What are extrema of a function: critical points of maximum and minimum. Extrema of the function

Passing in these inequalities to the limit at Δ x→ 0 and taking into account that the derivative f "(x 0) exists, and therefore the limit on the left does not depend on how Δ x→ 0, we get: at Δ x → 0 – 0 f" (x 0) ≥ 0 a at Δ x → 0 + 0 f" (x 0) ≤ 0. Since f" (x 0) defines a number, then these two inequalities are compatible only if f" (x 0) = 0.

The proven theorem states that the maximum and minimum points can only be among those values ​​of the argument at which the derivative becomes zero.

We considered the case when a function has a derivative at all points of a certain segment. What is the situation in cases where the derivative does not exist? Let's look at examples.

y =|x |.

The function has no derivative at the point x=0 (at this point the graph of the function does not have a defined tangent), but at this point the function has a minimum, since y(0)=0, and for all x ≠ 0y > 0.

Thus, from the given examples and the formulated theorem it is clear that a function can have an extremum only in two cases: 1) at points where the derivative exists and is equal to zero; 2) at the point where the derivative does not exist.

However, if at some point x 0 we know that f "(x 0 ) =0, then one cannot conclude from this that at the point x 0 the function has an extremum.

For example.

But period x=0 is not an extremum point, since to the left of this point the function values ​​are located below the axis Ox, and on the right above.

Values ​​of an argument from the domain of a function at which the derivative of the function vanishes or does not exist are called critical points .

From all of the above it follows that the extremum points of the function are among the critical points, and, however, not every critical point is an extremum point. Therefore, to find the extremum of a function, you need to find all the critical points of the function, and then examine each of these points separately for maximum and minimum. The following theorem serves this purpose.

Theorem 2. (A sufficient condition for the existence of an extremum.) Let the function be continuous on some interval containing the critical point x 0, and is differentiable at all points of this interval (except, perhaps, the point itself x 0). If, when moving from left to right through this point, the derivative changes sign from plus to minus, then at the point x = x 0 function has a maximum. If, when passing through x 0 from left to right, the derivative changes sign from minus to plus, then the function has a minimum at this point.

Thus, if

f "(x)>0 at x <x 0 and f "(x)< 0 at x> x 0, then x 0 – maximum point;

Proof. Let us first assume that when passing through x 0 the derivative changes sign from plus to minus, i.e. in front of everyone x, close to the point x 0 f "(x)> 0 for x< x 0 , f "(x)< 0 for x> x 0 . Let's apply Lagrange's theorem to the difference f(x) - f(x 0 ) = f "(c)(x- x 0), where c lies between x And x 0 .

Let x< x 0 . Then c< x 0 and f "(c)> 0. That's why f "(c)(x- x 0)< 0 and therefore

f(x) - f(x 0 )< 0, i.e. f(x)< f(x 0 ).

Let x > x 0 . Then c>x 0 and f "(c)< 0. Means f "(c)(x- x 0)< 0. That's why f(x) - f(x 0 ) <0,т.е.f(x) < f(x 0 ) .

Thus, for all values x close enough to x 0 f(x) < f(x 0 ) . And this means that at the point x 0 function has a maximum.

The second part of the minimum theorem is proved in a similar way.

Let us illustrate the meaning of this theorem in the figure. Let f "(x 1 ) =0 and for any x, close enough to x 1, the inequalities are satisfied

f "(x)< 0 at x< x 1 , f "(x)> 0 at x> x 1 .

Then to the left of the point x 1 the function increases and decreases on the right, therefore, when x = x 1 function goes from increasing to decreasing, that is, it has a maximum.

Similarly, we can consider points x 2 and x 3 .

All of the above can be schematically depicted in the picture:

Rule for studying the function y=f(x) for extremum

Find the domain of a function f(x).

Find the first derivative of a function f "(x) .

Determine critical points for this:

find the real roots of the equation f "(x) =0;

find all values x for which the derivative f "(x) does not exist.

Determine the sign of the derivative to the left and right of the critical point. Since the sign of the derivative remains constant between two critical points, it is sufficient to determine the sign of the derivative at one point to the left and one point to the right of the critical point.

Calculate the value of the function at the extremum points.

To determine the nature of a function and talk about its behavior, it is necessary to find intervals of increase and decrease. This process is called function research and graphing. The extremum point is used when finding the largest and smallest values ​​of a function, since at them the function increases or decreases from the interval.

