The largest and smallest values ​​of a function of two variables in a closed region. The largest and smallest value of the function

Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let the given function have finite partial derivatives of the first order in this region (with the possible exception of a finite number of points). To find the largest and smallest values ​​of a function of two variables in a given closed region, three steps of a simple algorithm are required.

Algorithm for finding the largest and smallest values ​​of the function $z=f(x,y)$ in the closed domain $D$.

1. Find the critical points of the function $z=f(x,y)$ that belong to the region $D$. Compute function values ​​at critical points.
2. Investigate the behavior of the function $z=f(x,y)$ on the boundary of the domain $D$ by finding the points of possible maximum and minimum values. Calculate the function values ​​at the obtained points.
3. From the function values ​​obtained in the previous two paragraphs, choose the largest and smallest.

What are critical points? show/hide

Under critical points imply points where both first-order partial derivatives are equal to zero (i.e. $\frac(\partial z)(\partial x)=0$ and $\frac(\partial z)(\partial y)=0 $) or at least one partial derivative does not exist.

Often the points at which the first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.

Example #1

Find the maximum and minimum values ​​of the function $z=x^2+2xy-y^2-4x$ in the closed region bounded by the lines $x=3$, $y=0$ and $y=x+1$.

We will follow the above, but first we will deal with the drawing of a given area, which we will denote by the letter $D$. We are given the equations of three straight lines, which limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the y-axis (axis Oy). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct a straight line $y=x+1$ let's find two points through which we draw this straight line. You can, of course, substitute a couple of arbitrary values ​​instead of $x$. For example, substituting $x=10$, we get: $y=x+1=10+1=11$. We have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points where the line $y=x+1$ intersects with the lines $x=3$ and $y=0$. Why is it better? Because we will lay down a couple of birds with one stone: we will get two points for constructing the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines that bound the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ - at the point $(-1;0)$. In order not to clutter up the course of the solution with auxiliary explanations, I will put the question of obtaining these two points in a note.

How were the points $(3;4)$ and $(-1;0)$ obtained? show/hide

Let's start from the point of intersection of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second lines, so to find unknown coordinates, you need to solve the system of equations:

$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$

The solution of such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.

Now let's find the point of intersection of the lines $y=x+1$ and $y=0$. Again, we compose and solve the system of equations:

$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$

Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (abscissa axis).

Everything is ready to build a drawing that will look like this:

The question of the note seems obvious, because everything can be seen from the figure. However, it is worth remembering that the drawing cannot serve as evidence. The figure is just an illustration for clarity.

Our area was set using the equations of lines that limit it. It's obvious that these lines define a triangle, don't they? Or not quite obvious? Or maybe we are given a different area, bounded by the same lines:

Of course, the condition says that the area is closed, so the picture shown is wrong. But to avoid such ambiguities, it is better to define regions by inequalities. We are interested in the part of the plane located under the line $y=x+1$? Ok, so $y ≤ x+1$. Our area should be located above the line $y=0$? Great, so $y ≥ 0$. By the way, the last two inequalities are easily combined into one: $0 ≤ y ≤ x+1$.

$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$

These inequalities define the domain $D$, and define it uniquely, without any ambiguities. But how does this help us in the question at the beginning of the footnote? It will also help :) We need to check if the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:

$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right.$$

Both inequalities are true. The point $M_1(1;1)$ belongs to the region $D$.

Now it is the turn to investigate the behavior of the function on the boundary of the domain, i.e. go to. Let's start with the straight line $y=0$.

The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. The resulting substitution function of one variable $x$ will be denoted as $f_1(x)$:

$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$

Now for the function $f_1(x)$ we need to find the largest and smallest values ​​on the interval $-1 ≤ x ≤ 3$. Find the derivative of this function and equate it to zero:

$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$

The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we also add $M_2(2;0)$ to the list of points. In addition, we calculate the values ​​of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at the points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.

