The largest and smallest value of a function are examples of solutions. The smallest and largest values ​​of a function on a segment. Algorithm for examining convexity and inflection 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 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.

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 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 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)= a function of general form (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 equal to

(10; 20) – minimum extremum.

7) examine the convexity and inflection point of the function graph

With this service you can find the largest and smallest value of a function one variable f(x) with the solution formatted in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables. You can also find the intervals of increasing and decreasing functions.

Rules for entering functions:

Necessary condition for the extremum of a function of one variable

The equation f" 0 (x *) = 0 is necessary condition extremum of a function of one variable, i.e. at point x * the first derivative of the function must vanish. It identifies stationary points x c at which the function does not increase or decrease.

Sufficient condition for the extremum of a function of one variable

Let f 0 (x) be twice differentiable with respect to x belonging to the set D. If at point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then point x * is the local (global) minimum point of the function.

If at point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

Then point x * is a local (global) maximum.

Example No. 1. Find the largest and smallest values ​​of the function: on the segment.
Solution.

The critical point is one x 1 = 2 (f’(x)=0). This point belongs to the segment. (The point x=0 is not critical, since 0∉).
We calculate the values ​​of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 at x=2; f max =9 at x=1

Example No. 2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let's find the critical points: 1-cos(x)=2, cos(x)=½, x=± π / 3 +2πk, k∈Z. We find y’’=2sin(x), calculate , which means x= π / 3 +2πk, k∈Z are the minimum points of the function; , which means x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example No. 3. Investigate the extremum function in the vicinity of the point x=0.
Solution. Here it is necessary to find the extrema of the function. If the extremum x=0, then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it can happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides the derivative changes sign. At these points it is necessary to use other methods to study functions at an extremum.

Example No. 4. Divide the number 49 into two terms whose product will be greatest.
Solution. Let us denote x as the first term. Then (49-x) is the second term.
The product will be maximum: x·(49-x) → max

Often we have to solve problems in which it is necessary to find the largest or smallest value from the set of values ​​that a function takes on a segment.

Let us turn, for example, to the graph of the function f(x) = 1 + 2x 2 – x 4 on the segment [-1; 2]. To work with a function, we need to build its graph.

From the plotted graph it is clear that the function takes the greatest value on this segment, equal to 2, at the points: x = -1 and x = 1; the function takes the smallest value equal to -7 at x = 2.

Point x = 0 is the minimum point of the function f(x) = 1 + 2x 2 – x 4. This means that there is a neighborhood of the point x = 0, for example, the interval (-1/2; 1/2) - such that in this neighborhood the function takes its smallest value at x = 0. However, on a larger interval, for example, on the segment [ -1; 2], the function takes its smallest value at the end of the segment, and not at the minimum point.

Thus, to find the smallest value of a function on a certain segment, it is necessary to compare its values ​​at the ends of the segment and at the minimum points.

In general, assume that the function f(x) is continuous on an interval and that the function has a derivative at each interior point of this interval.

To find the largest and smallest values ​​of a function on a segment, you need to:

1) find the values ​​of the function at the ends of the segment, i.e. numbers f(a) and f(b);

2) find the values ​​of the function at stationary points that belong to the interval (a; b);

3) select the largest and smallest values ​​from the found values.

We will apply the acquired knowledge in practice and consider the problem.

Find the largest and smallest values ​​of the function f(x) = x 3 + x/3 on the segment.

Solution.

1) f(1/2) = 6 1/8, f(2) = 9 ½.

2) f´(x) = 3x 2 – 3/x 2 = (3x 4 – 3)/x 2, 3x 4 – 3 = 0; x 1 = 1, x 2 = -1.

The interval (1/2; 2) contains one stationary point x 1 = 1, f(1) = 4.

3) Of the numbers 6 1/8, 9 ½ and 4, the largest is 9 ½, the smallest is 4.

Answer. The largest value of the function is 9 ½, the smallest value of the function is 4.

Often, when solving problems, it is necessary to find the largest and smallest values ​​of a function not on a segment, but on an interval.

