Sometimes in problems B14 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 monotony. Definition 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

Definition. 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 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 smaller f(x).

Examples. The logarithm increases monotonically if the base a > 1, and monotonically decreases if 0 0. f (x) = log a x (a > 0; a 1; x > 0) 1, and monotonically decreases if 0 0. f (x) = log a x (a > 0; a 1; x > 0)"> 1, and monotonically decreases if 0 0. f (x) = log a x (a > 0; a 1; x > 0)"> 1, and decreases monotonically if 0 0. f (x) = log a x (a > 0; a 1; x > 0)" title="Examples . The logarithm increases monotonically if the base a > 1, and monotonically decreases if 0 0. f (x) = log a x (a > 0; a 1; x > 0)"> title="Examples. The logarithm increases monotonically if the base a > 1, and monotonically decreases if 0 0. f (x) = log a x (a > 0; a 1; x > 0)"> !}

Examples. The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0 0: 1 and decreases at 0 0:"> 1 and decreases at 0 0:"> 1 and decreases at 0 0:" title="Examples. The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0 0:"> title="Examples. The exponential function behaves similarly to the logarithm: it increases when a > 1 and decreases when 0 0:"> !}

0) or down (a 0) or down (a 9 Coordinates of the vertex of the parabola Most often, the argument of the function is replaced by a square trinomial of the form Its graph is a standard parabola, in which we are interested in the branches: The branches of the parabola can go up (for a > 0) or down (a 0) or the greatest (a 0) or down (a 0) or down (a 0) or greatest (a 0) or down (a 0) or down (a title="Coordinates of the vertex of a parabola Most often, the argument of the function is replaced by a quadratic trinomial of the form Its graph is a standard parabola, in which we are interested in the branches: The branches of a parabola can go up (for a > 0) or down (a

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; But there is only one such point - the vertex of the parabola x 0, the coordinates of which are calculated literally verbally and without any derivatives.

Thus, solving the problem is greatly simplified and comes down to just two steps: Write out the equation of the parabola and find its vertex using the formula: Find the value of the original function at this point: f (x 0). If no additional conditions no, that will be the answer.

0. Vertex of a parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3" title="Find smallest value functions: Solution: Under the root stands quadratic function The graph of this parabola function has upward branches, since the coefficient a = 1 > 0. Top of the parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3" class="link_thumb"> 18 Find the smallest value of the function: Solution: Under the root there is a quadratic function. 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 0. Top of the parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3"> 0. Top of the parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3"> 0. Vertex of the parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3" title="Find the smallest value of the function: Solution: Under the root there is a quadratic function. The graph of this function is a parabola with branches upward, since the coefficient a = 1 > 0. The vertex of the parabola: x 0 = b/(2a) = 6/(2 1) = 6/2 = 3"> title="Find the smallest value of the function: Solution: Under the root there is a quadratic function. 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"> !}

Find the smallest value of the function: Solution Under the logarithm, the quadratic function is again. The graph of the parabola has branches upward, because a = 1 > 0. Vertex of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1 0. Top of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1"> 0. Top of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1"> 0. Vertex of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1" title="Find the smallest value of the function: Solution Under the logarithm is again a quadratic function. The graph of the parabola has upward branches, since a = 1 > 0. Vertex of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1"> title="Find the smallest value of the function: Solution Under the logarithm, the quadratic function is again. The graph of the parabola has branches upward, because a = 1 > 0. Vertex of the parabola: x 0 = b/(2a) = 2/(2 1) = 2/2 = 1"> !}

Find the largest value of the function: Solution: The exponent contains a quadratic function Let's rewrite it in normal form: Obviously, the graph of this function is a parabola, branches down (a = 1

Corollaries from the domain of the function Sometimes to solve Problem B14 it is not enough to simply find the vertex of the parabola. The desired value may lie at the end of the segment, and not at all at the extremum point. If the problem does not specify a segment at all, we look at the range of permissible values ​​of the original function. Namely:

