From a practical point of view, the greatest interest is in using the derivative to find the largest and smallest values of a function. What is this connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life we have to solve problems of optimizing some parameters. And these are the tasks of finding the largest and smallest values of a function.
It should be noted that the largest and smallest value functions are usually searched for 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 segment, an open interval 
, an infinite interval.
In this article we will talk about finding the largest and smallest values of an explicitly specified function of one variable y=f(x) .
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The largest and smallest value of a function - definitions, illustrations.
Let's briefly look at the main definitions.
The largest value of the function 
that for anyone 
inequality is true.
The smallest value of the function y=f(x) on the interval X is called such a value 
that for anyone inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) accepted value on the interval under consideration at the abscissa.
Stationary points– these are the values of the argument at which the derivative of the function becomes zero.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. From this theorem it follows that if a differentiable function has an extremum ( local minimum or local maximum) at a certain point, then this point is stationary. Thus, the function often takes its largest (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take on its largest and smallest values at points at which 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 definition of the function, or the interval X is infinite. And some functions at infinity and at the boundaries of the domain of definition can take on 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 will give a graphic illustration. Look at the pictures and a lot will become clearer.
On the segment
In the first figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside the segment [-6;6].
Consider the case depicted in the second figure. Let's change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest at the point with the abscissa corresponding to the right boundary of the interval.
In Figure 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.
On an open interval
In the fourth figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside 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 highest value(max y) at a stationary point with abscissa x=1, and the smallest value (min y) is achieved on the right boundary of the interval. At minus infinity, the function values asymptotically approach y=3.
Over the interval, the function reaches neither the smallest nor the largest value. As x=2 approaches from the right, the function values tend to minus infinity (the straight line x=2 is 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 a segment.
Let us write an algorithm that allows us to find the largest and smallest values of a function on a segment.
- We find the domain of definition of the function and check whether it contains the entire segment.
- We find all the points at which the first derivative does not exist and which are contained in the segment (usually such points are found in functions with an argument under the modulus sign and in power functions with a fractional-rational exponent). If there are no such points, then move on to the next point.
- We determine all stationary points falling within the segment. To do this, we equate it to zero, solve the resulting equation and select suitable roots. If there are no stationary points or none of them fall into the segment, then move on to the next point.
- We calculate the values of the function at selected stationary points (if any), at points at which the first derivative does not exist (if any), as well as at x=a and x=b.
- From the obtained values of the function, we select the largest and smallest - they will be the required largest and smallest values of the function, respectively.
Let's analyze the algorithm for solving an example to find 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 segment [-4;-1] .
Solution.
The domain of a function is the entire set real numbers, except for zero, that is . Both segments fall within the definition domain.
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].
We determine stationary points 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 the stationary point, that is, for x=1, x=2 and x=4: 
Therefore, the greatest value of the function 
is achieved at x=1, and the smallest value 
– at x=2.
For the second case, we calculate the function values only at the ends of the segment [-4;-1] (since it does not contain a single stationary point): 
Let the function y =f(X) is continuous on the interval [ a, b]. As is known, such a function reaches its maximum and minimum values on this segment. The function can take these values either internal point 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, when x=A and x = b;
4) from all calculated values of the function, select 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.
Study of a function for convexity and inflection point.
Function y = f (x) called convexup in between (a, b) , if its graph lies under the tangent drawn at any point in this interval, and is called convex down (concave), if its graph lies above the tangent.
The point through which convexity is replaced by concavity or vice versa is called inflection point.
Algorithm for examining convexity and inflection point:
1. Find critical points of the second kind, that is, points at which the second derivative is equal to zero or does not exist.
2. Plot critical points on the number line, dividing it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upward, if, then the function is convex downward.
3. If, when passing through a critical point of the second kind, the sign changes 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. Study of a function for asymptotes.
Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point on the graph to this line tends to zero as the point on the graph moves indefinitely from the origin.
There are three types of asymptotes: vertical, horizontal and inclined.
Definition. The straight line is called vertical asymptote function graphics 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 – break point.
Definition. Straight y =A called horizontal asymptote function graphics y = f(x) at , if
Example.
|
x | |||
|
y |
Definition. Straight y =kx +b (k≠ 0) is called oblique asymptote function graphics y = f(x) at , where
General scheme for studying functions and constructing graphs.
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 (if x= 0 and at y = 0).
3. Examine for evenness and oddness of the function ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).
4. Find the asymptotes of the graph of the function.
5. Find the 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 function graph.
8. Based on the research conducted, construct a graph of the function.
Example. Explore the function and construct its graph.
1) D (y) =
x= 4 – break point.
2) When x = 0,
(0; ‒ 5) – point of intersection with oh.
At y = 0,
3)
y(‒
x)=
function general view(neither even nor odd).
4) We examine for asymptotes.
a) vertical
b) horizontal
c) find the oblique asymptotes where
‒oblique asymptote equation
5) B given equation there is no need to find intervals of monotonicity of the function.
6)
These critical points divide the entire domain of definition of the function into the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the results obtained in the form of the following table.
