How to solve extremum point equations. How to find the extremum (minimum and maximum points) of a function. Decreasing function definition

The function y = f(x) is called increasing (waning) in some interval if for x 1< x 2 выполняется неравенство(f(x 1) < f (x 2) (f(x 1) >f(x2)).

If a differentiable function y = f(x) on a segment increases (decreases), then its derivative on this segment f "(x) > 0, (f "(x)< 0).

Dot xabout called local maximum point (minimum) of the function f(x) if there is a neighborhood of the point x o, for all points of which the inequality f(x) ≤ f(x o), (f(x) ≥f(x o)) is true.

The maximum and minimum points are called extremum points, and the values ​​of the function at these points are its extrema.

extremum points

Necessary conditions for an extremum. If point xabout is an extremum point of the function f (x), then either f "(x o) \u003d 0, or f (x o) does not exist. Such points are called critical, where the function itself is defined at the critical point. The extrema of a function should be sought among its critical points.

The first sufficient condition. Let xabout- critical point. If f "(x) when passing through a point xabout changes the plus sign to minus, then at the point x o the function has a maximum, otherwise it has a minimum. If the derivative does not change sign when passing through a critical point, then at the point xabout there is no extremum.

The second sufficient condition. Let the function f(x) have f " (x) in a neighborhood of the point xabout and the second derivative f "" (x 0) at the very point x o. If f "(x o) \u003d 0, f "" (x 0)> 0, (f "" (x 0)<0), то точкаx o is a local minimum (maximum) point of the function f(x). If f "" (x 0) = 0, then you must either use the first sufficient condition, or involve higher ones.

On a segment, the function y =f(x) can reach its minimum or maximum value either at critical points or at the ends of the segment.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x - 2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. So as when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum.When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Having calculated the values ​​of the function at the points x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Tasks for finding the extremum of a function

Example 3.23.a

Solution. x and y. The area of ​​the site is equal to S =xy. Let y is the length of the side adjacent to the wall. Then, by the condition, the equality 2x + y = a must hold. Therefore, y = a - 2x and S =x(a - 2x), where 0 ≤x ≤a/2 (the length and width of the pad cannot be negative). S " = a - 4x, a - 4x = 0 for x = a/4, whence y = a - 2×a/4 = a/2. Since x = a/4 is the only critical point, check if the sign changes derivative as we pass through this point, for x< a/4, S " >0, and for x > a/4, S "< 0, значит, в точке x = a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед). Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24.

Solution.
R = 2, H = 16/4 = 4.

Example 3.22. Find the extrema of the function f(x) = 2x 3 - 15x 2 + 36x - 14.

Solution. Since f "(x) \u003d 6x 2 - 30x +36 \u003d 6 (x - 2) (x - 3), then the critical points of the function x 1 \u003d 2 and x 2 \u003d 3. Extreme points can only be at these points. So as when passing through the point x 1 \u003d 2, the derivative changes sign from plus to minus, then at this point the function has a maximum.When passing through the point x 2 \u003d 3, the derivative changes sign from minus to plus, therefore, at the point x 2 \u003d 3, the function has a minimum. Having calculated the values ​​of the function at the points x 1 = 2 and x 2 = 3, we find the extrema of the function: maximum f (2) = 14 and minimum f (3) = 13.

Example 3.23. It is necessary to build a rectangular area near the stone wall so that it is fenced off with wire mesh on three sides, and adjoins the wall on the fourth side. For this there is a linear meters of the grid. At what aspect ratio will the site have the largest area?

Solution. Denote the sides of the site through x and y. The area of ​​the site is S = xy. Let y is the length of the side adjacent to the wall. Then, by the condition, the equality 2x + y = a must hold. Therefore y = a - 2x and S = x(a - 2x), where
0 ≤x ≤a/2 (the length and width of the site cannot be negative). S "= a - 4x, a - 4x = 0 for x = a/4, whence
y = a - 2a/4 = a/2. Since x = a/4 is the only critical point, let's check whether the sign of the derivative changes when passing through this point. At x< a/4, S " >0, and for x >a/4 S "< 0, значит, в точке x=a/4 функция S имеет максимум. Значение функции S(a/4) = a/4(a - a/2) = a 2 /8 (кв. ед). Поскольку S непрерывна на и ее значения на концах S(0) и S(a/2) равны нулю, то найденное значение будет наибольшим значением функции. Таким образом, наиболее выгодным соотношением сторон площадки при данных условиях задачи является y = 2x.

Example 3.24. It is required to make a closed cylindrical tank with a capacity of V=16p ≈ 50 m 3 . What should be the dimensions of the tank (radius R and height H) in order to use the least amount of material for its manufacture?

