Find the smallest value of the function f x x. Largest and Smallest Function Values

The concept of the largest and smallest values \u200b\u200bof a function.

The concept of highest and lowest values \u200b\u200bis closely related to the concept of the critical point of a function.

Definition 1

$ x_0 $ is called a critical point of the function $ f (x) $ if:

1) $ x_0 $ - inner point of the domain of definition;

2) $ f "\\ left (x_0 \\ right) \u003d 0 $ or does not exist.

Let us now introduce the definitions of the largest and smallest values \u200b\u200bof a function.

Definition 2

The function $ y \u003d f (x) $, defined on the interval $ X $, reaches its maximum value if there is a point $ x_0 \\ in X $ such that for all $ x \\ in X $ the inequality

Definition 3

The function $ y \u003d f (x) $, defined on the interval $ X $, reaches its smallest value if there is a point $ x_0 \\ in X $ such that for all $ x \\ in X $ the inequality

Weierstrass' theorem on a continuous function on an interval

To begin with, we introduce the concept of a continuous function on an interval:

Definition 4

A function $ f \\ left (x \\ right) $ is called continuous on the segment $$ if it is continuous at each point of the interval $ (a, b) $, and is also continuous on the right at the point $ x \u003d a $ and on the left at the point $ x \u003d b $.

Let us formulate a theorem on a continuous function on an interval.

Theorem 1

Weierstrass theorem

The function $ f \\ left (x \\ right) $ continuous on the segment $$ reaches its maximum and minimum value on this segment, that is, there are points $ \\ alpha, \\ beta \\ in $ such that for all $ x \\ in $ inequality $ f (\\ alpha) \\ le f (x) \\ le f (\\ beta) $.

The geometric interpretation of the theorem is shown in Figure 1.

Here the function $ f (x) $ reaches its smallest value at the point $ x \u003d \\ alpha $ and reaches its largest value at the point $ x \u003d \\ beta $.

Scheme for finding the largest and smallest values \u200b\u200bof the function $ f (x) $ on the segment $$

1) Find the derivative $ f "(x) $;

2) Find the points at which the derivative $ f "\\ left (x \\ right) \u003d 0 $;

3) Find the points at which the derivative $ f "(x) $ does not exist;

4) Select from the points obtained in points 2 and 3 those that belong to the segment $$;

5) Calculate the value of the function at the points obtained in step 4, as well as at the ends of the segment $$;

6) Select the highest and the lowest value from the obtained values.

Problems of finding the largest and smallest values \u200b\u200bof a function on a segment

Example 1

Find the largest and smallest value of a function on a segment: $ f (x) \u003d (2x) ^ 3-15x ^ 2 + 36x + 1 $

Decision.

1) $ f "\\ left (x \\ right) \u003d 6x ^ 2-30x + 36 $;

2) $ f "\\ left (x \\ right) \u003d 0 $;

\ \ \

4) $ 2 \\ in \\ left, \\ 3 \\ in $;

5) Values:

\ \ \ \

6) The largest found value is $ 33 $, the smallest value found is $ 1 $. Thus, we get:

Answer: $ max \u003d 33, \\ min \u003d 1 $.

Example 2

Find the largest and smallest function value on the segment: $ f \\ left (x \\ right) \u003d x ^ 3-3x ^ 2-45x + 225 $

Decision.

The solution will be carried out according to the above scheme.

1) $ f "\\ left (x \\ right) \u003d 3x ^ 2-6x-45 $;

2) $ f "\\ left (x \\ right) \u003d 0 $;

\ \ \

3) $ f "(x) $ exists at all points of the domain;

4) $ -3 \\ notin \\ left, \\ 5 \\ in $;

5) Values:

\ \ \

6) The largest found value is $ 225, the smallest value found is $ 50 $. Thus, we get:

Answer: $ max \u003d 225, \\ min \u003d $ 50.

Example 3

Find the largest and smallest function value on the segment [-2,2]: $ f \\ left (x \\ right) \u003d \\ frac (x ^ 2-6x + 9) (x-1) $

Decision.

The solution will be carried out according to the above scheme.

