A miniature and rather simple task of the kind that serves as a lifeline for a floating student. In nature, the sleepy realm of mid-July, so it's time to settle down with a laptop on the beach. Early in the morning, a sunbeam of theory began to play in order to soon focus on practice, which, despite its declared lightness, contains glass fragments in the sand. In this regard, I recommend conscientiously consider a few examples of this page. To solve practical tasks, you need to be able to find derivatives and understand the material of the article Intervals of monotonicity and extrema of a function.
First, briefly about the main thing. In a lesson about function continuity I gave the definition of continuity at a point and continuity on an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on a segment if:
1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.
The second paragraph deals with the so-called unilateral continuity functions at a point. There are several approaches to its definition, but I will stick to the line started earlier:
The function is continuous at a point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: 
. It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at that point: 
Imagine that the green dots are the nails on which the magic rubber band is attached:
Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited- a hedge above, a hedge below, and our product grazes in a paddock. Thus, a function continuous on a segment is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and rigorously proved Weierstrass' first theorem.… Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled the graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Indeed, how do you know what awaits us beyond the horizon? After all, once the Earth was considered flat, so today even ordinary teleportation requires proof =)
According to second Weierstrass theorem, continuous on the segmentfunction reaches its exact top edge and his exact bottom edge .
The number is also called the maximum value of the function on the segment and denoted by , and the number - the minimum value of the function on the segment marked .
In our case: 
Note
: in theory, records are common 
.
Roughly speaking, the largest value is located where the highest point of the graph, and the smallest - where the lowest point.
Important! As already pointed out in the article on extrema of the function, the largest value of the function and smallest function value – NOT THE SAME, what function maximum and function minimum. So, in this example, the number is the minimum of the function, but not the minimum value.
By the way, what happens outside the segment? Yes, even the flood, in the context of the problem under consideration, this does not interest us at all. The task involves only finding two numbers 
and that's it!
Moreover, the solution is purely analytical, therefore, no need to draw!
The algorithm lies on the surface and suggests itself from the above figure:
1) Find the function values in critical points, that belong to this segment.
Catch one more goodie: there is no need to check a sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum not yet guaranteed what is the minimum or maximum value. The demonstration function reaches its maximum and, by the will of fate, the same number is the largest value of the function on the interval . But, of course, such a coincidence does not always take place.
So, at the first step, it is faster and easier to calculate the values of the function at critical points belonging to the segment, without bothering whether they have extrema or not.
2) We calculate the values of the function at the ends of the segment.
3) Among the values of the function found in the 1st and 2nd paragraphs, select the smallest and largest number, write down the answer.
We sit on the shore of the blue sea and hit the heels in shallow water:
Example 1
Find the largest and smallest values of a function on a segment
Decision:
1) Calculate the values of the function at critical points belonging to this segment:
Let us calculate the value of the function at the second critical point:
2) Calculate the values of the function at the ends of the segment: 
3) "Bold" results were obtained with exponentials and logarithms, which significantly complicates their comparison. For this reason, we will arm ourselves with a calculator or Excel and calculate the approximate values, not forgetting that: 
Now everything is clear.
Answer:
Fractional-rational instance for independent solution:
Example 6
Find the maximum and minimum values of a function on a segment
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.
- We find the domain of the function and check if it contains the entire segment .
- 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.
- 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.
- 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 .
- 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 a single stationary point): 
Decision.
Let's start with the scope of the function. The square trinomial in the denominator of a fraction must not vanish: 
It is easy to check that all intervals from the condition of the problem belong to the domain of the function.
Let's differentiate the function: 
Obviously, the derivative exists on the entire domain of the function.
Let's find stationary points. The derivative vanishes at . This stationary point falls within the intervals (-3;1] and (-3;2) .
And now you can compare the results obtained at each point with the graph of the function. The blue dotted lines indicate the asymptotes.
This can end with finding the largest and smallest value of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest value of the function on any interval. This gives a clearer picture and a rigorous justification of the results.
Problem Statement 2:
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.
Decision:
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.