Solution. The total surface area of ​​the cylinder is S = 2pR(R+H). We know the volume of the cylinder V = pR 2 H Þ H = V/pR 2 =16p/ pR 2 = 16/ R 2 . Hence, S(R) = 2p(R 2 +16/R). We find the derivative of this function:
S "(R) \u003d 2p (2R- 16 / R 2) \u003d 4p (R- 8 / R 2). S " (R) \u003d 0 for R 3 \u003d 8, therefore,
R = 2, H = 16/4 = 4.

Consider two teeth of a well-known saw profile. Let's direct the axis along the flat side of the saw, and the axis - perpendicular to it. Let's get a graph of some function, shown in Fig. one.

It is quite obvious that both at the point and at the point, the values ​​of the function turn out to be the largest in comparison with the values ​​at the neighboring points on the right and left, and at the point - the smallest in comparison with the neighboring points. The points are called the extremum points of the function (from the Latin extremum - “extreme”), the points and are the maximum points, and the point is the minimum point (from the Latin maximum and minimum - “greatest” and “smallest”).

Let us refine the definition of an extremum.

A function at a point is said to have a maximum if there is an interval containing the point and belonging to the domain of the function, such that for all points of this interval it turns out to be . Accordingly, the function at a point has a minimum if the condition is satisfied for all points of a certain interval.

On fig. Figures 2 and 3 show graphs of functions that have an extremum at a point.

Let us pay attention to the fact that, by definition, the extremum point must lie inside the interval of setting the function, and not at its end. Therefore, for the function shown in Fig. 1, it cannot be assumed that it has a minimum at the point.

If in this definition of the maximum (minimum) of a function, we replace the strict inequality with a non-strict one , then we obtain the definition of a non-strict maximum (non-strict minimum). Consider, for example, the profile of a mountain top (Fig. 4). Each point of a flat area - a segment is a non-strict maximum point.

In differential calculus, the study of a function for extrema is very effective and quite simply carried out using a derivative. One of the main theorems of differential calculus, which establishes a necessary condition for the extremum of a differentiable function, is Fermat's theorem (see Fermat's theorem). Let the function at a point have an extremum. If there is a derivative at this point, then it is equal to zero.

In geometric language, Fermat's theorem means that at the extremum point the tangent to the graph of the function is horizontal (Fig. 5). The converse statement, of course, is not true, which is shown, for example, by the graph in Fig. 6.

The theorem is named after the French mathematician P. Fermat, who was one of the first to solve a number of extremum problems. He did not yet have at his disposal the concept of a derivative, but applied a method in his research, the essence of which is expressed in the statement of the theorem.

A sufficient condition for the extremum of a differentiable function is a change in the sign of the derivative. If at a point the derivative changes sign from minus to plus, i.e. its decrease is replaced by an increase, then the point will be the minimum point. On the contrary, the point will be the maximum point if the derivative changes sign from plus to minus, i.e. goes from ascending to descending.

The point where the derivative of the function is equal to zero is called stationary. If a differentiable function is investigated for an extremum, then all its stationary points should be found and the signs of the derivative should be considered to the left and to the right of them.

We investigate the function for an extremum.

Let's find its derivative: .

Before learning how to find the extrema of a function, it is necessary to understand what an extremum is. The most general definition of an extremum says that it is the smallest or largest value of a function used in mathematics on a certain set of a number line or graph. In the place where the minimum is, the extremum of the minimum appears, and where the maximum is, the extremum of the maximum appears. Also in such a discipline as mathematical analysis, local extrema of a function are distinguished. Now let's look at how to find extremums.

Extremes in mathematics are among the most important characteristics of a function, they show its largest and smallest value. The extrema are found mainly at the critical points of the found functions. It is worth noting that it is at the extremum point that the function radically changes its direction. If we calculate the derivative of the extremum point, then, according to the definition, it must be equal to zero or it will be completely absent. Thus, to learn how to find the extremum of a function, you need to perform two sequential tasks:

• find the derivative for the function that needs to be determined by the task;
• find the roots of the equation.

The sequence of finding the extremum

1. Write down the function f(x) that is given. Find its first-order derivative f "(x). Equate the resulting expression to zero.
2. Now you have to solve the equation that turned out. The resulting solutions will be the roots of the equation, as well as the critical points of the function being defined.
3. Now we determine which critical points (maximum or minimum) are the found roots. The next step, after we learned how to find the extremum points of a function, is to find the second derivative of the desired function f "(x). It will be necessary to substitute the values ​​​​of the found critical points into a specific inequality and then calculate what happens. If this happens, that the second derivative turns out to be greater than zero at the critical point, then it will be the minimum point, and otherwise it will be the maximum point.
4. It remains to calculate the value of the initial function at the required maximum and minimum points of the function. To do this, we substitute the obtained values ​​into the function and calculate. However, it should be noted that if the critical point turned out to be a maximum, then the extremum will also be maximum, and if it is a minimum, then it will be minimum by analogy.

