Study of functions for monotonicity and extremum.

Root

Extrema and convexity.

Asymptotes of the graph of a functionDefinition. Critical point functions = at(f X

) is the point at which the derivative is zero or does not exist. Theorem. If in the interval (a; b) the derivative

) is the point at which the derivative is zero or does not exist. positive/negative, then the function increases/decreases in this interval. If, upon passing through the critical point, the derivative

Asymptotes of the graph of a function changes sign from “+” to “−” (from “−” to “+”), then − is the maximum (minimum) point of the function Function called convex up(down) in the interval (a; b), if in this interval the points of the graph lie under (above) the tangents constructed at these points. Inflection point

is a point in the graph of a function that divides it into parts with different directions of convexity.

Example 2.3. Explore function

for monotony and extrema, convexity.

1. We examine the function for monotonicity and extrema. Let's make a drawing ().

rice. 2.1

y′′

−

issue down

issue up

Rice. 2.2. Study of a function for convexity

Let's calculate the ordinates of the inflection points of the graph:

Coordinates of inflection points: (0; 0), (1; −1).

1) 2) 3)

4) 5) 6)

2.32. Examine the function for monotonicity and extrema:

1) 2.33. Find the smallest and largest values of the function:

2) on the interval;

3) on the interval [−1; 1];

4) on the interval [−4; 4];

on the interval [−2; 1]. f 2.34. Production costs C (cu) depend on the volume of output f(units): Find the highest production costs if f changes over the interval . Find value

, at which the profit will be maximum if the revenue from the sale of a unit of production is equal to 15 c.u. e.

2.35. It is required to allocate a rectangular plot of land of 512 m2, fence it and divide it with a fence into three equal parts parallel to one of the sides of the site. What should be the size of the site so that the least amount of material is used for the fencing?

2.36. Given the perimeter of a rectangular window, find its dimensions such that it lets in the greatest amount of light. f 2.37. Find the maximum profit if income R and costs C are determined by the formulas: where

− quantity of goods sold. 2.38. Dependence of production volume W from capital costs TO
determined by the function from capital costs Find the change interval

, where increasing capital costs is ineffective.

2.40. The dependence of the volume of output (in monetary units) on capital costs is determined by the function Find the interval of values at which increasing capital costs is ineffective.

2.41. It is believed that the increase in sales from advertising costs (million rubles) is determined by the ratio Income from the sale of a unit of production is equal to 20 thousand rubles. Find the level of advertising costs at which the company will receive maximum profit.

2.42. The income from the production of products using resource units is equal to The cost of a resource unit is 10 den. units How much of a resource should be purchased in order for the profit to be greatest?

2.43. The cost function has the form The income from the sale of a unit of production is 50. Find the maximum profit value that the manufacturer can receive.

2.44. The dependence of the monopoly's income on the quantity of output is defined as: The cost function in this interval has the form Find the optimal output value for the monopoly.

2.45. The price for the products of a monopoly producer is set in accordance with the ratio identified as . At what value of product output will the income from its sales be greatest?

2.46. The cost function has the following form at at . Currently the level of production Under what condition on the parameter p Is it profitable for a company to reduce output if the income from the sale of a unit of output is 50?

Knowledge hypermarket >>Mathematics >>Mathematics grade 10 >> Application of the derivative to study functions for monotonicity and extrema

§ 35. Application of the derivative to study functions for monotonicity and extrema

1. Study of functions for monotonicity

In Fig. 129 shows a graph of some increasing differentiable function y = f(x). Let's draw tangents to graphics at points x= x 1 and x- x 2. What do the constructed lines have in common? What they have in common is that they make an acute angle with the x-axis, which means that both straight lines have a positive angular coefficient. But the angular coefficient of the tangent is equal to the value of the derivative in the abscissa of the point of tangency. Thus, at the point x = x 3 the tangent is parallel to the x axis, at this point the equality f"(X 3) = 0 holds. In general, at any point x from the domain of definition of the increasing differentiable function the inequality holds

In Fig. 130 shows a graph of some decreasing differentiable functions y = f(x). Let's draw tangents to the graph at points x= x 1 and x= x 2. At the constructed straight lines? What they have in common is that they both make an obtuse angle with the x-axis, which means both straight lines have a negative slope. But the angular coefficient of the tangent is equal to the value of the derivative in the abscissa of the point of tangency. Thus, at the point x = x 3 the tangent is parallel to the x axis, at this point the equality f "(x 3) = 0 holds. In general, at any point x from the domain of definition of the decreasing differentiable function, the inequality
These considerations show that there is a certain connection between the nature of the monotonicity of a function and the sign of its derivative:

if a function increases on an interval and has a derivative on it, then the derivative is non-negative; if a function decreases on an interval and has a derivative on it, then the derivative is non-positive.
For practice, it is much more important that the converse theorems are also true, showing how the sign of the derivative can be used to establish the nature of the monotonicity of a function on an interval. In this case, in order to avoid misunderstandings, only open spaces are taken, i.e. intervals or open beams. The fact is that for a function defined on the interval [a, b], it is not very correct to pose the question of the existence and value of the derivative at the end point (at the point x = a or at the point x = b), since at the point x = and the increment of the argument can only be positive, and at the point x = b - only negative. The definition of a derivative does not provide for such restrictions.

