The largest and smallest value of a function. The largest and smallest values of the function The smallest value of the function f x
Sometimes in problems B15 there are “bad” functions for which it is difficult to find a derivative. Previously, this only happened during sample tests, but now these tasks are so common that they can no longer be ignored when preparing for the real Unified State Exam.
In this case, other techniques work, one of which is monotone.
A function f (x) is said to be monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following holds:
x 1< x 2 ⇒ f (x 1 ) < f (x 2 ).
A function f (x) is said to be monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following holds:
x 1< x 2 ⇒ f (x 1 ) >f(x2).
In other words, for an increasing function, the larger x, the larger f(x). For a decreasing function the opposite is true: the larger x, the less f(x).
For example, the logarithm increases monotonically if the base a > 1, and monotonically decreases if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.
f (x) = log a x (a > 0; a ≠ 1; x > 0)
The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:
The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, exponential function defined for all numbers, not just x > 0:
f (x) = a x (a > 0)
Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where the monotony is broken.
All these functions are never found in their pure form. They add polynomials, fractions and other nonsense, which makes it difficult to calculate the derivative. Let's look at what happens in this case.
Parabola vertex coordinatesMost often the function argument is replaced with quadratic trinomial of the form y = ax 2 + bx + c. Its graph is a standard parabola in which we are interested in:
Of greatest interest is vertex of parabola, the abscissa of which is calculated by the formula:
So, we have found the extremum point of the quadratic function. But if the original function is monotonic, for it the point x 0 will also be an extremum point. Thus, let us formulate the key rule:
Extremum points of a quadratic trinomial and complex function, which it is included in, coincide. Therefore, you can look for x 0 for a quadratic trinomial, and forget about the function.
From the above reasoning, it remains unclear which point we get: maximum or minimum. However, the tasks are specifically designed so that this does not matter. Judge for yourself:
Thus, solving the problem is greatly simplified and comes down to just two steps:
At first glance, this algorithm and its rationale may seem complex. I deliberately do not post a “bare” solution diagram, since thoughtless application of such rules is fraught with errors.
Let's look at real problems from the test Unified State Examination in mathematics - that's where this technique occurs most often. At the same time, we will make sure that in this way many B15 problems become almost oral.
Stands under the root quadratic function y = x 2 + 6x + 13. The graph of this function is a parabola with branches upward, since the coefficient a = 1 > 0.
Vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 1) = −6/2 = −3
Since the branches of the parabola are directed upward, at the point x 0 = −3 the function y = x 2 + 6x + 13 takes on its minimum value.
The root increases monotonically, which means x 0 is the minimum point of the entire function. We have:
Task. Find the smallest value of the function:
y = log 2 (x 2 + 2x + 9)
Under the logarithm there is again a quadratic function: y = x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.
Vertex of the parabola:
x 0 = −b /(2a ) = −2/(2 1) = −2/2 = −1
So, at the point x 0 = −1 the quadratic function takes on its minimum value. But the function y = log 2 x is monotonic, so:
y min = y (−1) = log 2 ((−1) 2 + 2 · (−1) + 9) = ... = log 2 8 = 3
The exponent contains the quadratic function y = 1 − 4x − x 2 . Let's rewrite it in normal form: y = −x 2 − 4x + 1.
Obviously, the graph of this function is a parabola, branches down (a = −1< 0). Поэтому вершина будет точкой максимума:
x 0 = −b /(2a ) = −(−4)/(2 · (−1)) = 4/(−2) = −2
The original function is exponential, it is monotonic, therefore highest value will be at the found point x 0 = −2:
An attentive reader will probably notice that we did not write out the range of permissible values of the root and logarithm. But this was not required: inside there are functions whose values are always positive.
Corollaries from the domain of a functionSometimes simply finding the vertex of the parabola is not enough to solve Problem B15. The value you are looking for may lie at the end of the segment, and not at all at the extremum point. If the problem does not indicate a segment at all, look at range of acceptable values original function. Namely:
Please note again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how this works with specific examples:
Task. Find the largest value of the function:
Under the root is again a quadratic function: y = 3 − 2x − x 2 . Its graph is a parabola, but branches down because a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический Square root of a negative number does not exist.
We write out the range of permissible values (APV):
3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; 1]
Now let's find the vertex of the parabola:
x 0 = −b /(2a ) = −(−2)/(2 · (−1)) = 2/(−2) = −1
The point x 0 = −1 belongs to the ODZ segment - and this is good. Now we calculate the value of the function at the point x 0, as well as at the ends of the ODZ:
y(−3) = y(1) = 0
So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.
Task. Find the smallest value of the function:
y = log 0.5 (6x − x 2 − 5)
Inside the logarithm there is a quadratic function y = 6x − x 2 − 5. This is a parabola with branches down, but there cannot be negative numbers in the logarithm, so we write out the ODZ:
6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)
Please note: the inequality is strict, so the ends do not belong to the ODZ. This differs the logarithm from the root, where the ends of the segment suit us quite well.
