What does it mean to find the largest value of a function. The smallest and largest values ​​of a function on a segment

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 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 get the points at which we 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've got two points of possibility 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: $z_(min)=-75; \; z_(max)=125$.

The standard algorithm for solving such tasks involves, after finding the zeros of the function, the determination of the signs of the derivative on the intervals. Then the calculation of the values ​​at the found maximum (or minimum) points and on the border of the interval, depending on what question is in the condition.

I advise you to do things a little differently. Why? Wrote about it.

I propose to solve such tasks as follows:

1. Find the derivative.
2. Find the zeros of the derivative.
3. Determine which of them belong to the given interval.
4. We calculate the values ​​of the function on the boundaries of the interval and points of item 3.
5. We draw a conclusion (we answer the question posed).

In the course of solving the presented examples, the solution was not considered in detail. quadratic equations, you should be able to do this. They should also know.

Consider examples:

77422. Find the largest value of the function y=x 3 –3x+4 on the segment [–2;0].

Let's find the zeros of the derivative:

The point x = –1 belongs to the interval specified in the condition.

We calculate the function values ​​at points –2, –1 and 0:

The largest value of the function is 6.

Answer: 6

77425. Find the smallest value of the function y \u003d x 3 - 3x 2 + 2 on the segment.

Find the derivative of the given function:

Let's find the zeros of the derivative:

The point x = 2 belongs to the interval specified in the condition.

We calculate the function values ​​at points 1, 2 and 4:

The smallest value of the function is -2.

Answer: -2

77426. Find the largest value of the function y \u003d x 3 - 6x 2 on the segment [-3; 3].

Find the derivative of the given function:

Let's find the zeros of the derivative:

The point x = 0 belongs to the interval specified in the condition.

We calculate the function values ​​at points –3, 0 and 3:

The smallest value of the function is 0.

Answer: 0

77429. Find the smallest value of the function y \u003d x 3 - 2x 2 + x + 3 on the segment.

Find the derivative of the given function:

3x 2 - 4x + 1 = 0

We get the roots: x 1 \u003d 1 x 1 \u003d 1/3.

Only x = 1 belongs to the interval specified in the condition.

Find the function values ​​at points 1 and 4:

We found that the smallest value of the function is 3.

Answer: 3

77430. Find the largest value of the function y \u003d x 3 + 2x 2 + x + 3 on the segment [- 4; -one].

Find the derivative of the given function:

Find the zeros of the derivative, solve the quadratic equation:

3x 2 + 4x + 1 = 0

Let's get the roots:

The root х = –1 belongs to the interval specified in the condition.

Find the function values ​​at points –4, –1, –1/3 and 1:

We found that the largest value of the function is 3.

Answer: 3

77433. Find the smallest value of the function y \u003d x 3 - x 2 - 40x +3 on the segment.

Find the derivative of the given function:

Find the zeros of the derivative, solve the quadratic equation:

3x 2 - 2x - 40 = 0

Let's get the roots:

The root x = 4 belongs to the interval specified in the condition.

We find the values ​​of the function at points 0 and 4:

We found that the smallest value of the function is -109.

Answer: -109

Consider a method for determining the largest and smallest values ​​of functions without a derivative. This approach can be used if you have big problems with the definition of the derivative. The principle is simple - we substitute all integer values ​​from the interval into the function (the fact is that in all such prototypes the answer is an integer).

77437. Find the smallest value of the function y \u003d 7 + 12x - x 3 on the segment [-2; 2].

We substitute points from -2 to 2: View Solution

77434. Find the largest value of the function y \u003d x 3 + 2x 2 - 4x + 4 on the segment [-2; 0].

That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell about the site in social networks.

The largest (smallest) value of the function is the largest (smallest) accepted value of the ordinate in the considered interval.

To find the largest or smallest value of a function, you need to:

1. Check which stationary points are included in the given segment.
2. Calculate the value of the function at the ends of the segment and at stationary points from step 3
3. Choose from the results obtained the largest or smallest value.

To find the maximum or minimum points, you need to:

1. Find the derivative of the function $f"(x)$
2. Find stationary points by solving the equation $f"(x)=0$
3. Factorize the derivative of a function.
4. Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the obtained intervals, using the notation of clause 3.
5. Find the maximum or minimum points according to the rule: if at a point the derivative changes sign from plus to minus, then this will be the maximum point (if from minus to plus, then this will be the minimum point). In practice, it is convenient to use the image of arrows on the intervals: on the interval where the derivative is positive, the arrow is drawn upwards and vice versa.