1) $ f "\\ left (x \\ right) \u003d \\ frac (\\ left (2x-6 \\ right) \\ left (x-1 \\ right) - (x ^ 2-6x + 9)) (((x- 1)) ^ 2) \u003d \\ frac (x ^ 2-2x-3) (((x-1)) ^ 2) $;

2) $ f "\\ left (x \\ right) \u003d 0 $;

\\ [\\ frac (x ^ 2-2x-3) (((x-1)) ^ 2) \u003d 0 \\] \\ \\

3) $ f "(x) $ does not exist at the point $ x \u003d 1 $

4) $ 3 \\ notin \\ left [-2,2 \\ right], \\ -1 \\ in \\ left [-2,2 \\ right], \\ 1 \\ in \\ left [-2,2 \\ right] $, but 1 does not belong to the domain of definition;

5) Values:

\ \ \

6) The largest found value is $ 1 $, the smallest value found is $ -8 \\ frac (1) (3) $. Thus, we get: \\ end (enumerate)

Answer: $ max \u003d 1, \\ min \u003d\u003d - 8 \\ frac (1) (3) $.

Let the function y \u003df (x) is continuous on the segment [ a, b]. As is known, such a function reaches its maximum and minimum values \u200b\u200bon this segment. The function can take these values \u200b\u200beither at the inner point of the segment [ a, b], or on the segment boundary.

To find the largest and smallest values \u200b\u200bof the function on the segment [ a, b] you need:

1) find the critical points of the function in the interval ( a, b);

2) calculate the values \u200b\u200bof the function at the found critical points;

3) calculate the values \u200b\u200bof the function at the ends of the segment, that is, for x= and and x \u003d b;

4) from all calculated values \u200b\u200bof the function, select the largest and the smallest.

Example. Find the largest and smallest function values

on the segment.

Find the critical points:

These points lie inside the line segment; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;

at the point x\u003d 3 and at the point x= 0.

Investigation of the function for convexity and inflection point.

Function y = f (x) called convex up in between (a, b) if its graph lies under the tangent drawn at any point of this interval, and is called convex down (concave)if its graph lies above the tangent.

The point, when passing through which the convexity is replaced by concavity or vice versa, is called inflection point.

Research algorithm for convexity and inflection point:

1. Find the critical points of the second kind, that is, the points at which the second derivative is zero or does not exist.

2. Draw critical points on the number line, dividing it into intervals. Find the sign of the second derivative at 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, changes sign and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find her ordinate.

Asymptotes of the graph of a function. Investigation of the 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 straight line tends to zero with an unlimited distance from the origin of the graph point.

There are three types of asymptotes: vertical, horizontal and inclined.

Definition. The straight line is called vertical asymptotefunction graphics y \u003d 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\u003d 2 - break point.

Definition.Straight y \u003dA called horizontal asymptote function graphics y \u003d f (x) at, if

Example.

x
y

Definition.Straight y \u003dkx +b (k≠ 0) is called oblique asymptote function graphics y \u003d f (x) at, where

General scheme for the study of functions and plotting.

Function research algorithmy \u003d 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 (at x \u003d 0 and for y = 0).

3. Investigate for evenness and oddness of the function ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ oddness).

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 points of inflection of the graph of the function.

8. Based on the research carried out, build a graph of the function.

Example.Examine the function and plot it.

1) D (y) =

x \u003d 4 - break point.

2) When x = 0,

(0; - 5) - intersection point with oy.

When y = 0,

3) y(‒ x)= general function (neither even nor odd).

4) Investigate for asymptotes.

a) vertical

b) horizontal

c) find oblique asymptotes where

‒ Oblique asymptote equation

5) In this equation, it is not required to find the intervals of monotonicity of the function.

6)

These critical points split the entire domain of the function on 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

The table shows that the point x \u003d ‒2 ‒ maximum point, at the point x \u003d 4 ‒ no extremum, x \u003d 10 ‒ minimum point.

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 extreme.

The minimum of this function is

(10; 20) - minimum extremum.

7) investigate for convexity and inflection point of the graph of the function

In practice, it is quite common to use a derivative to compute 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 those cases when it is necessary to determine the optimal value of any parameter. To solve such problems correctly, you need to understand well what the largest and smallest function values \u200b\u200bare.