We are looking for the vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 · (−1)) = −6/(−2) = 3
The vertex of the parabola fits according to the ODZ: x 0 = 3 ∈ (1; 5). But since we are not interested in the ends of the segment, we calculate the value of the function only at the point x 0:
y min = y (3) = log 0.5 (6 3 − 3 2 − 5) = log 0.5 (18 − 9 − 5) = log 0.5 4 = −2
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 at the internal point of the segment [ a, b], or on the boundary of the segment.
To find the largest and smallest values of a function on the segment [ a, b] necessary:
1) find the critical points of the function in the interval ( a, b);
2) calculate the values of the function at the found critical points;
3) calculate the values of the function at the ends of the segment, that is, 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.
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
Function Research Algorithm y = 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 the 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 build 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:
|
no extras |
From the table it is clear that the point X= ‒2‒maximum point, at point X= 4‒no extremum, X= 10 – minimum point.
Let's substitute the value (‒ 3) into the equation:
9 + 24 ‒ 20 > 0
25 ‒ 40 ‒ 20 < 0
121 ‒ 88 ‒ 20 > 0
The maximum of this function is 
(‒ 2; ‒ 4) – maximum extremum.
The minimum of this function is 
(10; 20) – minimum extremum.
7) examine the convexity and inflection point of the function graph
In practice, it is quite common to use the derivative in order to calculate the largest and smallest value of a function. We perform this action when we figure out how to minimize costs, increase profits, calculate the optimal load on production, etc., that is, in cases where we need to determine the optimal value of a parameter. To solve such problems correctly, you need to have a good understanding of what the largest and smallest values of a function are.
Typically we define these values within a certain interval x, which in turn may correspond to the entire domain of the function or part of it. It can be like a segment [a; b ] , and open interval (a ; b), (a ; b ], [ a ; b), infinite interval (a ; b), (a ; b ], [ a ; b) or infinite interval - ∞ ; a , (- ∞ ; a ] , [ a ; + ∞) , (- ∞ ; + ∞) .
In this material we will tell you how to calculate the largest and smallest values of an explicitly defined function with one variable y=f(x) y = f (x) .
Basic definitionsLet's start, as always, with the formulation of basic definitions.
Definition 1
The largest value of the function y = f (x) on a certain interval x is the value m a x y = f (x 0) x ∈ X, which for any value x x ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x) valid 0) .
Definition 2
Lowest value function y = f (x) on a certain interval x is the value m i n x ∈ X y = f (x 0), which for any value x ∈ X, x ≠ x 0 makes the inequality f(X f (x) ≥ f ( x 0) .
These definitions are quite obvious. Even simpler, we can say this: the greatest value of a function is its most great importance on a known interval at abscissa x 0, and the smallest is the smallest accepted value on the same interval at x 0.
Definition 3
Stationary points are those values of the argument of a function at which its derivative becomes 0.
Why do we need to know what stationary points are? To answer this question, we need to remember Fermat's theorem. It follows from it that a stationary point is the point at which the extremum of the differentiable function is located (i.e., its local minimum or maximum). Consequently, the function will take the smallest or largest value on a certain interval precisely at one of the stationary points.
A function can also take on the largest or smallest value at those points at which the function itself is defined and its first derivative does not exist.
The first question that arises when studying this topic: in all cases can we determine the largest or smallest value of a function on a given interval? No, we cannot do this when the boundaries of a given interval coincide with the boundaries of the definition area, or if we are dealing with an infinite interval. It also happens that a function in a given segment or at infinity will take infinitely small or infinitely large values. In these cases, it is not possible to determine the largest and/or smallest value.
These points will become clearer after being depicted on the graphs:
The first figure shows us a function that takes the largest and smallest values (m a x y and m i n y) at stationary points located on the segment [ - 6 ; 6].
Let us examine in detail the case indicated in the second graph. Let's change the value of the segment to [ 1 ; 6 ] and we find that the maximum value of the function will be achieved at the point with the abscissa at the right boundary of the interval, and the minimum - at the stationary point.
In the third figure, the abscissas of the points represent the boundary points of the segment [ - 3 ; 2]. They correspond to the largest and smallest value of a given function.
Now let's look at the fourth picture. In it, the function takes m a x y (the largest value) and m i n y (the smallest value) at stationary points on the open interval (- 6; 6).
If we take the interval [ 1 ; 6), then we can say that the smallest value of the function on it will be achieved at a stationary point. The greatest value will be unknown to us. The function could take its maximum value at x equal to 6 if x = 6 belonged to the interval. This is exactly the case shown in graph 5.