This article reveals the definitions, formulates a sufficient sign of increase and decrease on the interval and a condition for the existence of an extremum. This applies to solving examples and problems. The section on differentiating functions should be repeated, because the solution will need to use finding the derivative.

Yandex.RTB R-A-339285-1 Definition 1

The function y = f (x) will increase on the interval x when, for any x 1 ∈ X and x 2 ∈ X, x 2 > x 1, the inequality f (x 2) > f (x 1) is satisfied. In other words, a larger value of the argument corresponds to a larger value of the function.

Definition 2

The function y = f (x) is considered to be decreasing on the interval x when, for any x 1 ∈ X, x 2 ∈ X, x 2 > x 1, the equality f (x 2) > f (x 1) is considered true. In other words, a larger function value corresponds to a smaller argument value. Consider the figure below.

Comment: When the function is definite and continuous at the ends of the interval of increasing and decreasing, that is (a; b), where x = a, x = b, the points are included in the interval of increasing and decreasing. This does not contradict the definition; it means that it takes place on the interval x.

The main properties of elementary functions of the type y = sin x are certainty and continuity for real values ​​of the arguments. From here we get that the sine increases over the interval - π 2; π 2, then the increase on the segment has the form - π 2; π 2.

The point x 0 is called maximum point for the function y = f (x), when for all values ​​of x the inequality f (x 0) ≥ f (x) is valid. Maximum function is the value of the function at a point, and is denoted by y m a x .

The point x 0 is called the minimum point for the function y = f (x), when for all values ​​of x the inequality f (x 0) ≤ f (x) is valid. Minimum functions is the value of the function at a point, and has a designation of the form y m i n .

Neighborhoods of the point x 0 are considered extremum points, and the value of the function that corresponds to the extremum points. Consider the figure below.

Extrema of a function with the largest and smallest value of the function. Consider the figure below.

The first figure says that it is necessary to find the largest value of the function from the segment [a; b ] . It is found using maximum points and is equal to the maximum value of the function, and the second figure is more like finding the maximum point at x = b.

Sufficient conditions for a function to increase and decrease

To find the maxima and minima of a function, it is necessary to apply signs of extremum in the case when the function satisfies these conditions. The first sign is considered the most frequently used.

The first sufficient condition for an extremum

Let a function y = f (x) be given, which is differentiable in an ε neighborhood of the point x 0, and has continuity at the given point x 0. From here we get that

• when f " (x) > 0 with x ∈ (x 0 - ε ; x 0) and f " (x)< 0 при x ∈ (x 0 ; x 0 + ε) , тогда x 0 является точкой максимума;
• when f "(x)< 0 с x ∈ (x 0 - ε ; x 0) и f " (x) >0 for x ∈ (x 0 ; x 0 + ε), then x 0 is the minimum point.

In other words, we obtain their conditions for setting the sign:

• when the function is continuous at the point x 0, then it has a derivative with a changing sign, that is, from + to -, which means the point is called a maximum;
• when the function is continuous at the point x 0, then it has a derivative with a changing sign from - to +, which means the point is called a minimum.

To correctly determine the maximum and minimum points of a function, you must follow the algorithm for finding them:

• find the domain of definition;
• find the derivative of the function on this area;
• identify zeros and points where the function does not exist;
• determining the sign of the derivative on intervals;
• select points where the function changes sign.

Let's consider the algorithm by solving several examples of finding extrema of a function.

Example 1

Find the maximum and minimum points of the given function y = 2 (x + 1) 2 x - 2 .

Solution

The domain of definition of this function is all real numbers except x = 2. First, let's find the derivative of the function and get:

y " = 2 x + 1 2 x - 2 " = 2 x + 1 2 " (x - 2) - (x + 1) 2 (x - 2) " (x - 2) 2 = = 2 2 (x + 1) (x + 1) " (x - 2) - (x + 1) 2 1 (x - 2) 2 = 2 2 (x + 1) (x - 2 ) - (x + 2) 2 (x - 2) 2 = = 2 · (x + 1) · (x - 5) (x - 2) 2

From here we see that the zeros of the function are x = - 1, x = 5, x = 2, that is, each bracket must be equated to zero. Let's mark it on the number axis and get:

Now we determine the signs of the derivative from each interval. It is necessary to select a point included in the interval and substitute it into the expression. For example, points x = - 2, x = 0, x = 3, x = 6.

