Find the largest or smallest value of a function. The largest and smallest value of the function

The largest and smallest values ​​of the function

concepts mathematical analysis. The value taken by a function at some point of the set on which this function is defined is called the largest (smallest) value on this set if the function does not have a larger (smaller) value at any other point in the set. N. and n. h. f. in comparison with its values ​​at all sufficiently close points are called extrema (respectively, maxima and minima) of the function. N. and n. h. f., given on a segment, can be achieved either at points where the derivative is equal to zero, or at points where it does not exist, or at the ends of the segment. A continuous function given on a segment necessarily reaches its maximum and minimum values; if a continuous function is considered on an interval (that is, a segment with excluded ends), then among its values ​​on this interval there may not be a maximum or a minimum. For example, the function at = x, given on the interval , reaches the largest and smallest values, respectively, at x= 1 and x= 0 (i.e., at the ends of the segment); if we consider this function on the interval (0; 1), then among its values ​​on this interval there is neither the largest nor the smallest, since for each x0 there is always a point of this interval lying to the right (to the left) x0, and such that the value of the function at this point will be greater (respectively, less) than at the point x0. Similar statements are valid for functions of several variables. See also Extreme.

Big soviet encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .

See what "The largest and smallest values ​​of a function" are in other dictionaries:

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Concepts of mathematical analysis. The value taken by the function at some point of the set on which this function is defined is called the largest (smallest) on this set, if at no other point the function has a greater (less) ... ... encyclopedic Dictionary

The concepts of mathematics. analysis. The value taken by the function at a particular point of the set, pa rum this function is given, called. largest (smallest) on this set, if at no other point does the function have a larger (smaller) value ... Natural science. encyclopedic Dictionary

MAXIMUM AND MINIMUM FUNCTION- respectively, the largest and smallest values ​​of the function in comparison with its values ​​at all sufficiently close points. High and low points are called extreme points... Great Polytechnic Encyclopedia

The largest and, accordingly, the smallest values ​​of a function that takes real values. The point of the domain of definition of the function in question, in which it takes a maximum or minimum, is called. respectively the maximum point or the minimum point ... ... Mathematical Encyclopedia

A ternary function in the theory of functional systems and ternary logic is a function of type, where is a ternary set, and is a non-negative integer, which is called the arity or locality of the function. The elements of the set are digital ... ... Wikipedia

Representation of Boolean functions by normal forms (see Normal forms of Boolean functions). the simplest with respect to some measure of complexity. Usually, the complexity of a normal form is understood as the number of letters in it. In this case, the simplest form is called ... ... Mathematical Encyclopedia

A function that receives infinitesimal increments as the argument increments infinitesimally. A single-valued function f (x) is called continuous for the value of the argument x0, if for all values ​​of the argument x that differ sufficiently little from x0 ... Great Soviet Encyclopedia

- (Latin maximum and minimum, literally the largest and smallest) (Math.), the largest and smallest values ​​​​of a function compared to its values ​​\u200b\u200bin sufficiently close points. In the figure, the function y \u003d f (x) has a maximum at the points x1 and x3, and at the point x2 ... ... encyclopedic Dictionary

- (from the Latin maximum and minimum, the largest and smallest) (mathematical), the largest and smallest values ​​​​of a function compared to its values ​​\u200b\u200bin sufficiently close points. High and low points are called extreme points... Modern Encyclopedia

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 region $D$ 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 entries, and in the future we will begin to write down 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: $z_(min)=-75; \; z_(max)=125$.

In this article I will talk about algorithm for finding the largest and smallest value function, minimum and maximum points.

From theory, we will definitely need derivative table and differentiation rules. It's all in this board:

Algorithm for finding the largest and smallest values.

I find it easier to explain specific example. Consider:

Example: Find highest value functions y=x^5+20x^3–65x on the segment [–4;0].

Step 1. We take the derivative.