The proofs of these theorems are usually carried out in a course of higher mathematics. We will limit ourselves to the “hands-on” reasoning carried out above and, for greater persuasiveness, we will also give a physical interpretation of the theorems formulated.

Let a material point move in a straight line, s = s(t) is the law of motion. If the speed is always positive, then the point is constantly moving away from the origin, i.e. the function s = s(t) increases. If the speed is always negative, then the point is constantly approaching the origin, i.e. the function s = s(t) is decreasing. If the speed of movement was positive, then at some particular moment in time it turned to zero, and then became positive again, then the moving body at the specified moment in time seems to slow down, but still continues to move away from the starting point. So in this case the function s = s(t) increases. What is speed? This is the derivative of the path with respect to time. This means that the nature of the monotonicity of the function—in this case, the function s = s(t)—depends on the sign of the derivative (speed). This is exactly what both theorems formulated say.

Example 1. Prove that the function is increasing on the entire number line.
Solution. Let's find the derivative of the given function:

Obviously, for all x the inequality . This means, according to Theorem 1, the function increases on the entire number line.

Example 2. a) Prove that the function y = 5cos x + 3m4x - 10x decreases on the entire number line;
b) decide the equation 5cos x + sin4x - 10x = x 3 + 5.

Solution, a) Find the derivative of the given function:

The resulting expression is always negative. In fact, for all values of x the following inequalities hold:

This inequality holds for all values of x. This means, by Theorem 2, the function decreases on the entire number line.

b) Consider the equation 5cosx + sin4x - 10x = x 3 + 5. As was just established, y = 5cosx + sin4x-10x is a decreasing function. At the same time, y = x 3 +5 is an increasing function. The following statement holds: if one of the functions y = f(x) or y = s(x) increases, and the other decreases, and if the equation f(x) = g(x) has a root, then only one (Fig. 131 clearly illustrates this statement). It is not difficult to find the root of a given equation - this is the number x = 0 (with this value, the equation turns into the correct numerical equality 5 = 5).
So, x = 0 is the only root of the given equation.

Example 3. a) Examine the monotonicity of the function y = 2x 3 + 3x 2 -1; b) construct a graph of this function.

Solution, a) To study a function for monotonicity means to find out at which intervals of the domain of definition the function increases and at which it decreases. According to Theorems 1 and 2, this is related to the sign of the derivative.

Let's find the derivative of this function: f"(x)=6x 2 +6x and then f"(x)=6x(x + 1).

In Fig. 132 the signs of the derivative over intervals of the domain of definition are schematically indicated: on the ray (-oo, -1) the derivative is positive, on the interval (-1,0) - negative, on the ray (0,+ - positive. This means that on the first of the indicated intervals the function increases, decreases in the second, increases in the third.

Usually, if a function is continuous not only on an open interval, but also at its end points, these end points are included in the monotonicity interval of the function.

Thus, the given function increases on the ray, increases on the ray and decreases on the segment [-1,0].

b) Graphs of functions are built “point by point”. To do this, you need to make a table of the values of the function y = 2x3 +3x 2 -1, where you must include the values of the function at the end points of the monotonicity intervals x = -1 and x = 0 and a couple of other values:

Let's mark these points on the coordinate plane. Let us take into account the intervals of increase and decrease of the function found in part a) and also the fact that at the points x = -1 and x = 0 the derivative of the function is equal to zero, i.e. the tangent to the graph of the function at the indicated points is parallel to the abscissa axis, moreover, at the point (-1; 0) it even coincides with the abscissa axis. Finally, let us take into account that the function is continuous, i.e. its graph is a solid line. The graph of the function specified in the condition is shown in Fig. 133.