Usually, we define these values \u200b\u200bwithin a certain interval x, which may in turn correspond to the entire domain of the function or its part. It can be like a segment [a; b] and an open interval (a; b), (a; b], [a; b), an infinite interval (a; b), (a; b], [a; b) or an infinite interval - ∞; a, (- ∞; a], [a; + ∞), (- ∞; + ∞).

In this article, we will describe how the largest and smallest value of an explicitly given function with one variable y \u003d f (x) y \u003d f (x) is calculated.

Basic definitions

Let's start, as always, by formulating the basic definitions.

Definition 1

The largest value of the function y \u003d f (x) on some interval x is the value maxy \u003d f (x 0) x ∈ X, which for any value xx ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x 0).

Definition 2

The smallest value of the function y \u003d f (x) on some interval x is the value minx ∈ X y \u003d 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 fairly obvious. It is even simpler to say this: the largest value of a function is its largest value in a known interval at x 0, and the smallest is the smallest accepted value in the same interval at x 0.

Definition 3

Stationary points are those values \u200b\u200bof the argument of a function at which its derivative vanishes.

Why do we need to know what stationary points are? To answer this question, one must recall Fermat's theorem. It follows from it that a stationary point is a 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 over a certain interval precisely at one of the stationary points.

Another function can take the largest or smallest value at those points at which the function itself is definite, and its first derivative does not exist.

The first question that arises when studying this topic: in all cases, we can 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 will coincide with the boundaries of the domain of definition, 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 highest and / or lowest value.

These moments will become clearer after being shown on the graphs:

The first figure shows us a function that takes the largest and smallest values \u200b\u200b(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 largest value of the function will be achieved at a point with an abscissa in the right boundary of the interval, and the smallest - at a stationary point.

In the third figure, the abscissas of the points represent the boundary points of the segment [- 3; 2]. They correspond to the highest and lowest values \u200b\u200bof the given function.

Now let's look at the fourth figure. 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. We will not know the highest value. The function could take its largest value at x equal to 6 if x \u003d 6 belonged to the interval. This particular case is depicted in graph 5.

On graph 6, this function acquires the smallest value in the right border 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 with an abscissa equal to 1. The function will reach its smallest value at the border of the interval on the right side. At minus infinity, the values \u200b\u200bof the function will asymptotically approach y \u003d 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 \u200b\u200bof the function will tend to minus infinity, since the straight line x \u003d 2 is the vertical asymptote. If the abscissa tends to plus infinity, then the values \u200b\u200bof the function will asymptotically approach y \u003d 3. It is this case that is depicted in Figure 8.

In this section, we present a sequence of actions that must be performed to find the largest or smallest value of a function on a certain segment.

1. First, let's find the domain of the function. Let us check whether the segment specified in the condition is included in it.
2. Now let's calculate the points contained in this segment, where 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, let's find out which stationary points fall into 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 choose the appropriate roots. If we do not get any stationary points or they do not fall into the given segment, then we proceed to the next step.
4. We determine what values \u200b\u200bthe function will take at the given stationary points (if any), or at those points where the first derivative does not exist (if any), or we calculate the values \u200b\u200bfor x \u003d a and x \u003d b.
5. 5. We have got a series of function values, from which we now need to select the largest and the smallest. These will be the largest and smallest values \u200b\u200bof 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 \u003d x 3 + 4 x 2 is given. Determine its largest and smallest value on the segments [1; 4] and [- 4; - 1 ] .

Decision:

Let's start by finding the domain of this 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 for differentiating the fraction:

y "\u003d x 3 + 4 x 2" \u003d x 3 + 4 "x 2 - x 3 + 4 x 2" x 4 \u003d \u003d 3 x 2 x 2 - (x 3 - 4) 2 xx 4 \u003d x 3 - 8 x 3

We learned that the derivative of the function will exist at all points of the segments [1; 4] and [- 4; - 1 ] .

Now we need to define the stationary points of the function. We do this using the equation x 3 - 8 x 3 \u003d 0. It only has one valid root, which is 2. It will be a stationary point of the function and fall into the first segment [1; 4 ] .