Y" = (x^5+20x^3–65x)" = 5x^4 + 20*3x^2 - 65 = 5x^4 + 60x^2 - 65

Step 2 Finding extremum points.

extremum point we name such points at which the function reaches its maximum or minimum value.

To find the extremum points, it is necessary to equate the derivative of the function to zero (y "= 0)

5x^4 + 60x^2 - 65 = 0

Now we solve this biquadratic equation and the found roots are our extremum points.

I solve such equations by replacing t = x^2, then 5t^2 + 60t - 65 = 0.

Reduce the equation by 5, we get: t^2 + 12t - 13 = 0

D = 12^2 - 4*1*(-13) = 196

T_(1) = (-12 + sqrt(196))/2 = (-12 + 14)/2 = 1

T_(2) = (-12 - sqrt(196))/2 = (-12 - 14)/2 = -13

We make the reverse substitution x^2 = t:

X_(1 and 2) = ±sqrt(1) = ±1
x_(3 and 4) = ±sqrt(-13) (we exclude, under the root cannot be negative numbers(unless, of course, we are talking about complex numbers)

Total: x_(1) = 1 and x_(2) = -1 - these are our extremum points.

Step 3 We determine the largest and smallest value.

Substitution method.

In the condition, we were given the segment [b][–4;0]. The point x=1 is not included in this segment. So we don't consider it. But in addition to the point x=-1, we also need to consider the left and right borders of our segment, that is, the points -4 and 0. To do this, we substitute all these three points into the original function. Notice the original one is the one given in the condition (y=x^5+20x^3–65x), some start substituting into the derivative...

Y(-1) = (-1)^5 + 20*(-1)^3 - 65*(-1) = -1 - 20 + 65 = [b]44
y(0) = (0)^5 + 20*(0)^3 - 65*(0) = 0
y(-4) = (-4)^5 + 20*(-4)^3 - 65*(-4) = -1024 - 1280 + 260 = -2044

This means that the maximum value of the function is [b]44 and it is reached at the points [b]-1, which is called the maximum point of the function on the segment [-4; 0].

We decided and got an answer, we are great, you can relax. But stop! Don't you think that counting y(-4) is somehow too complicated? In conditions of limited time, it is better to use another method, I call it like this:

Through intervals of constancy.

These gaps are found for the derivative of the function, that is, for our biquadratic equation.

I do it in the following way. I draw a directional line. I set the points: -4, -1, 0, 1. Despite the fact that 1 is not included in the given segment, it should still be noted in order to correctly determine the intervals of constancy. Let's take some number many times greater than 1, let's say 100, mentally substitute it into our biquadratic equation 5(100)^4 + 60(100)^2 - 65. Even without counting anything, it becomes obvious that at the point 100 the function has plus sign. This means that for intervals from 1 to 100 it has a plus sign. When passing through 1 (we go from right to left), the function will change sign to minus. When passing through the point 0, the function will retain its sign, since this is only the boundary of the segment, and not the root of the equation. When passing through -1, the function will again change sign to plus.

From theory, we know that where the derivative of the function is (and we drew this for it) changes sign from plus to minus (point -1 in our case) function reaches its local maximum (y(-1)=44 as calculated earlier) on this segment (this is logically very clear, the function has ceased to increase, since it reached its maximum and began to decrease).

Accordingly, where the derivative of the function changes sign from minus to plus, achieved local minimum of a function. Yes, yes, we also found the local minimum point, which is 1, and y(1) is the minimum value of the function on the interval, let's say from -1 to +∞. Please note that this is only a LOCAL MINIMUM, that is, a minimum on a certain segment. Since the actual (global) minimum function will reach somewhere there, in -∞.

In my opinion, the first method is easier theoretically, and the second one is easier in terms of arithmetic operations, but much more difficult in terms of theory. Indeed, sometimes there are cases when the function does not change sign when passing through the root of the equation, and indeed you can get confused with these local, global maxima and minima, although you will have to master this well anyway if you plan to enter technical university(and why otherwise hand over profile exam and solve this problem). But practice and only practice will teach you how to solve such problems once and for all. And you can train on our website. Here .