So, let's calculate the values ​​of the function $z$ at the points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points in the original expression $z=x^2+2xy-y^2-4x$. For example, for the point $M_2$ we get:

$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$

However, the calculations can be simplified a bit. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll spell it out in detail:

\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)

Of course, there is usually no need for such detailed records, and in the future we will begin to write all calculations in a shorter way:

$$z_2=f_1(2)=2^2-4\cdot 2=-4;\; z_3=f_1(-1)=(-1)^2-4\cdot (-1)=5;\; z_4=f_1(3)=3^2-4\cdot 3=-3.$$

Now let's turn to the straight line $x=3$. This line bounds the domain $D$ under the condition $0 ≤ y ≤ 4$. Substitute $x=3$ into the given function $z$. As a result of such a substitution, we get the function $f_2(y)$:

$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$

For the function $f_2(y)$, you need to find the largest and smallest values ​​on the interval $0 ≤ y ≤ 4$. Find the derivative of this function and equate it to zero:

$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$

The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we add $M_5(3;3)$ to the points found earlier. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at the points $M_4(3;0)$ and $M_6(3;4)$. At the point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at the points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:

\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; &z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)

And, finally, consider the last boundary of $D$, i.e. line $y=x+1$. This line bounds the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:

$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$

Once again we have a function of one variable $x$. And again, you need to find the largest and smallest values ​​of this function on the segment $-1 ≤ x ≤ 3$. Find the derivative of the function $f_(3)(x)$ and equate it to zero:

$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$

The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. The points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we have already found the value of the function in them.

$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$

The second step of the solution is completed. We got seven values:

$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$

Let's turn to. Choosing the largest and smallest values ​​from those numbers that were obtained in the third paragraph, we will have:

$$z_(min)=-4; \; z_(max)=6.$$

The problem is solved, it remains only to write down the answer.

Answer: $z_(min)=-4; \; z_(max)=6$.

Example #2

Find the largest and smallest values ​​of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.

Let's build a drawing first. The equation $x^2+y^2=25$ (this is the boundary line of the given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ 25$ satisfy all points inside and on the mentioned circle.

We will act on. Let's find partial derivatives and find out the critical points.

$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$

There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. find stationary points.

$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8.\end(aligned) \right.$$

We got a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check if the inequality $x^2+y^2 ≤ 25$, which defines our domain $D$, holds. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ is not satisfied. Conclusion: the point $(6;-8)$ does not belong to the region $D$.

Thus, there are no critical points inside $D$. Let's move on, to. We need to investigate the behavior of the function on the boundary of the given area, i.e. on the circle $x^2+y^2=25$. You can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the circle equation we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:

$$ z=x^2+y^2-12x+16y=x^2+25-x^2-12x+16\sqrt(25-x^2)=25-12x+16\sqrt(25-x ^2); \;\; -5≤ x ≤ 5. $$

The further solution will be completely identical to the study of the behavior of the function on the boundary of the region in the previous example No. 1. However, it seems to me more reasonable in this situation to apply the Lagrange method. We are only interested in the first part of this method. After applying the first part of the Lagrange method, we will get points at which and examine the function $z$ for the minimum and maximum values.

We compose the Lagrange function:

$$ F=z(x,y)+\lambda\cdot(x^2+y^2-25)=x^2+y^2-12x+16y+\lambda\cdot (x^2+y^2 -25). $$

We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:

$$ F_(x)^(")=2x-12+2\lambda x; \;\; F_(y)^(")=2y+16+2\lambda y.\\ \left \( \begin (aligned) & 2x-12+2\lambda x=0;\\ & 2y+16+2\lambda y=0;\\ & x^2+y^2-25=0.\end(aligned) \ right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$

To solve this system, let's immediately indicate that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:

$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$

The resulting contradiction $0=6$ says that the value $\lambda=-1$ is invalid. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:

\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)

I believe that it becomes obvious here why we specifically stipulated the $\lambda\neq -1$ condition. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator is $1+\lambda\neq 0$.