Table of derivatives of some elementary functions:

Basic rules of differentiation

1. The derivative of the sum and difference is equal to the derivative of each term

$(f(x) ± g(x))′= f′(x)± g′(x)$

Find the derivative of the function $f(x) = 3x^5 – cosx + (1)/(x)$

The derivative of the sum and difference is equal to the derivative of each term

$f′(x)=(3x^5)′–(cosx)′+((1)/(x))"=15x^4+sinx-(1)/(x^2)$

2. Derivative of a product.

$(f(x)∙g(x))′=f′(x)∙g(x)+f(x)∙g(x)′$

Find the derivative $f(x)=4x∙cosx$

$f′(x)=(4x)′∙cosx+4x∙(cosx)′=4∙cosx-4x∙sinx$

3. Derivative of the quotient

$((f(x))/(g(x)))"=(f^"(x)∙g(x)-f(x)∙g(x)")/(g^2(x) )$

Find the derivative $f(x)=(5x^5)/(e^x)$

$f"(x)=((5x^5)"∙e^x-5x^5∙(e^x)")/((e^x)^2)=(25x^4∙e^x- 5x^5∙e^x)/((e^x)^2)$

4. The derivative of a complex function is equal to the product of the derivative of the external function and the derivative of the internal function

$f(g(x))′=f′(g(x))∙g′(x)$

$f′(x)=cos′(5x)∙(5x)′= - sin(5x)∙5= -5sin(5x)$

Find the minimum point of the function $y=2x-ln⁡(x+11)+4$

1. Find the ODZ of the function: $x+11>0; x>-11$

2. Find the derivative of the function $y"=2-(1)/(x+11)=(2x+22-1)/(x+11)=(2x+21)/(x+11)$

3. Find stationary points by equating the derivative to zero

$(2x+21)/(x+11)=0$

A fraction is zero if the numerator is zero and the denominator is not zero

$2x+21=0; x≠-11$

4. Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the obtained intervals. To do this, we substitute into the derivative any number from the extreme right region, for example, zero.

$y"(0)=(2∙0+21)/(0+11)=(21)/(11)>0$

5. At the minimum point, the derivative changes sign from minus to plus, therefore, the $-10.5$ point is the minimum point.

Answer: $-10.5$

Find the maximum value of the function $y=6x^5-90x^3-5$ on the segment $[-5;1]$

1. Find the derivative of the function $y′=30x^4-270x^2$

2. Equate the derivative to zero and find stationary points

$30x^4-270x^2=0$

Let's take the common factor $30x^2$ out of brackets

$30x^2(x^2-9)=0$

$30x^2(x-3)(x+3)=0$

Set each factor equal to zero

$x^2=0 ; x-3=0; x+3=0$

$x=0;x=3;x=-3$

3. Choose stationary points that belong to the given segment $[-5;1]$

Stationary points $x=0$ and $x=-3$ are suitable for us

4. Calculate the value of the function at the ends of the segment and at stationary points from item 3

In task B14 from the USE in mathematics, you need to find the smallest or largest value of a function of one variable. This is a rather trivial task from mathematical analysis, and it is for this reason that every graduate can and should learn how to solve it normally. high school. Let's analyze a few examples that schoolchildren solved at the diagnostic work in mathematics, which took place in Moscow on December 7, 2011.

Depending on the interval on which you want to find the maximum or minimum value of the function, one of the following standard algorithms is used to solve this problem.

I. Algorithm for finding the largest or smallest value of a function on a segment:

• Find the derivative of a function.
• Select from the points suspected of an extremum those that belong to a given segment and the domain of the function.
• Calculate values functions(not a derivative!) at these points.
• Among the obtained values, choose the largest or smallest, it will be the desired one.

Example 1 Find the smallest value of a function
y = x 3 – 18x 2 + 81x+ 23 on the segment .

Decision: we act according to the algorithm for finding the smallest value of a function on a segment:

• The scope of the function is not limited: D(y) = R.
• The derivative of the function is: y' = 3x 2 – 36x+ 81. The scope of the derivative of a function is also not limited: D(y') = R.
• Zeros of the derivative: y' = 3x 2 – 36x+ 81 = 0, so x 2 – 12x+ 27 = 0, whence x= 3 and x= 9, our interval includes only x= 9 (one point suspicious for an extremum).
• We find the value of the function at a point suspicious of an extremum and at the edges of the interval. For the convenience of calculations, we represent the function in the form: y = x 3 – 18x 2 + 81x + 23 = x(x-9) 2 +23:
  • y(8) \u003d 8 (8-9) 2 +23 \u003d 31;
  • y(9) = 9 (9-9) 2 +23 = 23;
  • y(13) = 13 (13-9) 2 +23 = 231.