We get that

y " (- 2) = 2 · (x + 1) · (x - 5) (x - 2) 2 x = - 2 = 2 · (- 2 + 1) · (- 2 - 5) (- 2 - 2) 2 = 2 · 7 16 = 7 8 > 0, which means that the interval - ∞ ; - 1 has a positive derivative. Similarly, we find that

y " (0) = 2 · (0 + 1) · 0 - 5 0 - 2 2 = 2 · - 5 4 = - 5 2< 0 y " (3) = 2 · (3 + 1) · (3 - 5) (3 - 2) 2 = 2 · - 8 1 = - 16 < 0 y " (6) = 2 · (6 + 1) · (6 - 5) (6 - 2) 2 = 2 · 7 16 = 7 8 > 0

Since the second interval turned out to be less than zero, it means that the derivative on the interval will be negative. The third with a minus, the fourth with a plus. To determine continuity, you need to pay attention to the sign of the derivative; if it changes, then this is an extremum point.

We find that at the point x = - 1 the function will be continuous, which means that the derivative will change sign from + to -. According to the first sign, we have that x = - 1 is a maximum point, which means we get

y m a x = y (- 1) = 2 (x + 1) 2 x - 2 x = - 1 = 2 (- 1 + 1) 2 - 1 - 2 = 0

The point x = 5 indicates that the function is continuous, and the derivative will change sign from – to +. This means that x = -1 is the minimum point, and its determination has the form

y m i n = y (5) = 2 (x + 1) 2 x - 2 x = 5 = 2 (5 + 1) 2 5 - 2 = 24

Graphic image

Answer: y m a x = y (- 1) = 0, y m i n = y (5) = 24.

It is worth paying attention to the fact that the use of the first sufficient criterion for an extremum does not require the differentiability of the function at the point x 0, this simplifies the calculation.

Example 2

Find the maximum and minimum points of the function y = 1 6 x 3 = 2 x 2 + 22 3 x - 8.

Solution.

The domain of a function is all real numbers. This can be written as a system of equations of the form:

1 6 x 3 - 2 x 2 - 22 3 x - 8 , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 , x ≥ 0

Then you need to find the derivative:

y " = 1 6 x 3 - 2 x 2 - 22 3 x - 8 " , x< 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 " , x >0 y " = - 1 2 x 2 - 4 x - 22 3 , x< 0 1 2 x 2 - 4 x + 22 3 , x > 0

The point x = 0 does not have a derivative, because the values ​​of the one-sided limits are different. We get that:

lim y "x → 0 - 0 = lim y x → 0 - 0 - 1 2 x 2 - 4 x - 22 3 = - 1 2 (0 - 0) 2 - 4 (0 - 0) - 22 3 = - 22 3 lim y " x → 0 + 0 = lim y x → 0 - 0 1 2 x 2 - 4 x + 22 3 = 1 2 (0 + 0) 2 - 4 (0 + 0) + 22 3 = + 22 3

It follows that the function is continuous at the point x = 0, then we calculate

lim y x → 0 - 0 = lim x → 0 - 0 - 1 6 x 3 - 2 x 2 - 22 3 x - 8 = = - 1 6 · (0 - 0) 3 - 2 · (0 - 0) 2 - 22 3 (0 - 0) - 8 = - 8 lim y x → 0 + 0 = lim x → 0 - 0 1 6 x 3 - 2 x 2 + 22 3 x - 8 = = 1 6 (0 + 0) 3 - 2 · (0 + 0) 2 + 22 3 · (0 + 0) - 8 = - 8 y (0) = 1 6 x 3 - 2 x 2 + 22 3 x - 8 x = 0 = 1 6 · 0 3 - 2 0 2 + 22 3 0 - 8 = - 8

It is necessary to perform calculations to find the value of the argument when the derivative becomes zero:

1 2 x 2 - 4 x - 22 3 , x< 0 D = (- 4) 2 - 4 · - 1 2 · - 22 3 = 4 3 x 1 = 4 + 4 3 2 · - 1 2 = - 4 - 2 3 3 < 0 x 2 = 4 - 4 3 2 · - 1 2 = - 4 + 2 3 3 < 0