We calculate the values \u200b\u200bof the function at the ends of the first segment and at a given point, i.e. for x \u003d 1, x \u003d 2 and x \u003d 4:

y (1) \u003d 1 3 + 4 1 2 \u003d 5 y (2) \u003d 2 3 + 4 2 2 \u003d 3 y (4) \u003d 4 3 + 4 4 2 \u003d 4 1 4

We have obtained that the largest value of the function m a x y x ∈ [1; 4] \u003d y (2) \u003d 3 will be achieved at x \u003d 1, and the least m i n y x ∈ [1; 4] \u003d y (2) \u003d 3 - for x \u003d 2.

The second segment does not include any stationary points, so we need to calculate the values \u200b\u200bof the function only at the ends of the given segment:

y (- 1) \u003d (- 1) 3 + 4 (- 1) 2 \u003d 3

Hence, m a x y x ∈ [- 4; - 1] \u003d y (- 1) \u003d 3, m i n y x ∈ [- 4; - 1] \u003d y (- 4) \u003d - 3 3 4.

Answer:For the segment [1; 4] - m a x y x ∈ [1; 4] \u003d y (2) \u003d 3, m i n y x ∈ [1; 4] \u003d y (2) \u003d 3, for the segment [- 4; - 1] - m a x y x ∈ [- 4; - 1] \u003d y (- 1) \u003d 3, m i n y x ∈ [- 4; - 1] \u003d y (- 4) \u003d - 3 3 4.

See picture:

Before studying this method, we advise you to repeat 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 the smallest value of a function on an open or infinite interval, perform the following steps in sequence.

1. First, you need to check whether the specified interval will be a subset of the scope of this function.
2. Let us determine all the points that are contained in the required interval and in which the first derivative does not exist. Usually they are for functions where the argument is enclosed in the modulus sign, and for power functions with fractionally rational exponent. If these points are missing, then you can proceed to the next step.
3. Now we will determine which stationary points fall into the given interval. First, we equate the derivative to 0, solve the equation and find 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 [a; b), then we need to calculate the value of the function at the point x \u003d 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 \u003d 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 [a; + ∞), then it is necessary to calculate the value at the point x \u003d a and the limit at plus infinity lim x → + ∞ f (x).
• If the interval looks like (- ∞; b], calculate the value at the point x \u003d b and the limit at minus infinity lim x → - ∞ f (x).
• If - ∞; b, then we assume 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 at minus and plus infinity lim x → + ∞ f (x), lim x → - ∞ f (x).

1. Finally, you need to draw a conclusion based on the obtained function values \u200b\u200band limits. There are many possibilities 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 value of the function. Below we will analyze 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 a function y \u003d 3 e 1 x 2 + x - 6 - 4. Calculate its highest and lowest values \u200b\u200bin the intervals - ∞; - 4, - ∞; - 3, (- 3; 1], (- 3; 2), [1; 2), 2; + ∞, [4; + ∞).

Decision

The first step is to find the domain of the function. The denominator of the fraction contains a square trinomial, which should not vanish:

x 2 + x - 6 \u003d 0 D \u003d 1 2 - 4 1 (- 6) \u003d 25 x 1 \u003d - 1 - 5 2 \u003d - 3 x 2 \u003d - 1 + 5 2 \u003d 2 ⇒ D (y): x ∈ (- ∞; - 3) ∪ (- 3; 2) ∪ (2; + ∞)

We got the domain of the function, to which all the intervals specified in the condition belong.

Now let's differentiate the function and get:

y "\u003d 3 e 1 x 2 + x - 6 - 4" \u003d 3 e 1 x 2 + x - 6 "\u003d 3 e 1 x 2 + x - 6 1 x 2 + x - 6" \u003d \u003d 3 · E 1 x 2 + x - 6 · 1 "· x 2 + x - 6 - 1 · x 2 + x - 6" (x 2 + x - 6) 2 \u003d - 3 · (2 \u200b\u200bx + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2

Consequently, the derivatives of the function exist over the entire domain of its definition.

Let's move on to finding stationary points. The derivative of the function vanishes at x \u003d - 1 2. This is a stationary point located in the intervals (- 3; 1] and (- 3; 2).