Let us substitute the obtained expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:

$$ \left(\frac(6)(1+\lambda) \right)^2+\left(\frac(-8)(1+\lambda) \right)^2=25;\\ \frac( 36)((1+\lambda)^2)+\frac(64)((1+\lambda)^2)=25;\\ \frac(100)((1+\lambda)^2)=25 ; \; (1+\lambda)^2=4. $$

It follows from the resulting equality that $1+\lambda=2$ or $1+\lambda=-2$. Hence, we have two values ​​of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values ​​$x$ and $y$:

\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)

So, we got two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Find the values ​​of the function $z$ at the points $M_1$ and $M_2$:

\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)

We should choose the largest and smallest values ​​from those that we obtained in the first and second steps. But in this case, the choice is small :) We have:

$$z_(min)=-75; \; z_(max)=125. $$

Answer: $z_(min)=-75; \; z_(max)=125$.

Lesson on the topic: "Finding the largest and smallest values ​​of a continuous function on a segment"

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What will we study:

1. Finding the largest and smallest values ​​according to the graph of the function.
2. Finding the largest and smallest value using the derivative.
3. Algorithm for finding the largest and smallest values ​​of a continuous function y=f(x) on the segment .
4. The largest and smallest value of a function on an open interval.
5. Examples.

Finding the largest and smallest value from the graph of a function

Guys, we have found the largest and smallest values ​​of the function before. We looked at the graph of a function and concluded where the function reaches its maximum value, and where it reaches its minimum.
Let's repeat:

The graph of our function shows that the largest value is reached at the point x= 1, it is equal to 2. The smallest value is reached at the point x= -1, and it is equal to -2. In this way, it is quite easy to find the largest and smallest values, but it is not always possible to plot a function graph.

Finding the largest and smallest value using the derivative

Guys, what do you think, how can you find the largest and smallest value using the derivative?

The answer can be found in the topic extrema of the function. There you and I found the maximum and minimum points, aren't the terms similar. However, one should not confuse the maximum and minimum values ​​with the maximum and minimum of a function, these are different concepts.

So let's introduce the rules:
a) If a function is continuous on an interval, then it reaches its maximum and minimum values ​​on this interval.
b) The function can reach the maximum and minimum values ​​both at the ends of the segments and inside it. Let's look at this point in more detail.

In figure a, the function reaches its maximum and minimum values ​​at the ends of the segments.
In figure b, the function reaches its maximum and minimum values ​​inside the interval . In figure c, the minimum point is inside the segment, and the maximum point is at the end of the segment, at point b.
c) If the largest and smallest values ​​are reached inside the segment, then only at stationary or critical points.

Algorithm for finding the largest and smallest values ​​of a continuous function y= f(x) on a segment

• Find the derivative f "(x).
• Find stationary and critical points inside the segment .
• Calculate the value of the function at stationary and critical points, as well as at f(a) and f(b). Choose the smallest and largest values, these will be the points of the smallest and largest values ​​of the function.

The largest and smallest value of a function on an open interval

Guys, how to find the largest and smallest value of a function on an open interval? To do this, we use an important theorem, which is proved in the course of higher mathematics.

Theorem. Let the function y= f(x) be continuous on the interval x, and have inside this interval the only stationary or critical point x= x0, then:
a) if x= x0 is the maximum point, then y is max. = f(x0).
b) if x= x0 is the minimum point, then y min. = f(x0).

Example

Find the largest and smallest value of the function y= $\frac(x^3)(3)$ + 2x 2 + 4x - 5 on the segment
a) [-9;-1], b) [-3; 3], c) .
Solution: Find the derivative: y "= x 2 + 4x + 4.
The derivative exists on the entire domain of definition, then we need to find stationary points.
y"= 0, with x= -2.
Further calculations will be carried out for the required segments.
a) Find the values ​​of the function at the ends of the segment and at the stationary point.
Then y nam. = -122, at x= -9; y max. = y = -7$\frac(1)(3)$, for x= -1.
b) Find the values ​​of the function at the ends of the segment and at the stationary point. The largest and smallest value is reached at the ends of the segment.
Then y nam. = -8, at x= -3, y max. = 34, at x= 3.
c) The stationary point does not fall on our segment, we find the values ​​at the ends of the segment.
Then y nam. = 34, at x= 3, y max. = 436, at x= 9.

Example

Find the largest and smallest value of the function y= x 2 - 3x + 5 + |1-x| on the segment.
Solution: Expand the module and transform our function:
y= x 2 - 3x + 5 + 1 - x, for x ≤ 1.
y= x 2 - 3x + 5 - 1 + x, for x ≥ 1.