1 2 x 2 - 4 x + 22 3 , x > 0 D = (- 4) 2 - 4 1 2 22 3 = 4 3 x 3 = 4 + 4 3 2 1 2 = 4 + 2 3 3 > 0 x 4 = 4 - 4 3 2 1 2 = 4 - 2 3 3 > 0

All obtained points must be marked on a straight line to determine the sign of each interval. Therefore, it is necessary to calculate the derivative at arbitrary points for each interval. For example, we can take points with values ​​x = - 6, x = - 4, x = - 1, x = 1, x = 4, x = 6. We get that

y " (- 6) = - 1 2 x 2 - 4 x - 22 3 x = - 6 = - 1 2 · - 6 2 - 4 · (- 6) - 22 3 = - 4 3< 0 y " (- 4) = - 1 2 x 2 - 4 x - 22 3 x = - 4 = - 1 2 · (- 4) 2 - 4 · (- 4) - 22 3 = 2 3 >0 y " (- 1) = - 1 2 x 2 - 4 x - 22 3 x = - 1 = - 1 2 · (- 1) 2 - 4 · (- 1) - 22 3 = 23 6< 0 y " (1) = 1 2 x 2 - 4 x + 22 3 x = 1 = 1 2 · 1 2 - 4 · 1 + 22 3 = 23 6 >0 y "(4) = 1 2 x 2 - 4 x + 22 3 x = 4 = 1 2 4 2 - 4 4 + 22 3 = - 2 3< 0 y " (6) = 1 2 x 2 - 4 x + 22 3 x = 6 = 1 2 · 6 2 - 4 · 6 + 22 3 = 4 3 > 0

The image on the straight line looks like

This means that we come to the conclusion that it is necessary to resort to the first sign of an extremum. Let us calculate and find that

x = - 4 - 2 3 3 , x = 0 , x = 4 + 2 3 3 , then from here the maximum points have the values ​​x = - 4 + 2 3 3 , x = 4 - 2 3 3

Let's move on to calculating the minimums:

y m i n = y - 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 0 = - 8 y m i n = y 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 + 2 3 3 = - 8 27 3

Let's calculate the maxima of the function. We get that

y m a x = y - 4 + 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 1 6 x 3 - 2 2 + 22 3 x - 8 x = 4 - 2 3 3 = 8 27 3

Graphic image

Answer:

y m i n = y - 4 - 2 3 3 = - 8 27 3 y m i n = y (0) = - 8 y m i n = y 4 + 2 3 3 = - 8 27 3 y m a x = y - 4 + 2 3 3 = 8 27 3 y m a x = y 4 - 2 3 3 = 8 27 3

If a function f " (x 0) = 0 is given, then if f "" (x 0) > 0, we obtain that x 0 is a minimum point if f "" (x 0)< 0 , то точкой максимума. Признак связан с нахождением производной в точке x 0 .

Example 3

Find the maxima and minima of the function y = 8 x x + 1.

Solution

First, we find the domain of definition. We get that

D(y) : x ≥ 0 x ≠ - 1 ⇔ x ≥ 0

It is necessary to differentiate the function, after which we get

y " = 8 x x + 1 " = 8 x " (x + 1) - x (x + 1) " (x + 1) 2 = = 8 1 2 x (x + 1) - x 1 (x + 1) 2 = 4 x + 1 - 2 x (x + 1) 2 x = 4 - x + 1 (x + 1) 2 x

At x = 1, the derivative becomes zero, which means that the point is a possible extremum. To clarify, it is necessary to find the second derivative and calculate the value at x = 1. We get:

y "" = 4 - x + 1 (x + 1) 2 x " = = 4 (- x + 1) " (x + 1) 2 x - (- x + 1) x + 1 2 x " (x + 1) 4 x = = 4 (- 1) (x + 1) 2 x - (- x + 1) x + 1 2 " x + (x + 1) 2 x " (x + 1) 4 x = = 4 - (x + 1) 2 x - (- x + 1) 2 x + 1 (x + 1) " x + (x + 1) 2 2 x (x + 1) 4 x = = - (x + 1) 2 x - (- x + 1) x + 1 2 x + x + 1 2 x (x + 1) 4 x = = 2 3 x 2 - 6 x - 1 x + 1 3 x 3 ⇒ y "" (1) = 2 3 1 2 - 6 1 - 1 (1 + 1) 3 (1) 3 = 2 · - 4 8 = - 1< 0

This means that using the 2 sufficient condition for an extremum, we obtain that x = 1 is a maximum point. Otherwise, the entry looks like y m a x = y (1) = 8 1 1 + 1 = 4.