In graph 6, this function acquires its smallest value at the right boundary of the interval (- 3; 2 ], and we cannot draw definite conclusions about the largest value.
In Figure 7 we see that the function will have m a x y at a stationary point having an abscissa equal to 1. The function will reach its minimum value at the boundary of the interval on the right side. At minus infinity, the function values will asymptotically approach y = 3.
If we take the interval x ∈ 2 ; + ∞ , then we will see that the given function will take neither the smallest nor the largest value on it. If x tends to 2, then the values of the function will tend to minus infinity, since the straight line x = 2 is a vertical asymptote. If the abscissa tends to plus infinity, then the function values will asymptotically approach y = 3. This is exactly the case shown in Figure 8.
In this paragraph we will present the sequence of actions that need to be performed to find the largest or smallest value of a function on a certain segment.
Let's see how to correctly apply this algorithm when solving problems.
Example 1
Condition: the function y = x 3 + 4 x 2 is given. Determine its largest and smallest values on the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .
Solution:
Let's start by finding the domain of definition of a given function. In this case, it will be the set of all real numbers except 0. In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; + ∞ . Both segments specified in the condition will be inside the definition area.
Now we calculate the derivative of the function according to the rule of fraction differentiation:
y " = x 3 + 4 x 2 " = x 3 + 4 " x 2 - x 3 + 4 x 2 " x 4 = = 3 x 2 x 2 - (x 3 - 4) 2 x x 4 = x 3 - 8 x 3
We learned that the derivative of a function will exist at all points of the segments [ 1 ; 4 ] and [ - 4 ; - 1 ] .
Now we need to determine the stationary points of the function. Let's do this using the equation x 3 - 8 x 3 = 0. It has only one real root, which is 2. It will be a stationary point of the function and will fall into the first segment [1; 4 ] .
Let us calculate the values of the function at the ends of the first segment and at this point, i.e. for x = 1, x = 2 and x = 4:
y (1) = 1 3 + 4 1 2 = 5 y (2) = 2 3 + 4 2 2 = 3 y (4) = 4 3 + 4 4 2 = 4 1 4
We found that the largest value of the function m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 will be achieved at x = 1, and the smallest m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – at x = 2.
The second segment does not include a single stationary point, so we need to calculate the function values only at the ends of the given segment:
y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3
This means m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
Answer: For the segment [ 1 ; 4 ] - m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 , for the segment [ - 4 ; - 1 ] - m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
See picture:
Before you study this method, we advise you to review how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and/or smallest value of a function on an open or infinite interval, perform the following steps sequentially.
- If the interval is of the form [ a ; b) , then we need to calculate the value of the function at the point x = a and the one-sided limit lim x → b - 0 f (x) .
- If the interval has the form (a; b ], then we need to calculate the value of the function at the point x = b and the one-sided limit lim x → a + 0 f (x).
- If the interval has the form (a; b), then we need to calculate the one-sided limits lim x → b - 0 f (x), lim x → a + 0 f (x).
- If the interval is of the form [ a ; + ∞), then we need to calculate the value at the point x = a and the limit at plus infinity lim x → + ∞ f (x) .
- If the interval looks like (- ∞ ; b ] , we calculate the value at the point x = b and the limit at minus infinity lim x → - ∞ f (x) .
- If - ∞ ; b , then we consider the one-sided limit lim x → b - 0 f (x) and the limit at minus infinity lim x → - ∞ f (x)
- If - ∞; + ∞ , then we consider the limits on minus and plus infinity lim x → + ∞ f (x) , lim x → - ∞ f (x) .
Condition: given function y = 3 e 1 x 2 + x - 6 - 4 . Calculate its largest and smallest value in the intervals - ∞ ; - 4, - ∞; - 3 , (- 3 ; 1 ] , (- 3 ; 2) , [ 1 ; 2) , 2 ; + ∞ , [ 4 ; + ∞) .
Solution
First of all, we find the domain of definition of the function. The denominator of the fraction contains a quadratic trinomial, which should not turn to 0:
x 2 + x - 6 = 0 D = 1 2 - 4 1 (- 6) = 25 x 1 = - 1 - 5 2 = - 3 x 2 = - 1 + 5 2 = 2 ⇒ D (y) : x ∈ (- ∞ ; - 3) ∪ (- 3 ; 2) ∪ (2 ; + ∞)
We have obtained the domain of definition of the function to which all the intervals specified in the condition belong.
Now let's differentiate the function and get:
y" = 3 e 1 x 2 + x - 6 - 4 " = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " = = 3 · e 1 x 2 + x - 6 · 1 " · x 2 + x - 6 - 1 · x 2 + x - 6 " (x 2 + x - 6) 2 = - 3 · (2 x + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2
Consequently, derivatives of a function exist throughout its entire domain of definition.