If you have any questions, or something is not clear, be sure to ask. I will be happy to answer you, and make changes, additions to the article. Remember we are making this site together!

The figures below show where the function can reach its smallest and largest value. In the left figure, the smallest and largest values ​​are fixed at the points of the local minimum and maximum of the function. In the right figure - at the ends of the segment.

If the function y = f(x) continuous on the interval [ a, b] , then it reaches on this segment least and highest values . This, as already mentioned, can happen either in extremum points or at the ends of the segment. Therefore, to find least and the largest values ​​of the function , continuous on the segment [ a, b] , you need to calculate its values ​​in all critical points and at the ends of the segment, and then choose the smallest and largest of them.

Let, for example, it is required to determine the maximum value of the function f(x) on the segment [ a, b] . To do this, find all its critical points lying on [ a, b] .

critical point is called the point at which function defined, and her derivative is either zero or does not exist. Then you should calculate the values ​​of the function at critical points. And, finally, one should compare the values ​​of the function at critical points and at the ends of the segment ( f(a) and f(b) ). The largest of these numbers will be the largest value of the function on the segment [a, b] .

The problem of finding the smallest values ​​of the function .

We are looking for the smallest and largest values ​​​​of the function together

Example 1. Find the smallest and largest values ​​of a function on the segment [-1, 2] .

Solution. We find the derivative of this function. Equate the derivative to zero () and get two critical points: and . To find the smallest and largest values ​​of a function on a given segment, it is enough to calculate its values ​​at the ends of the segment and at the point , since the point does not belong to the segment [-1, 2] . These function values ​​are the following: , , . It follows that smallest function value(marked in red on the graph below), equal to -7, is reached at the right end of the segment - at the point , and greatest(also red on the graph), is equal to 9, - at the critical point .

If the function is continuous in a certain interval and this interval is not a segment (but is, for example, an interval; the difference between an interval and a segment: the boundary points of the interval are not included in the interval, but the boundary points of the segment are included in the segment), then among the values ​​of the function there may not be be the smallest and largest. So, for example, the function depicted in the figure below is continuous on ]-∞, +∞[ and does not have the largest value.

However, for any interval (closed, open, or infinite), the following property of continuous functions holds.

For self-checking during calculations, you can use online derivatives calculator .

Example 4. Find the smallest and largest values ​​of a function on the segment [-1, 3] .

Solution. We find the derivative of this function as the derivative of the quotient:

.

We equate the derivative to zero, which gives us one critical point: . It belongs to the interval [-1, 3] . To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Let's compare these values. Conclusion: equal to -5/13, at the point and the greatest value equal to 1 at the point .

We continue to search for the smallest and largest values ​​​​of the function together

There are teachers who, on the topic of finding the smallest and largest values ​​of a function, do not give students examples more complicated than those just considered, that is, those in which the function is a polynomial or a fraction, the numerator and denominator of which are polynomials. But we will not limit ourselves to such examples, since among teachers there are lovers of making students think in full (table of derivatives). Therefore, the logarithm and the trigonometric function will be used.

Example 8. Find the smallest and largest values ​​of a function on the segment .

Solution. We find the derivative of this function as derivative of the product :

We equate the derivative to zero, which gives one critical point: . It belongs to the segment. To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

The result of all actions: the function reaches its minimum value, equal to 0, at a point and at a point and the greatest value equal to e² , at the point .

For self-checking during calculations, you can use online derivatives calculator .

Example 9. Find the smallest and largest values ​​of a function on the segment .

Solution. We find the derivative of this function:

Equate the derivative to zero:

The only critical point belongs to the segment . To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Conclusion: the function reaches its minimum value, equal to , at the point and the greatest value, equal to , at the point .