So, from the obtained values, the smallest is 23. Answer: 23.

II. The algorithm for finding the largest or smallest value of a function:

• Find the scope of the function.
• Find the derivative of a function.
• Determine the points that are suspicious of an extremum (those points at which the derivative of the function vanishes, and the points at which there is no two-sided finite derivative).
• Mark these points and the domain of the function on the number line and determine the signs derivative(not functions!) on the resulting intervals.
• Define values functions(not a derivative!) at the minimum points (those points at which the sign of the derivative changes from minus to plus), the smallest of these values ​​will be the smallest value of the function. If there are no minimum points, then the function does not have a minimum value.
• Define values functions(not a derivative!) at the maximum points (those points at which the sign of the derivative changes from plus to minus), the largest of these values ​​will be the largest value of the function. If there are no maximum points, then the function does not have a maximum value.

Example 2 Find the largest value of the function.

The algorithm for finding the largest and smallest values ​​of a continuous function on a segment:

1) Find all critical points of the function that belong to the interval ;

2) Calculate the values ​​of the function at these points and at the ends of the segment;

3) From the obtained values, choose the largest and smallest.

Example 8.1. Find the largest and smallest values ​​of a function
on the segment
.

Decision. 1) Find the critical points of the function.

,




.

On the segment
the denominator does not vanish. Therefore, a fraction is zero if and only if the numerator is zero:




.

Means,
is the critical point of the function. It belongs to this segment.

Find the value of the function at the critical point:

2) Find the values ​​of the function at the ends of the segment:

, .

3) From the obtained values, we select the largest and smallest:

,
.

9. Problems of finding the largest and smallest values ​​of quantities

When solving problems for calculating the smallest and largest values ​​of quantities, it is necessary first of all to determine for which quantity in the problem it is required to find the smallest or largest value. This value will be the function under study. Then one of the quantities, on the change of which the application of the function depends, should be taken as an independent variable and the function should be expressed through it. In this case, it is necessary to choose as an independent variable the value through which the function under study is expressed most simply. After that, the problem is solved to find the smallest and largest values ​​of the obtained function in a certain interval of change of the independent variable, which is usually set from the very essence of the problem.

Example 9.1. Find the height of the largest cone that can be inscribed in a sphere of radius .

R solution. Denoting the radius of the base, height and volume of the cone, respectively ,and , write
. This equality expresses the dependence on two variables and ; Let us exclude one of these quantities, namely . To do this, from a right triangle
we derive (by the theorem on the square of the perpendicular dropped from the vertex of the right angle to the hypotenuse):

Figure 6 - Illustration for example 9.1.

or
.

Substituting the value into the formula for the volume of a cone, we get:

.

We see that the volume cone inscribed in a ball of radius , is a function of the height of this cone . To find the height at which the inscribed cone has a large volume, it means to find such , for which the function has a maximum. We are looking for the maximum function:

1)
,

2)
,
,
, where
or
,

3)
.

Substituting instead of at first
, and then
, we get:

In the first case, we have a minimum (
at
), in the second, the desired maximum (since
at
).

Therefore, at
cone inscribed in a sphere of radius , has the largest volume.

P Example 9.2. Required to be fenced with wire mesh length 60 m a rectangular area adjacent to the wall of the house (Fig. 7). What should be the length and width of the plot so that it has the largest area?

Decision. Let the plot width m, and the area m 2 , then:

Figure 7 - Illustration for ex. 9.2.

Values and cannot be negative, so the multiplier
, a
.

Square there is a function , we define the intervals of its increase and decrease:

.
, and the function is increasing when
;
, and the function is decreasing when
. Hence the point
is the maximum point. Since this is the only point that belongs to the interval
, then at the point
function has the highest value.

Therefore, the plot area is largest (maximum) if the width
m, and the length m.

Example 9.3. What should be the dimensions of a rectangular room, the area of ​​\u200b\u200bwhich 36 m 2 so that its perimeter is the smallest?

Decision. Let the length be m, then the width of the rectangle m, and the perimeter:

.

Perimeter is the length function , defined for all positive values :
.

We define the intervals of its increase and decrease:

The sign of the derivative is determined by the sign of the difference
. In the interim

, and in between

.

Hence the point
is the minimum point. Since this is the only point that belongs to the interval:
, then at the point
function has the smallest value.

Therefore, the perimeter of a rectangle has the smallest value (minimum) if its length 6 m and width m = 6 m, i.e. when it is a square.