Let's move on to finding stationary points. The derivative of the function becomes 0 at x = - 1 2 . This is a stationary point that lies in the intervals (- 3 ; 1 ] and (- 3 ; 2) .
Let's calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ], as well as the limit at minus infinity:
y (- 4) = 3 e 1 (- 4) 2 + (- 4) - 6 - 4 = 3 e 1 6 - 4 ≈ - 0 . 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1
Since 3 e 1 6 - 4 > - 1, it means that m a x y x ∈ (- ∞ ; - 4 ] = y (- 4) = 3 e 1 6 - 4. This does not allow us to uniquely determine the smallest value of the function. We can only conclude that there is a constraint below - 1, since it is to this value that the function approaches asymptotically at minus infinity.
The peculiarity of the second interval is that there is not a single stationary point and not a single strict boundary in it. Consequently, we will not be able to calculate either the largest or smallest value of the function. Having defined the limit at minus infinity and as the argument tends to - 3 on the left side, we get only an interval of values:
lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 = 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
This means that the function values will be located in the interval - 1; +∞
To find the greatest value of the function in the third interval, we determine its value at the stationary point x = - 1 2 if x = 1. We will also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e 4 25 - 4 ≈ - 1 . 444 y (1) = 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1 . 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 = = 3 e 1 (- 0) - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
It turned out that the function will take the greatest value at a stationary point m a x y x ∈ (3; 1 ] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. Everything we know , is the presence of a lower limit to - 4 .
For the interval (- 3 ; 2), take the results of the previous calculation and once again calculate what the one-sided limit is equal to when tending to 2 on the left side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e - 4 25 - 4 ≈ - 1 . 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 = = 3 e 1 - 0 - 4 = 3 e - ∞ - 4 = 3 · 0 - 4 = - 4
This means that m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4, and the smallest value cannot be determined, and the values of the function are limited from below by the number - 4.
Based on what we got in the two previous calculations, we can say that on the interval [ 1 ; 2) the function will take the largest value at x = 1, but it is impossible to find the smallest.
On the interval (2 ; + ∞) the function will not reach either the largest or the smallest value, i.e. it will take values from the interval - 1 ; + ∞ .
lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
Having calculated what the value of the function will be equal to at x = 4, we find out that m a x y x ∈ [ 4 ; + ∞) = y (4) = 3 e 1 14 - 4 , and the given function at plus infinity will asymptotically approach the straight line y = - 1 .
Let's compare what we got in each calculation with the graph of the given function. In the figure, the asymptotes are shown by dotted lines.
That's all we wanted to tell you about finding the largest and smallest values of a function. The sequences of actions that we have given will help you make the necessary calculations as quickly and simply as possible. But remember that it is often useful to first find out at which intervals the function will decrease and at which it will increase, after which you can draw further conclusions. This way you can more accurately determine the largest and smallest values of the function and justify the results obtained.
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Another New Year's Eve... frosty weather and snowflakes on the window glass... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. There is an interesting article on this subject, which contains examples of two-dimensional fractal structures. Here we will look at more complex examples of three-dimensional fractals.
A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in in this case, a set of points), the details of which have the same shape as the original figure itself. That is, this is a self-similar structure, examining the details of which when magnified, we will see the same shape as without magnification. Whereas in the case of ordinary geometric figure(not a fractal), when zoomed in we will see details that have more simple form than the original figure itself. For example, at a high enough magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them we will again see the same complex shape, which will be repeated again and again with each increase.
Benoit Mandelbrot, the founder of the science of fractals, wrote in his article Fractals and Art in the Name of Science: “Fractals are geometric shapes, which in equally complex in their details as well as in their general form. That is, if part of a fractal is enlarged to the size of the whole, it will appear as the whole, either exactly, or perhaps with a slight deformation."
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 values of a function are usually sought on a certain interval X, which is either the entire domain of the function or part of the domain of definition. 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 defined function of one variable y=f(x) .
Page navigation.
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 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 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 some point, then this point is stationary. Thus, the function often takes its greatest (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take its greatest and minimum 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 presented in the seventh figure, the function takes the largest 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 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 value continuous function on the segment.Let us write an algorithm that allows us to find the largest and smallest values of a function on a segment.
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 definition of a function is the entire set of real numbers, with the exception of 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): 
Solution.
Let's start with the domain of the function. The square trinomial in the denominator of the fraction must not vanish: 
It is easy to check that all intervals from the problem statement belong to the domain of definition of the function.
Let's differentiate the function: 
Obviously, the derivative exists throughout the entire domain of definition of the function.
Let's find stationary points. The derivative goes to zero at . This stationary point falls within the intervals (-3;1] and (-3;2).
Now you can compare the results obtained in each point with the graph of the function. Blue dotted lines indicate asymptotes.
At this point we can finish with finding the largest and smallest values 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 values of the function on any interval. This gives a clearer picture and rigorous justification for the results.