Graphic image

Answer: y m a x = y (1) = 4 ..

The function y = f (x) has its derivative up to the nth order in the ε neighborhood of a given point x 0 and its derivative up to the n + 1st order at the point x 0 . Then f " (x 0) = f "" (x 0) = f " " " (x 0) = . . . = f n (x 0) = 0 .

It follows that when n is an even number, then x 0 is considered an inflection point, when n is an odd number, then x 0 is an extremum point, and f (n + 1) (x 0) > 0, then x 0 is a minimum point, f (n + 1) (x 0)< 0 , тогда x 0 является точкой максимума.

Example 4

Find the maximum and minimum points of the function y y = 1 16 (x + 1) 3 (x - 3) 4.

Solution

The original function is a rational entire function, which means that the domain of definition is all real numbers. It is necessary to differentiate the function. We get that

y " = 1 16 x + 1 3 " (x - 3) 4 + (x + 1) 3 x - 3 4 " = = 1 16 (3 (x + 1) 2 (x - 3) 4 + (x + 1) 3 4 (x - 3) 3) = = 1 16 (x + 1) 2 (x - 3) 3 (3 x - 9 + 4 x + 4) = 1 16 (x + 1) 2 (x - 3) 3 (7 x - 5)

This derivative will go to zero at x 1 = - 1, x 2 = 5 7, x 3 = 3. That is, the points can be possible extremum points. It is necessary to apply the third sufficient condition for the extremum. Finding the second derivative allows you to accurately determine the presence of a maximum and minimum of a function. The second derivative is calculated at the points of its possible extremum. We get that

y "" = 1 16 x + 1 2 (x - 3) 3 (7 x - 5) " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) y "" (- 1) = 0 y "" 5 7 = - 36864 2401< 0 y "" (3) = 0

This means that x 2 = 5 7 is the maximum point. Applying the 3rd sufficient criterion, we obtain that for n = 1 and f (n + 1) 5 7< 0 .

It is necessary to determine the nature of the points x 1 = - 1, x 3 = 3. To do this, you need to find the third derivative and calculate the values ​​at these points. We get that

y " " " = 1 8 (x + 1) (x - 3) 2 (21 x 2 - 30 x - 3) " = = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) y " " " (- 1) = 96 ≠ 0 y " " " (3) = 0

This means that x 1 = - 1 is the inflection point of the function, since for n = 2 and f (n + 1) (- 1) ≠ 0. It is necessary to investigate the point x 3 = 3. To do this, we find the 4th derivative and perform calculations at this point:

y (4) = 1 8 (x - 3) (105 x 3 - 225 x 2 - 45 x + 93) " = = 1 2 (105 x 3 - 405 x 2 + 315 x + 57) y (4) ( 3) = 96 > 0

From what was decided above we conclude that x 3 = 3 is the minimum point of the function.

Graphic image

Answer: x 2 = 5 7 is the maximum point, x 3 = 3 is the minimum point of the given function.

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This is a rather interesting section of mathematics, which absolutely all graduates and students encounter. However, not everyone likes matan. Some cannot understand even basic things like a seemingly standard function study. This article is intended to correct such an oversight. Want to learn more about function analysis? Would you like to know what extremum points are and how to find them? Then this article is for you.

Studying the graph of a function

First, it’s worth understanding why you need to analyze the graph at all. There are simple functions that are not difficult to draw. A striking example of such a function is a parabola. It won't be difficult to draw a graph. All that is needed is, using a simple transformation, to find the numbers at which the function takes the value 0. And in principle, this is all you need to know in order to draw a graph of a parabola.

But what if the function we need to graph is much more complex? Since the properties of complex functions are not quite obvious, it is necessary to carry out a whole analysis. Only after this can the function be depicted graphically. How to do this? You can find the answer to this question in this article.

Function Analysis Plan

The first thing we need to do is to conduct a superficial study of the function, during which we find the domain of definition. So, let's start in order. The domain of definition is the set of values ​​by which the function is defined. Simply put, these are the numbers that can be used in a function instead of x. To determine the scope, you just need to look at the record. For example, it is obvious that the function y (x) = x 3 + x 2 - x + 43 has a domain of definition that is the set of real numbers. Well, with a function like (x 2 - 2x)/x everything is a little different. Since the number in the denominator must not equal 0, the domain of definition of this function will be all real numbers other than zero.