Then our function will take the form:
\begin(equation*)f(x)= \begin(cases) x^2 - 4x + 6,\quad if\quad x ≤ 1 \\ x^2 - 2x + 4,\quad if\quad x ≥ 1 \end(cases) \end(equation*) Find critical points: \begin(equation*)f"(x)= \begin(cases) 2x - 4,\quad for \quad x ≤ 1 \\ 2x - 2, \quad when\quad x ≥ 1 \end(cases) \end(equation*) \begin(equation*)f"(x)=0,\quad when\quad x= \begin(cases) 2,\quad when \quad x ≤ 1 \\ 1,\quad for \quad x ≥ 1 \end(cases) \end(equation*) So, we have two stationary points and let's not forget that our function consists of two functions for different x.
Let's find the largest and smallest values ​​of the function, for this we calculate the values ​​of the function at stationary points and at the ends of the segment:
Answer: The function reaches its minimum value at the stationary point x= 1, y at least. = 3. The function reaches its maximum value at the end of the segment at the point x= 4, y max. = 12.

Example

Find the maximum value of the function y= $\frac(3x)(x^2 + 3)$ on the ray: , b) , c) [-4;7].
b) Find the largest and smallest value of the function y= x 2 - 6x + 8 + |x - 2| on the interval [-1;5].
c) Find the largest and smallest value of the function y= $-2x-\frac(1)(2x)$ on the ray (0;+∞).

From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest value of a function. What is it connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life, one has to solve the problem of optimizing some parameters. And this is the problem of finding the largest and smallest values ​​of the function.

It should be noted that the largest and smallest value of a function is usually sought on some interval X , which is either the entire domain of the function or part of the domain. The interval X itself can be a line segment, an open interval , an infinite interval .

In this article, we will talk about finding the largest and smallest values ​​of an explicitly given function of one variable y=f(x) .

Page navigation.

The largest and smallest value of a function - definitions, illustrations.

Let us briefly dwell on the main definitions.

The largest value of the function , which for any the inequality is true.

The smallest value of the function y=f(x) on the interval X is called such a value , which for any the inequality is true.

These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) value accepted in the interval under consideration with the abscissa.

Stationary points are the values ​​of the argument at which the derivative of the function vanishes.

Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its maximum (smallest) value on the interval X at one of the stationary points from this interval.

Also, a function can often take on the largest and smallest values ​​at points where the first derivative of this function does not exist, and the function itself is defined.

Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.

For clarity, we give a graphic illustration. Look at the pictures - and much will become clear.

On the segment

In the first figure, the function takes the largest (max y ) and smallest (min y ) values ​​at stationary points inside the segment [-6;6] .

Consider the case shown in the second figure. Change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.

In figure No. 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.

In the open range

In the fourth figure, the function takes the largest (max y ) and smallest (min y ) values ​​at stationary points within the open interval (-6;6) .

On the interval , no conclusions can be drawn about the largest value.

At infinity

In the example shown in the seventh figure, the function takes the largest value (max y ) at a stationary point with the abscissa x=1 , and the smallest value (min y ) is reached at the right boundary of the interval. At minus infinity, the values ​​of the function asymptotically approach y=3 .

On the interval, the function does not reach either the smallest or the largest value. As x=2 tends to the right, the function values ​​tend to minus infinity (the straight line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values ​​asymptotically approach y=3 . A graphic illustration of this example is shown in Figure 8.

Algorithm for finding the largest and smallest values ​​of a continuous function on the segment .

We write an algorithm that allows us to find the largest and smallest value of a function on a segment.

1. We find the domain of the function and check if it contains the entire segment .
2. We find all points at which the first derivative does not exist and which are contained in the segment (usually such points occur in functions with an argument under the module sign and in power functions with a fractional-rational exponent). If there are no such points, then go to the next point.
3. We determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them fall into the segment, then go to the next step.
4. We calculate the values ​​of the function at the selected stationary points (if any), at points where the first derivative does not exist (if any), and also at x=a and x=b .
5. From the obtained values ​​of the function, we select the largest and smallest - they will be the desired maximum and smallest values ​​of the function, respectively.