Algorithm for finding an extremum

To summarize the knowledge gained, let's make a brief algorithm of how to find extremum points.

1. We find the domain of the given function and its intervals, which determine exactly on what intervals the function is continuous.
2. We find the derivative of the function f "(x).
3. We calculate the critical points of the equation y = f (x).
4. We analyze the changes in the direction of the function f (x), as well as the sign of the derivative f "(x) where the critical points separate the domain of definition of this function.
5. Now we determine whether each point on the graph is a maximum or a minimum.
6. We find the values ​​of the function at those points that are extremums.
7. We fix the result of this study - extrema and intervals of monotonicity. That's all. Now we have considered how to find an extremum on any interval. If you need to find an extremum on a certain interval of a function, then this is done in a similar way, only the boundaries of the research being performed are necessarily taken into account.

So, we have considered how to find the extremum points of a function. With the help of simple calculations, as well as knowledge about finding derivatives, you can find any extremum and calculate it, as well as graphically designate it. Finding extremes is one of the most important sections of mathematics, both at school and at a higher educational institution, therefore, if you learn how to determine them correctly, then learning will become much easier and more interesting.

Function extremes

Definition 2

A point $x_0$ is called a point of maximum of the function $f(x)$ if there exists a neighborhood of this point such that for all $x$ from this neighborhood the inequality $f(x)\le f(x_0)$ is satisfied.

Definition 3

A point $x_0$ is called a point of maximum of the function $f(x)$ if there exists a neighborhood of this point such that for all $x$ from this neighborhood the inequality $f(x)\ge f(x_0)$ is satisfied.

The concept of an extremum of a function is closely related to the concept of a critical point of a function. Let us introduce its definition.

Definition 4

$x_0$ is called a critical point of the function $f(x)$ if:

1) $x_0$ - internal point of the domain of definition;

2) $f"\left(x_0\right)=0$ or does not exist.

For the concept of an extremum, one can formulate theorems on sufficient and necessary conditions for its existence.

Theorem 2

Sufficient extremum condition

Let the point $x_0$ be critical for the function $y=f(x)$ and lie in the interval $(a,b)$. Let on each interval $\left(a,x_0\right)\ and\ (x_0,b)$ the derivative $f"(x)$ exist and keep a constant sign. Then:

1) If on the interval $(a,x_0)$ the derivative $f"\left(x\right)>0$, and on the interval $(x_0,b)$ the derivative $f"\left(x\right)

2) If the derivative $f"\left(x\right)0$ is on the interval $(a,x_0)$, then the point $x_0$ is the minimum point for this function.

3) If both on the interval $(a,x_0)$ and on the interval $(x_0,b)$ the derivative $f"\left(x\right) >0$ or the derivative $f"\left(x\right)

This theorem is illustrated in Figure 1.

Figure 1. Sufficient condition for the existence of extrema

Examples of extremes (Fig. 2).

Figure 2. Examples of extremum points

The rule for examining a function for an extremum

2) Find the derivative $f"(x)$;

7) Draw conclusions about the presence of maxima and minima on each interval, using Theorem 2.

Function Ascending and Decreasing

Let us first introduce the definitions of increasing and decreasing functions.

Definition 5

A function $y=f(x)$ defined on an interval $X$ is called increasing if for any points $x_1,x_2\in X$ for $x_1

Definition 6

A function $y=f(x)$ defined on an interval $X$ is called decreasing if for any points $x_1,x_2\in X$ for $x_1f(x_2)$.

Examining a Function for Increasing and Decreasing

You can investigate functions for increasing and decreasing using the derivative.

In order to examine a function for intervals of increase and decrease, you must do the following:

1) Find the domain of the function $f(x)$;

2) Find the derivative $f"(x)$;

3) Find the points where the equality $f"\left(x\right)=0$;

4) Find points where $f"(x)$ does not exist;

5) Mark on the coordinate line all the found points and the domain of the given function;

6) Determine the sign of the derivative $f"(x)$ on each resulting interval;

7) Conclude: on the intervals where $f"\left(x\right)0$ the function increases.

Examples of problems for the study of functions for increasing, decreasing and the presence of extremum points

Example 1

Investigate the function for increasing and decreasing, and the presence of points of maxima and minima: $f(x)=(2x)^3-15x^2+36x+1$

Since the first 6 points are the same, we will draw them first.

1) Domain of definition - all real numbers;

2) $f"\left(x\right)=6x^2-30x+36$;

3) $f"\left(x\right)=0$;

\ \ \

4) $f"(x)$ exists at all points of the domain of definition;

5) Coordinate line:

Figure 3

6) Determine the sign of the derivative $f"(x)$ on each interval:

\ \}