Next, you need to find the so-called zeros of the function. These are the argument values ​​at which the entire function takes the value zero. To do this, it is necessary to equate the function to zero, consider it in detail and perform some transformations. Let's take the already familiar function y(x) = (x 2 - 2x)/x. From the school course we know that a fraction is equal to 0 when the numerator is equal to zero. Therefore, we discard the denominator and start working with the numerator, equating it to zero. We get x 2 - 2x = 0 and put x out of brackets. Hence x (x - 2) = 0. As a result, we find that our function is equal to zero when x equals 0 or 2.

When examining the graph of a function, many people encounter problems in the form of extremum points. And it's strange. After all, extremes are a fairly simple topic. Don't believe me? See for yourself by reading this part of the article, in which we will talk about minimum and maximum points.

First, it’s worth understanding what an extremum is. An extremum is the limit value that a function reaches on a graph. It turns out that there are two extreme values ​​- maximum and minimum. For clarity, you can look at the picture above. In the studied area, point -1 is the maximum of the function y (x) = x 5 - 5x, and point 1, accordingly, is the minimum.

Also, do not confuse the concepts. The extremum points of a function are those arguments at which a given function acquires extreme values. In turn, the extremum is the value of the minimums and maximums of a function. For example, consider the figure above again. -1 and 1 are the extrema points of the function, and 4 and -4 are the extrema themselves.

Finding extremum points

But how do you find the extremum points of a function? Everything is quite simple. The first thing to do is find the derivative of the equation. Let’s say we received the task: “Find the extremum points of the function y (x), x is the argument. For clarity, let’s take the function y (x) = x 3 + 2x 2 + x + 54. Let’s differentiate and get the following equation: 3x 2 + 4x + 1. As a result, we have a standard quadratic equation. All we need to do next is to equate it to zero and find the roots. Since the discriminant is greater than zero (D = 16 - 12 = 4), this equation is determined by two roots. Find them and get two values: 1/3 and -1. These will be the extremum points of the function. However, how can you still determine who is who? Which point is the maximum and which is the minimum? To do this, you need to take the neighboring point and find out its value. For example , take the number -2, which is located to the left along the coordinate line from -1. Substitute this value into our equation y(-2) = 12 - 8 + 1 = 5. As a result, we get a positive number. This means that in the interval from The function increases from 1/3 to -1. This, in turn, means that on the intervals from minus infinity to 1/3 and from -1 to plus infinity the function decreases. Thus, we can conclude that the number 1/3 is the minimum point of the function on the studied interval, and -1 is the maximum point.

It is also worth noting that the Unified State Exam requires not only finding extremum points, but also performing some kind of operation with them (adding, multiplying, etc.). It is for this reason that it is worth paying special attention to the conditions of the problem. After all, due to inattention, you can lose points.

Lesson on the topic: "Finding extrema points of functions. Examples"

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What we will study:
1. Introduction.
2. Minimum and maximum points.

4. How to calculate extrema?
5. Examples.

Introduction to Function Extrema

Guys, let's look at the graph of a certain function:

Notice that the behavior of our function y=f (x) is largely determined by two points x1 and x2. Let's take a closer look at the graph of the function at and around these points. Up to point x2 the function increases, at point x2 there is an inflection, and immediately after this point the function decreases to point x1. At point x1 the function bends again, and after that it increases again. For now, we will call points x1 and x2 inflection points. Let's draw tangents at these points:

The tangents at our points are parallel to the x-axis, which means that the slope of the tangent is zero. This means that the derivative of our function at these points is equal to zero.

Let's look at the graph of this function:

It is impossible to draw tangent lines at points x2 and x1. This means that the derivative does not exist at these points. Now let's look again at our points on the two graphs. Point x2 is the point at which the function reaches its greatest value in some region (near point x2). Point x1 is the point at which the function reaches its smallest value in some region (near point x1).

Minimum and maximum points

Definition: The point x= x0 is called the minimum point of the function y=f(x) if there is a neighborhood of the point x0 in which the inequality holds: f(x) ≥ f(x0).