In practical problems, the function f(x) usually has only one stationary point on a given interval: either a maximum point or a minimum point. In these cases, the function f(x) takes the greatest value on a given interval at the maximum point, and at the minimum point it takes on the smallest value on a given interval. Let's turn to the problem.

Write the number 36 as the product of two positive numbers whose sum is the smallest.

Solution.

1) Let the first factor be x, then the second factor is 36/x.

2) The sum of these numbers is x + 36/x.

3) According to the conditions of the problem, x is a positive number. So, the problem comes down to finding the value of x - such that the function f(x) = x + 36/x takes the smallest value on the interval x > 0.

4) Let’s find the derivative: f´(x) = 1 – 36/x 2 =((x + 6)(x – 6)) / x 2.

5) Stationary points x 1 = 6, x 2 = -6. On the interval x > 0 there is only one stationary point x = 6. When passing through the point x = 6, the derivative changes the “–” sign to the “+” sign, and therefore x = 6 is the minimum point. Consequently, the function f(x) = x + 36/x takes its smallest value on the interval x > 0 at the point x = 6 (this value is f(6) = 12).

Answer. 36 = 6 ∙ 6.

When solving some problems where you need to find the largest and smallest values ​​of a function, it is useful to use the following statement:

if the values ​​of the function f(x) on a certain interval are non-negative, then this function and the function (f(x)) n, where n is natural number, take the largest (smallest) value at the same point.

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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 article we will tell you how to calculate the largest and smallest value explicitly given function with one variable y=f(x) y = f (x) .

Basic definitions

Let'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

The smallest value of the 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.

On graph 6 the lowest value this function acquires at the right boundary of the interval (- 3; 2 ], and we cannot draw definite conclusions about the greatest 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.

1. First, let's find the domain of definition of the function. Let's check whether the segment specified in the condition is included in it.
2. Now let's calculate the points contained in this segment at which the first derivative does not exist. Most often they can be found in functions whose argument is written under the modulus sign, or in power functions, the exponent of which is a fractionally rational number.
3. Next, we will find out which stationary points will fall in the given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then select the appropriate roots. If we don’t get a single stationary point or they don’t fall into the given segment, then we move on to the next step.
4. We determine what values ​​the function will take at given stationary points (if any), or at those points at which the first derivative does not exist (if there are any), or we calculate the values ​​for x = a and x = b.
5. 5. We have a number of function values, from which we now need to select the largest and smallest. These will be the largest and smallest values ​​of the function that we need to find.

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, she will have a lot of everyone 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 studying 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.

1. First, you need to check whether the given interval will be a subset of the domain of the given function.
2. Let us determine all points that are contained in the required interval and at which the first derivative does not exist. They usually occur for functions where the argument is enclosed in the modulus sign, and for power functions with a fractionally rational exponent. If these points are missing, then you can proceed to the next step.
3. Now let’s determine which stationary points will fall within the given interval. First, we equate the derivative to 0, solve the equation and select suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.

• 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) .

1. At the end, you need to draw a conclusion based on the obtained function values ​​and limits. There are many options available here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest values ​​of the function. Below we will look at one typical example. Detailed Descriptions will help you understand what's what. If necessary, you can return to Figures 4 - 8 in the first part of the material.

Example 2

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 two previous calculations, we can state 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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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 ( x 2).

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, показательная функция определена для всех чисел, а не только для 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 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 coordinates

Most 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:

1. The branches of a parabola can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
2. The vertex of a parabola is the extremum point of a quadratic function at which this function takes its minimum (for a > 0) or maximum (a< 0) значение.

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:

1. There is no segment in the problem statement. Therefore, there is no need to calculate f(a) and f(b). It remains to consider only the extremum points;
2. But there is only one such point - this is the vertex of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.

Thus, solving the problem is greatly simplified and comes down to just two steps:

1. Write out the equation of the parabola y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a ;
2. Find the value of the original function at this point: f (x 0). If no additional conditions no, that will be the answer.

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 trial Unified State Exam in mathematics - this is where this technique is found 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, so the largest 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 function

Sometimes simply finding the vertex of the parabola is not enough to solve Problem B15. The desired value 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 segment ODZ - 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