We calculate the value of the function at x \u003d - 4 for the interval (- ∞; - 4], as well as the limit at minus infinity:

y (- 4) \u003d 3 e 1 (- 4) 2 + (- 4) - 6 - 4 \u003d 3 e 1 6 - 4 ≈ - 0. 456 lim x → - ∞ 3 e 1 x 2 + x - 6 \u003d 3 e 0 - 4 \u003d - 1

Since 3 e 1 6 - 4\u003e - 1, it means that maxyx ∈ (- ∞; - 4] \u003d y (- 4) \u003d 3 e 1 6 - 4. This does not allow us to unambiguously determine the smallest value of the function. We can only to conclude that there is a restriction - 1 at the bottom, since it is to this value that the function approaches asymptotically at minus infinity.

A feature of the second interval is that it does not contain a single stationary point and not a single strict boundary. Therefore, we cannot calculate either the largest or the smallest value of the function. Having determined the limit at minus infinity and as the argument tends to - 3 on the left side, we will get only the range of values:

lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 \u003d lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 \u003d 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 \u003d \u003d 3 e 1 (+ 0) - 4 \u003d 3 e + ∞ - 4 \u003d + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 \u003d 3 e 0 - 4 \u003d - 1

This means that the values \u200b\u200bof the function will be located in the interval - 1; + ∞

To find the largest value of the function in the third interval, we determine its value at the stationary point x \u003d - 1 2 if x \u003d 1. We also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:

y - 1 2 \u003d 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 \u003d 3 e 4 25 - 4 ≈ - 1. 444 y (1) \u003d 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1. 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 \u003d lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 \u003d 3 e 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 \u003d \u003d 3 e 1 (- 0) - 4 \u003d 3 e - ∞ - 4 \u003d 3 0 - 4 \u003d - 4

We have found that the function will take the greatest value at the stationary point maxyx ∈ (3; 1] \u003d y - 1 2 \u003d 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. Everything that we know , Is the presence of a restriction from below to - 4.

For the interval (- 3; 2), we take the results of the previous calculation and again calculate what the one-sided limit is equal to when tending to 2 on the left side:

y - 1 2 \u003d 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 \u003d 3 e - 4 25 - 4 ≈ - 1. 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 \u003d - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 \u003d lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 \u003d 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 \u003d \u003d 3 e 1 - 0 - 4 \u003d 3 e - ∞ - 4 \u003d 3 0 - 4 \u003d - 4

Hence, m a x y x ∈ (- 3; 2) \u003d y - 1 2 \u003d 3 e - 4 25 - 4, and the smallest value cannot be determined, and the values \u200b\u200bof the function are bounded 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 \u003d 1, and it is impossible to find the smallest one.

On the interval (2; + ∞), the function will reach neither the largest nor the smallest value, i.e. it will take values \u200b\u200bfrom the interval - 1; + ∞.

lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 \u003d lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 \u003d 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 \u003d \u003d 3 e 1 (+ 0) - 4 \u003d 3 e + ∞ - 4 \u003d + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 \u003d 3 e 0 - 4 \u003d - 1

Having calculated what the value of the function will be for x \u003d 4, we find out that m a x y x ∈ [4; + ∞) \u003d y (4) \u003d 3 e 1 14 - 4, and the given function at plus infinity will asymptotically approach the line y \u003d - 1.

Let's compare what we got in each calculation with the graph of the given function. In the figure, the asymptotes are shown with a dotted line.

That's all we wanted to tell you about finding the largest and smallest function value. 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 figure out at what intervals the function will decrease and at what intervals it will increase, after which you can draw further conclusions. Thus, you can more accurately determine the largest and smallest value of the function and justify the results.

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In task B14 from the exam in mathematics, you need to find the smallest or largest value of a function of one variable. This is a fairly trivial problem from mathematical analysis, and it is for this reason that every high school graduate can and should learn to solve it normally. Let us examine several examples that schoolchildren solved during the diagnostic work in mathematics, which took place in Moscow on December 7, 2011.

Depending on the interval at which you want to find the maximum or minimum value of the function, one of the following standard algorithms is used to solve this problem.

I. Algorithm for finding the largest or smallest value of a function on a segment:

• Find the derivative of the function.
• Choose from the points suspicious of an extremum, those that belong to the given segment and the domain of the function.
• Calculate values function (not derivative!) at these points.
• Choose the largest or the smallest among the obtained values, it will be the desired one.