Concluding our discussion on the study of functions for monotonicity, let us pay attention to one circumstance. We have seen that if the inequality f"(x) > 0 is satisfied on the interval X, then the function y-f(x) increases on the interval X; if the inequality f"(x) is satisfied on the interval X< 0, то функция убывает на этом промежутке. А что будет, если на всем промежутке выполняется тождество (х) =0 ? Видимо, функция не должна ни возрастать, ни убывать. Что же это за функция? Ответ очевиден - это постоянная функция у = С (буква С - первая буква слова соп81ап1а, что означает «постоянная»). Справедлива следующая теорема, формальное доказательство которой мы не приводим, ограничиваясь приведенными выше правдоподобными рассуждениями.

In the future, we will use this theorem, i.e. we can be convinced of its usefulness for mathematics. And now we will give (for the more curious) an example of using Theorem 3 (from the category of mathematical entertainment). We will present a new way to prove the well-known identity sin 2 x + cos 2 x= 1.
Consider the function y = f(x), where f(x) = sin 2 x + cos 2 x. Let's find its derivative:

So, for all x the equality f"(x) = 0 holds, which means f(x) = C. To find the value of C, it is enough to calculate the value of the function at any point x, for example, x = 0. We have: f(0) = sin 2 0+cos2 0=0 + 1 = 1.

Thus, C = 1, i.e. sin 2 x + cos 2 x = 1

2. Extremum points of the function and finding them

Let's return to the graph of the function y = 2 x 3 + 3x 2 -1 (Fig. 133). There are two unique points on the graph that determine its structure - these are points (-1; 0) and (0; -1). At these points:

1) there is a change in the nature of the monotonicity of the function (to the left of the point x = -1 the function increases, to the right of it, but only up to the point x = 0, the function decreases; to the left of the point x = 0 the function decreases, to the right of it it increases);

2) the tangent to the graph of the function is parallel to the x axis, i.e. the derivative of the function at each of the indicated points is equal to zero;

3) f(-1) - the largest value of the function, but not in the entire domain of definition, but in the local sense, i.e. compared to the values of the function from a certain neighborhood of the point x = -1. In the same way, f(0) is the smallest value of the function, but not in the entire domain of definition, but in the local sense, i.e. compared to the values of the function from some neighborhood of the point x = 0.

Now look at fig. 134, which shows a graph of another function. Isn't it similar to the previous graph? It has the same two unique points, but one of the above three features of these points has changed: now the tangents to the graph at these points are not parallel to the x-axis. At the point x = -1 the tangent does not exist at all, and at the point x = 0 it is perpendicular to the x axis (more precisely, it coincides with the y axis).

You already know the further course of reasoning: if a new mathematical model or a new feature of a mathematical model appears, it must be specially studied, i.e. introduce a new term, new designations, formulate new properties.

Definition 1. A point x = x 0 is called a minimum point of the function y = f (x) if this point has a neighborhood for which all points (except the point x = x 0 itself) satisfy the following inequality:
f(x)>f(x0).

Thus, the functions whose graphs are shown in Fig. 133 and 134 have a minimum point x=0. Why? Because this point has a neighborhood, for example, or (-0.2, 0.2), for all
points of which, except for the point x = 0, the inequality f(x) > f(O) holds. This is true for both functions.
The value of the function at the minimum point is usually denoted by . Do not confuse this value (the smallest, but in a local sense) with i.e. with the smallest value of the function in the entire domain of definition under consideration (in the global sense). Look again at Fig. 133 and 134. You see that neither one nor the other function has the smallest value, but it does exist.

Definition 2. A point x = x 0 is called a maximum point of the function y = f (x) if this point has a neighborhood for which all points, except the point x = x 0 itself, satisfy the following inequality:
f(x)

Thus, the functions whose graphs are shown in Fig. 133 and 134 have a maximum point x = - 1. Why? Because at this point
there is a neighborhood, for example, for all points of which, except x=-1, the inequality f(x) is satisfied< f(-1). Это верно для обеих функций.
The value of the function at the maximum point is usually denoted by . Do not confuse this value (the largest, but in a local sense) with ., i.e. with the largest value of the function in the entire domain of definition under consideration (in the global sense). Look again at Fig. 133 and 134. You see that neither one nor the other function has the greatest value, but exists.

The minimum and maximum points of a function are combined by a common term - extremum points (from the Latin word extremum - “extreme”).

How to find extremum points of a function? We can find the answer to this question by once again analyzing the graphical models presented in Fig. 133 and 134.

Please note: for the function whose graph is shown in Fig. 133, at both extremum points the derivative vanishes (tangents are parallel to the x-axis). And for the function whose graph is shown in Fig. 134, at both extremum points the derivative does not exist. This is not accidental, since, as proven in the course of mathematical analysis, the following theorem is true.

Theorem 4. If the function y = f(x) has an extremum at the point x = x 0, then at this point the derivative of the function is either zero or does not exist.