Definition: The point x=x0 is called the maximum point of the function y=f(x) if there is a neighborhood of the point x0 in which the inequality holds: f(x) ≤ f(x0).

Guys, what is a neighborhood?

Definition: A neighborhood of a point is a set of points containing our point and those close to it.

We can set the neighborhood ourselves. For example, for a point x=2, we can define a neighborhood in the form of points 1 and 3.

Let's return to our graphs, look at point x2, it is larger than all other points from a certain neighborhood, then by definition it is a maximum point. Now let's look at point x1, it is smaller than all other points from a certain neighborhood, then by definition it is a minimum point.

Guys, let's introduce the notation:

Y min - minimum point,
y max - maximum point.

Important! Guys, do not confuse the maximum and minimum points with the smallest and largest value of the function. The minimum and maximum values ​​are sought over the entire domain of definition of a given function, and the minimum and maximum points are sought in a certain neighborhood.

Extrema of the function

For minimum and maximum points there is a common term - extremum points.

Extremum (lat. extremum – extreme) – the maximum or minimum value of a function on a given set. The point at which the extremum is reached is called the extremum point.

Accordingly, if a minimum is reached, the extremum point is called a minimum point, and if a maximum is reached, it is called a maximum point.

How to look for extrema of a function?

Let's go back to our charts. At our points, the derivative either vanishes (on the first graph) or does not exist (on the second graph).

Then we can make an important statement: If the function y= f(x) has an extremum at the point x=x0, then at this point the derivative of the function is either zero or does not exist.

The points at which the derivative is equal to zero are called stationary.

The points at which the derivative of a function does not exist are called critical.

How to calculate extremes?

Guys, let's go back to the first graph of the function:

Analyzing this graph, we said: up to point x2 the function increases, at point x2 an inflection occurs, and after this point the function decreases to point x1. At point x1 the function bends again, and after that the function increases again.

Based on such reasoning, we can conclude that the function at extremum points changes the nature of monotonicity, and therefore the derivative function changes sign. Recall: if a function decreases, then the derivative is less than or equal to zero, and if the function increases, then the derivative is greater than or equal to zero.

Let us summarize the knowledge gained with the following statement:

Theorem: A sufficient condition for an extremum: let the function y=f(x) be continuous on some interval X and have a stationary or critical point x= x0 inside the interval. Then:

• If this point has a neighborhood in which f’(x)>0 holds for x x0, then point x0 is the minimum point of the function y= f(x).
• If this point has a neighborhood in which f'(x) holds for x 0 and x> x0. If this point has a neighborhood in which both to the left and to the right of the point x0 the signs of the derivative are the same, then at the point x0 there is no extreme.

To solve problems, remember these rules: If the signs of derivatives are defined then:

Algorithm for studying a continuous function y= f(x) for monotonicity and extrema:

• Find the derivative of y'.
• Find stationary points (the derivative is zero) and critical points (the derivative does not exist).
• Mark stationary and critical points on the number line and determine the signs of the derivative on the resulting intervals.
• Based on the above statements, draw a conclusion about the nature of the extremum points.

Examples of finding extreme points

1) Find the extremum points of the function and determine their nature: y= 7+ 12*x - x 3

Solution: Our function is continuous, then we will use our algorithm:
a) y"= 12 - 3x 2,
b) y"= 0, at x= ±2,

Point x= -2 is the minimum point of the function, point x= 2 is the maximum point of the function.
Answer: x= -2 is the minimum point of the function, x= 2 is the maximum point of the function.

2) Find the extremum points of the function and determine their nature.

3) Find the extremum points of the function y= x - 2cos(x) and determine their nature, for -π ≤ x ≤ π.

Solution: Our function is continuous, let's use our algorithm:
a) y"= 1 + 2sin(x),
b) find the values ​​in which the derivative is equal to zero: 1 + 2sin(x)= 0, sin(x)= -1/2,
because -π ≤ x ≤ π, then: x= -π/6, -5π/6,
c) mark stationary points on the number line and determine the signs of the derivative: d) look at our figure, which shows the rules for determining extrema.
Point x= -5π/6 is the maximum point of the function.
Point x= -π/6 is the minimum point of the function.
Answer: x= -5π/6 is the maximum point of the function, x= -π/6 is the minimum point of the function.

4) Find the extremum points of the function and determine their nature:

Problems to solve independently