Let the function $z=f(x,y)$ be defined and continuous in some bounded closed area$D$. Let the given function in this region have finite partial derivatives of the first order (except, perhaps, for 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 a closed domain $D$.
- Find critical points of the function $z=f(x,y)$ belonging to the domain $D$. Calculate the function values at critical points.
- Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$, finding the points of possible maximum and minimum values. Calculate the function values at the obtained points.
- From the function values obtained in the previous two paragraphs, select the largest and smallest.
What are critical points? show\hide
Under critical points imply points at which 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 first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset critical points.
Example No. 1
Find the largest and smallest values of the function $z=x^2+2xy-y^2-4x$ in a closed region, limited by 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 equations of three straight lines that limit this area. The straight line $x=3$ passes through the point $(3;0)$ parallel to the ordinate axis (Oy axis). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct the line $y=x+1$, we will find two points through which we will draw this 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 at which the line $y=x+1$ intersects the lines $x=3$ and $y=0$. Why is this better? Because we will kill a couple of birds with one stone: we will get two points to construct the line $y=x+1$ and at the same time find out at what points this line intersects other lines that limit the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ intersects at the point $(-1;0)$. In order not to clutter up the progress 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 intersection point of the lines $y=x+1$ and $x=3$. The coordinates of the desired point belong to both the first and second straight lines, therefore, to find the unknown coordinates, you need to solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$
The solution to such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the intersection point of the lines $y=x+1$ and $y=0$. Let us again compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (x-axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything can be seen from the picture. However, it is worth remembering that a drawing cannot serve as evidence. The drawing is for illustrative purposes only.
Our area was defined using the equations of lines that bound it. Obviously, these lines define a triangle, right? Or is it not entirely obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is incorrect. But to avoid such ambiguities, it is better to define regions by inequalities. Are we interested in the part of the plane located under the straight line $y=x+1$? Ok, so $y ≤ x+1$. Should our area be located above the line $y=0$? Great, that means $y ≥ 0$. By the way, the last two inequalities can easily be combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the region $D$, and they define it unambiguously, without allowing any ambiguity. But how does this help us with the question stated at the beginning of the note? It will also help :) We need to check whether the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right $$.
Both inequalities are valid. Point $M_1(1;1)$ belongs to region $D$.
Now it’s the turn to study the behavior of the function at the boundary of the region, i.e. let's go to . Let's start with the straight line $y=0$.
The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Let's substitute $y=0$ into given function$z(x,y)=x^2+2xy-y^2-4x$. We denote the function of one variable $x$ obtained as a result of substitution as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we will also add $M_2(2;0)$ to the list of points. In addition, let us calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points into the original expression $z=x^2+2xy-y^2-4x$. For example, for point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a little. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll write it down in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, in such a detailed records Usually there is no need, and in the future we will write down all calculations briefly:
$$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.$$
Now let's turn to the straight line $x=3$. This straight line limits the region $D$ under the condition $0 ≤ y ≤ 4$. Let's substitute $x=3$ into the given function $z$. As a result of this substitution we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$ we need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Let's find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we will also add $M_5(3;3)$ to the previously found points. In addition, you need to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at points $M_4(3;0)$ and $M_6(3;4)$. At point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; & z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And finally, consider the last boundary of the region $D$, i.e. straight line $y=x+1$. This straight line limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again we need to find the largest and smallest values of this function on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. Points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We received seven values:
$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$
Let's turn to . Choosing the largest and smallest values from the numbers obtained in the third paragraph, we will have:
$$z_(min)=-4; \; z_(max)=6.$$
The problem is solved, all that remains is to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example No. 2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
First, let's build a drawing. The equation $x^2+y^2=25$ (this is the boundary line of a given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ $25 satisfy all points inside and on the mentioned circle.
We will act according to. Let's find partial derivatives and find out the critical points.
$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. let's find stationary points.
$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8. \end(aligned) \right $$.
We got stationary point$(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check whether the inequality $x^2+y^2 ≤ 25$ holds, which defines our region $D$. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ does not hold. Conclusion: point $(6;-8)$ does not belong to area $D$.
So, there are no critical points inside the region $D$. Let's move on to... We need to study the behavior of the function on the boundary of a given area, i.e. on the circle $x^2+y^2=25$. We can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the equation of a circle we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ 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. $$
The further solution will be completely identical to the study of the behavior of the function at the boundary of the region in the previous example No. 1. However, it seems to me more reasonable to apply the Lagrange method in this situation. We will be interested only in the first part of this method. After applying the first part of the Lagrange method, we will obtain points at which we will examine the function $z$ for minimum and maximum values.
We compose the Lagrange function:
$$ 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. right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$
To solve this system, let's immediately point out that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ indicates that the value $\lambda=-1$ is unacceptable. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the condition $\lambda\neq -1$. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator $1+\lambda\neq 0$.
Let us substitute the resulting expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \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. $$
From the resulting equality it follows that $1+\lambda=2$ or $1+\lambda=-2$. Hence we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we got two points of possibility conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Let's find the values of the function $z$ at points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should select the largest and smallest values from those we obtained in the first and second steps. But in in this case the choice is small :) We have:
$$ z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=$125.