For convenience, we will agree to call the internal points of the domain of definition of a function, in which the derivative of the function is equal to zero, stationary, and the internal points of the domain of definition of the function, in which the function is continuous, but the derivative of the function does not exist, critical.

Example 4. Construct a graph of the function y = 2x 2 -6x + 3.

Solution. You know that the graph of a given quadratic function is a parabola, and the branches of the parabola are directed upward, since the coefficient of xr is positive. But in this case, the vertex of the parabola is the minimum point of the function, the tangent to the parabola at its vertex is parallel to the x-axis, which means that at the vertex of the parabola the condition y"=0 must be satisfied. We have: y"=(2x 2 -6x + 3)"=4x -6.

Equating the derivative to zero, we get: 4x-6=0; x = 1.5.

Substituting the found value of x into the parabola equation, we get:

y = 21.52 - 6-1.5 + 3 = -1.5. So, the vertex of the parabola is the point (1.5; -1.5), and the axis of the parabola is the straight line x = 1.5 (Fig. 135). As control points, it is convenient to take the point (0; 3) and the point (3; 3) symmetrical to it relative to the parabola axis. In Fig. 136, using the three points found, a parabola is constructed - a graph of a given quadratic function.

Do you remember how we plotted the quadratic function y = ax 2 + bx + c in grades 8-9? Almost the same way, only the axis of the parabola was found not using a derivative, but using a formula that had to be memorized. The solution shown in Example 4 frees you from having to remember this formula. To find the abscissa of the vertex of the parabola y = ax 2 + bx + c or the equation of its axis of symmetry, it is enough to equate the derivative of the quadratic function to zero.

Now let's return to Theorem 4, which says that if at the point x = x 0 the function y = f(x) has an extremum, then x = x 0 is a stationary or critical point of the function. A natural question arises: is the converse theorem true, i.e. Is it true that if x = x 0 is a stationary or critical point, then at this point the function has an extremum? We answer: no, incorrect. Look at fig. 137, which shows a graph of an increasing function that does not have extremum points. This function has a stationary point x = x 1, at which the derivative vanishes (at this point the graph of the function has a tangent parallel to the x axis), but this is not an extremum point, but an inflection point, and there is a critical point x = x 2, in which the derivative does not exist, but this is also not an extremum point, but a break point in the graph. Therefore, let's say this: Theorem 4 provides only a necessary condition for an extremum (the direct theorem is valid), but it is not a sufficient condition (the inverse theorem is not true).

But what about a sufficient condition? How to find out whether there is an extremum at a stationary or critical point? To answer this question, consider again the function graphs presented in Fig. 133, 134, 136 and 137.
We note that when passing through the maximum point (we are talking about the point x = -1 in Fig. 133 and 134), the nature of the monotonicity of the function changes: to the left of the maximum point the function increases, to the right it decreases. The signs of the derivative change accordingly: to the left of the maximum point the derivative is positive, to the right it is negative.
We note that when passing through the minimum point (we are talking about the point x = 0 in Fig. 133 and 134 and the point x = 1.5 in Fig. 136), the nature of the monotonicity of the function also changes: to the left of the minimum point the function decreases, to the right it increases . The signs of the derivative change accordingly: to the left of the minimum point the derivative is negative, to the right it is positive.

If both to the left and to the right of the stationary or critical point the derivative has the same sign, then there is no extremum at this point, this is exactly the case with the function whose graph is shown in Fig. 137.
Our reasoning can serve as confirmation (but, of course, not proof - rigorous proofs are carried out in the course of mathematical analysis) of the validity of the following theorem.

Theorem 5 (sufficient conditions for an extremum). Let the function y=f(x) be continuous on the interval X and have a stationary or critical point x = x 0 inside the interval.

a) if this point has a neighborhood such that in it at x<х 0 выполняется неравенство f(x) < 0,а при x >x 0 - inequality f"x)>0, then x =x 0 - minimum point of the function Y=f(x);

b) if this point has a neighborhood such that in it for x< x 0 выполняется неравенство f"(x) >Oh, and for x > x 0 - inequality f(x)< О, то x = x 0 - точка максимума функции У=f(х);

c) if this point has a neighborhood such that both to the left and to the right of the point x 0 the signs of the derivative are the same, then at the point x = x 0 there is no extremum.

Example 5. a) Find the extremum points of the function
y = 3x 4 -16x 3 + 24x2 -11; b) construct a graph of this function.

Solution, a) Find the derivative of this function:

The derivative vanishes at the points x = O and x = 2 - these are two stationary points of the given function. In Fig. 138 schematically shows the signs of the derivative over intervals of the domain of definition: on the interval the derivative is negative, on the interval (0, 2) it is positive, on the interval it is positive.
This means that x = 0 is the minimum point of the function, and x = 2 is not the extremum point. At the first of the above intervals the function decreases, at the second and third it increases.