Let's analyze the algorithm when solving an example for finding the largest and smallest values ​​of a function on a segment.

Example.

Find the largest and smallest value of a function

• on the segment;
• on the interval [-4;-1] .

Decision.

The domain of the function is the entire set of real numbers, except for zero, that is, . Both segments fall within the domain of definition.

We find the derivative of the function with respect to:

Obviously, the derivative of the function exists at all points of the segments and [-4;-1] .

Stationary points are determined from the equation . The only real root is x=2 . This stationary point falls into the first segment.

For the first case, we calculate the values ​​of the function at the ends of the segment and at a stationary point, that is, for x=1 , x=2 and x=4 :

Therefore, the largest value of the function is reached at x=1 , and the smallest value – at x=2 .

For the second case, we calculate the values ​​of the function only at the ends of the segment [-4;-1] (since it does not contain any stationary points):

Let the function y=f(X) continuous on the interval [ a, b]. As is known, such a function reaches its maximum and minimum values ​​on this interval. The function can take these values ​​either at an interior point of the segment [ a, b], or on the boundary of the segment.

To find the largest and smallest values ​​of a function on the segment [ a, b] necessary:

1) find the critical points of the function in the interval ( a, b);

2) calculate the values ​​of the function at the found critical points;

3) calculate the values ​​of the function at the ends of the segment, that is, for x=a and x = b;

4) from all the calculated values ​​of the function, choose the largest and smallest.

Example. Find the largest and smallest values ​​of a function

on the segment.

Finding critical points:

These points lie inside the segment ; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;

at the point x= 3 and at the point x= 0.

Investigation of a function for convexity and an inflection point.

Function y = f (x) called convexup in between (a, b) , if its graph lies under a tangent drawn at any point of this interval, and is called convex down (concave) if its graph lies above the tangent.

The point at the transition through which the convexity is replaced by concavity or vice versa is called inflection point.

Algorithm for studying for convexity and inflection point:

1. Find the critical points of the second kind, that is, the points at which the second derivative is equal to zero or does not exist.

2. Put critical points on the number line, breaking it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upwards, if, then the function is convex downwards.

3. If, when passing through a critical point of the second kind, it changes sign and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find its ordinate.

Asymptotes of the graph of a function. Investigation of a function into asymptotes.

Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point of the graph to this line tends to zero with an unlimited removal of the graph point from the origin.

There are three types of asymptotes: vertical, horizontal and inclined.

Definition. Direct called vertical asymptote function graph y = f(x), if at least one of the one-sided limits of the function at this point is equal to infinity, that is

where is the discontinuity point of the function, that is, it does not belong to the domain of definition.

Example.

D( y) = (‒ ∞; 2) (2; + ∞)

x= 2 - breaking point.

Definition. Straight y=A called horizontal asymptote function graph y = f(x) at , if

Example.

x
y

Definition. Straight y=kx +b (k≠ 0) is called oblique asymptote function graph y = f(x) at , where

General scheme for the study of functions and plotting.

Function research algorithmy = f(x) :

1. Find the domain of the function D (y).

2. Find (if possible) the points of intersection of the graph with the coordinate axes (with x= 0 and at y = 0).

3. Investigate for even and odd functions ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).

4. Find the asymptotes of the graph of the function.

5. Find intervals of monotonicity of the function.

6. Find the extrema of the function.

7. Find the intervals of convexity (concavity) and inflection points of the graph of the function.

8. On the basis of the conducted research, construct a graph of the function.

Example. Investigate the function and plot its graph.

1) D (y) =

x= 4 - breaking point.

2) When x = 0,

(0; – 5) – point of intersection with oy.

At y = 0,

3) y(‒ x)= general function (neither even nor odd).

4) We investigate for asymptotes.

a) vertical

b) horizontal

c) find oblique asymptotes where

‒oblique asymptote equation

5) In this equation, it is not required to find intervals of monotonicity of the function.

6)

These critical points partition the entire domain of the function on the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the obtained results in the form of the following table.