In applied extremal problems, finding the smallest (largest) function values, as a rule, is reduced to finding the minimum (maximum). But it is not the minima or maxima themselves that are of greater practical interest, but the values ​​of the argument at which they are achieved. When solving applied problems, an additional difficulty arises - the compilation of functions that describe the phenomenon or process under consideration.

Example 10 A tank with a capacity of 4, having the shape of a parallelepiped with a square base and open at the top, must be tinned. What should be the dimensions of the tank in order to cover it with the least amount of material?

Solution. Let x- base side h- tank height, S- its surface area without cover, V- its volume. The surface area of ​​the tank is expressed by the formula , i.e. is a function of two variables. To express S as a function of one variable, we use the fact that , whence . Substituting the found expression h into the formula for S:

Let us examine this function for an extremum. It is defined and differentiable everywhere in ]0, +∞[ , and

.

We equate the derivative to zero () and find the critical point. In addition, at , the derivative does not exist, but this value is not included in the domain of definition and therefore cannot be an extremum point. So, - the only critical point. Let's check it for the presence of an extremum using the second sufficient sign. Let's find the second derivative. When the second derivative is greater than zero (). This means that when the function reaches a minimum . Because this minimum - the only extremum of this function, it is its smallest value. So, the side of the base of the tank should be equal to 2 m, and its height.

For self-checking during calculations, you can use

Given a function that is defined and continuous on some interval . It is required to find the largest (smallest) value of the function on this interval.

Theoretical basis.
Theorem (Second Weierstrass Theorem):

If a function is defined and continuous in a closed interval , then it reaches its maximum and minimum values ​​in this interval.

The function can reach its maximum and minimum values ​​either at the internal points of the interval or at its boundaries. Let's illustrate all possible options.

Explanation:
1) The function reaches its maximum value on the left border of the interval at the point , and its minimum value on the right border of the interval at the point .
2) The function reaches its maximum value at the point (this is the maximum point), and its minimum value at the right boundary of the interval at the point.
3) The function reaches its maximum value on the left border of the interval at the point , and its minimum value at the point (this is the minimum point).
4) The function is constant on the interval, i.e. it reaches its minimum and maximum values ​​at any point in the interval, and the minimum and maximum values ​​are equal to each other.
5) The function reaches its maximum value at the point , and its minimum value at the point (despite the fact that the function has both a maximum and a minimum on this interval).
6) The function reaches its maximum value at a point (this is the maximum point), and its minimum value at a point (this is the minimum point).
Comment:

"Maximum" and "maximum value" are different things. This follows from the definition of the maximum and the intuitive understanding of the phrase "maximum value".

Algorithm for solving problem 2.



4) Choose from the obtained values ​​the largest (smallest) and write down the answer.

Example 4:

Determine the largest and smallest value of a function on the segment.
Solution:
1) Find the derivative of the function.

2) Find stationary points (and points that are suspicious of an extremum) by solving the equation . Pay attention to the points where there is no two-sided finite derivative.

3) Calculate the values ​​of the function at stationary points and at the boundaries of the interval.

4) Choose from the obtained values ​​the largest (smallest) and write down the answer.

The function on this segment reaches its maximum value at the point with coordinates .

The function on this segment reaches its minimum value at the point with coordinates .

You can verify the correctness of the calculations by looking at the graph of the function under study.


Comment: The function reaches its maximum value at the maximum point, and the minimum value at the boundary of the segment.

Special case.

Suppose you want to find the maximum and minimum value of some function on a segment. After the execution of the first paragraph of the algorithm, i.e. calculation of the derivative, it becomes clear that, for example, it takes only negative values ​​on the entire segment under consideration. Remember that if the derivative is negative, then the function is decreasing. We found that the function is decreasing on the entire interval. This situation is shown in the chart No. 1 at the beginning of the article.

The function decreases on the interval, i.e. it has no extremum points. It can be seen from the picture that the function will take the smallest value on the right border of the segment, and the largest value on the left. if the derivative on the interval is everywhere positive, then the function is increasing. The smallest value is on the left border of the segment, the largest is on the right.