At the minimum point x = 0 we have f(0) = -11 (substituted the value x = 0 into the analytical specification of the function), which means = -11.

b) To construct a graph of a function, you need to know the particularly important points of the graph. These include:
- found minimum point (0; -11);

Stationary point x = 2; at this point

Points of intersection with coordinate axes; in this example, this is the already found point (0; -11) - the point of intersection of the graph with the y-axis. And one more thing: you can guess that f(1)=0, which means that the point of intersection of the graph with the x-axis has been found - this is point (1; 0).

So, we have a minimum point (0; -11), the point of intersection of the graph with the x-axis - point (1; 0) and a stationary point (2; 5). At this point, the tangent to the graph of the function is horizontal, but this is not an extremum point, but an inflection point.

The function graph is shown schematically in Fig. 139. Note that there is one more point of intersection of the graph with the x-axis, but we were not able to find it.

Concluding this point, we note that we have actually developed

Algorithm for studying the continuous function "y = f(x)"on monotonicity and extrema

1. Find the derivative f"(x).
2. Find stationary and critical points.
3. Mark stationary and critical points on the number line and determine the signs of the derivative on the resulting intervals.
4. Based on the theorems from § 35, draw conclusions about the monotonicity of the function and its extremum points.
Note that if a given function has the form then the poles of the function, i.e. the points at which the denominator q(x) becomes zero are also marked on the number line, and this is done before determining the signs of the derivative. But, of course, the poles cannot be extremum points.
Example 6. Examine the function for monotonicity and extrema.
Solution. Note that the function is continuous everywhere, except at the point x = 0. Let's use the above algorithm.
1) Find the derivative of the given function:

2) The derivative vanishes at the points x = 2 and x = -2 - these are stationary points. The derivative does not exist at the point x = 0, but this is not a critical point, it is a discontinuity point of the function (pole).

3) Mark the points -2, 0 and 2 on the number line and place the signs of the derivative on the resulting intervals (Fig. 140).

4) We draw conclusions: on the ray (-°°, -2] the function decreases, on the half-interval [-2, 0] the function increases, on the half-interval (0, 2] the function decreases, on the ray the function increases, on the interval (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(x 2).

So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2). 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 - отрицательные числа. Тогда - х 1 >- x 2, and both sides of the last inequality are positive numbers, and therefore (we again used the inequality proven in example 1 from § 33). Next we have, where we get from.

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

Usually the terms “increasing function” and “decreasing function” are combined under the general name monotonic function, and the study of a function for increasing and decreasing is called the study of a function for monotonicity.

Solution.

1) Let’s plot the function y = 2x2 and take the branch of this parabola at x< 0 (рис. 130).

2) Construct and select its part on the segment (Fig. 131).

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

1. The domain of definition of the function is the entire number line.

2. y = 0 at x = 0; y > 0 for x > 0.

3. The function decreases on the ray (-oo, 0], increases on the segment, decreases on the ray, is convex upward on the segment, convex downward on the ray, then the function decreases. But on the numerical interval (Fig. 128).

3. Function y

1. Consider a function on the interval (0, + 00).
Let x1< х 2 . Так как х 1 и х 2 — положительные числа, то из х 1 < x 2 следует (см. пример 1 из § 33), т. е. f(x 1) >f(x 2).
So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2). 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 — отрицательные числа. Тогда - х 1 >- x 2, and both parts of the latter are unequal
properties are positive numbers, and therefore (we again used the inequality proven in Example 1 from § 33). Next we have, where we get from.
So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2) i.e. function decreases on the open ray (- 00 , 0)
Usually the terms “increasing function” and “decreasing function” are combined under the general name monotonic function, and the study of a function for increasing and decreasing is called the study of a function for monotonicity.

Solution.

1) Let’s plot the function y = 2x2 and take the branch of this parabola at x< 0 (рис. 130).

2) Let's build a graph of the function and highlight its part on the segment (Fig. 131).

3) Let's construct a hyperbola and select its part on the open ray (4, + 00) (Fig. 132).
4) Let us depict all three “pieces” in one coordinate system - this is the graph of the function y = f(x) (Fig. 133).
Let's read the graph of the function y = f(x).
1. The domain of definition of the function is the entire number line.

2. y = 0 at x = 0; y > 0 for x > 0.

3. The function decreases on the ray (-oo, 0], increases on the segment, decreases on the ray, is convex upward on the segment, convex downward on the ray)