Example 1. Find the smallest function value
y = x 3 – 18x 2 + 81x + 23 on the segment.

Decision:we act according to the algorithm for finding the smallest value of a function on a segment:

• The scope of the function is not limited: D (y) = R.
• The derivative of the function is: y ’ = 3x 2 – 36x + 81. The domain of definition of the derivative of the function is also not limited: D (y ') = R.
• Derivative zeros: y ’ = 3x 2 – 36x + 81 \u003d 0, so x 2 – 12x + 27 \u003d 0, whence x \u003d 3 and x \u003d 9, our interval includes only x \u003d 9 (one point suspicious of an extremum).
• Find the value of the function at a point suspicious of an extremum and at the edges of the interval. For the convenience of calculations, we represent the function as: y = x 3 – 18x 2 + 81x + 23 = x(x-9) 2 +23:
  • y(8) \u003d 8 (8-9) 2 +23 \u003d 31;
  • y(9) \u003d 9 (9-9) 2 +23 \u003d 23;
  • y(13) \u003d 13 (13-9) 2 +23 \u003d 231.

So, of the obtained values, the smallest is 23. Answer: 23.

II. Algorithm for finding the largest or smallest function value:

• Find the domain of the function.
• Find the derivative of the function.
• Determine points suspicious of an extremum (those points at which the derivative of the function vanishes and points at which there is no two-sided finite derivative).
• Mark these points and the domain of the function on the number line and determine the signs derivative (not functions!) on the resulting intervals.
• Define values function (not the derivative!) at the minimum points (those points at which the sign of the derivative changes from minus to plus), the smallest of these values \u200b\u200bwill be the smallest value of the function. If there are no minimum points, then the function does not have the smallest value.
• Define values function (not the derivative!) at the maximum points (those points at which the sign of the derivative changes from plus to minus), the largest of these values \u200b\u200bwill be the largest value of the function. If there are no maximum points, then the function has no maximum value.

Example 2. Find the largest value of the function.

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

Function entry rules:

A necessary condition for the extremum of a function of one variable

The equation f "0 (x *) \u003d 0 is a necessary condition for the extremum of a function of one variable, i.e. at the point x * the first derivative of the function must vanish. It selects 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 satisfied:

F "0 (x *) \u003d 0
f "" 0 (x *)\u003e 0

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

If at point x * the condition is satisfied:

F "0 (x *) \u003d 0
f "" 0 (x *)< 0

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

Example # 1. Find the largest and smallest values \u200b\u200bof the function: on the segment.
Decision.

One critical point x 1 \u003d 2 (f '(x) \u003d 0). This point belongs to the line segment. (The point x \u003d 0 is not critical, since 0∉).
We calculate the values \u200b\u200bof the function at the ends of the segment and at the critical point.
f (1) \u003d 9, f (2) \u003d 5/2, f (3) \u003d 3 8/81
Answer: f min \u003d 5/2 at x \u003d 2; f max \u003d 9 at x \u003d 1

Example # 2. Using the derivatives of higher orders, find the extremum of the function y \u003d x-2sin (x).
Decision.
Find the derivative of the function: y ’\u003d 1-2cos (x). Find the critical points: 1-cos (x) \u003d 2, cos (x) \u003d ½, x \u003d ± π / 3 + 2πk, k∈Z. We find y ’’ \u003d 2sin (x), calculate, so x \u003d π / 3 + 2πk, k∈Z are the minimum points of the function; , so x \u003d - π / 3 + 2πk, k∈Z are the maximum points of the function.

Example No. 3. Explore the extremum function in the vicinity of the point x \u003d 0.
Decision. Here it is necessary to find the extrema of the function. If the extremum is x \u003d 0, then find out its type (minimum or maximum). If there is no x \u003d 0 among the found points, then calculate the value of the function f (x \u003d 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 may 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, one has to apply other methods to study functions for extremum.

Example No. 4. Divide the number 49 into two terms, the product of which will be the largest.
Decision. Let us denote x as the first term. Then (49-x) is the second term.
The product will be the maximum: x (49-x) → max