?
a necessary condition for an extremum:
: “...we take the first derivative and equate it to zero: ...So, the solution to our equation: - it is at this point that the vertex of the parabola is located...”. Now, I think, everyone understands why the vertex of the parabola is located exactly at this point =) In general, we should start with a similar example here, but it is too simple (even for a teapot). In addition, there is an analogue at the very end of the lesson about derivative of a function. Therefore, let's increase the degree:
I remind you that you need to take some point in the interval and calculate the value of the derivative at it 
and determine its sign. It’s more profitable not to even count, but to “estimate” verbally. Let's take, for example, a point belonging to the interval and perform the substitution: 
. 
Two “pluses” and one “minus” give a “minus”, therefore, which means that the derivative is negative over the entire interval.
and decreases by . It is convenient to connect intervals of the same type with the join icon.
At the point the function reaches a minimum: 
and decreases by At the point the maximum of the function is reached: 
, and at the point – the minimum: .
Try once again to correlate the results of the study with the graph of this function.
There is no extremum at the critical point, but there is graph inflection(which, as a rule, happens in similar cases).
Soooo, who in the gallery offered to drink for this? =)
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What is monotonicity of a function? What are even, periodic, monotonic functions