0 2. Arithmetic Square root exists only from non-negative numbers: 3. The denominator of the fraction must not be zero:" title="1. The argument of the logarithm must be positive: y = log a f (x) f (x) > 0 2. The arithmetic square root exists only from non-negative numbers: 3. The denominator of the fraction must not be equal to zero:" class="link_thumb"> 26 !} 1. The argument of the logarithm must be positive: y = log a f (x) f (x) > 0 2. The arithmetic square root exists only from non-negative numbers: 3. The denominator of the fraction must not be zero: 0 2. The arithmetic square root exists only from non-negative numbers: 3. The denominator of a fraction must not be equal to zero: "> 0 2. The arithmetic square root exists only from non-negative numbers: 3. The denominator of a fraction must not be equal to zero: "> 0 2. Arithmetic the square root exists only of non-negative numbers: 3. The denominator of the fraction must not be zero:" title="1. The argument of the logarithm must be positive: y = log a f (x) f (x) > 0 2. Arithmetic square the root exists only from non-negative numbers: 3. The denominator of the fraction must not be equal to zero:"> title="1. The argument of the logarithm must be positive: y = log a f (x) f (x) > 0 2. The arithmetic square root exists only from non-negative numbers: 3. The denominator of the fraction must not be zero:"> !}

Solution Under the root is again a quadratic function. Its graph is parabolic, but the branches are directed downwards, since a = 1
Now let’s find the vertex of the parabola: x 0 = b/(2a) = (2)/(2 · (1)) = 2/(2) = 1 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 number 2. Answer: 2

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 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 ) = = log 0.5 (18 9 5) = log 0.5 4 = 2 Answer: -2

What is an extremum of a function and what is the necessary condition for an extremum?

The extremum of a function is the maximum and minimum of the function.

The necessary condition for the maximum and minimum (extremum) of a function is the following: if the function f(x) has an extremum at the point x = a, then at this point the derivative is either zero, or infinite, or does not exist.

This condition is necessary, but not sufficient. The derivative at the point x = a can go to zero, infinity, or not exist without the function having an extremum at this point.

What is the sufficient condition for the extremum of a function (maximum or minimum)?

First condition:

If, in sufficient proximity to the point x = a, the derivative f?(x) is positive to the left of a and negative to the right of a, then at the point x = a the function f(x) has maximum

If, in sufficient proximity to the point x = a, the derivative f?(x) is negative to the left of a and positive to the right of a, then at the point x = a the function f(x) has minimum provided that the function f(x) here is continuous.

Instead, you can use the second sufficient condition for the extremum of a function:

Let at the point x = a the first derivative f?(x) vanish; if the second derivative f??(a) is negative, then the function f(x) has a maximum at the point x = a, if it is positive, then it has a minimum.

What is the critical point of a function and how to find it?

This is the value of the function argument at which the function has an extremum (i.e. maximum or minimum). To find it you need find the derivative function f?(x) and, equating it to zero, solve the equation f?(x) = 0. The roots of this equation, as well as those points at which the derivative of this function does not exist, are critical points, i.e., values ​​of the argument at which there can be an extremum. They can be easily identified by looking at derivative graph: we are interested in those values ​​of the argument at which the graph of the function intersects the abscissa axis (Ox axis) and those at which the graph suffers discontinuities.

For example, let's find extremum of a parabola.

Function y(x) = 3x2 + 2x - 50.

Derivative of the function: y?(x) = 6x + 2

Solve the equation: y?(x) = 0

6x + 2 = 0, 6x = -2, x = -2/6 = -1/3

IN in this case the critical point is x0=-1/3. It is with this argument value that the function has extremum. To him find, substitute the found number in the expression for the function instead of “x”:

y0 = 3*(-1/3)2 + 2*(-1/3) - 50 = 3*1/9 - 2/3 - 50 = 1/3 - 2/3 - 50 = -1/3 - 50 = -50.333.

How to determine the maximum and minimum of a function, i.e. its largest and smallest values?

If the sign of the derivative when passing through the critical point x0 changes from “plus” to “minus”, then x0 is maximum point; if the sign of the derivative changes from minus to plus, then x0 is minimum point; if the sign does not change, then at point x0 there is neither a maximum nor a minimum.

For the example considered:

We take an arbitrary value of the argument to the left of the critical point: x = -1

At x = -1, the value of the derivative will be y?(-1) = 6*(-1) + 2 = -6 + 2 = -4 (i.e. the sign is “minus”).

Now we take an arbitrary value of the argument to the right of the critical point: x = 1

At x = 1, the value of the derivative will be y(1) = 6*1 + 2 = 6 + 2 = 8 (i.e. the sign is “plus”).

As you can see, the derivative changed sign from minus to plus when passing through the critical point. This means that at the critical value x0 we have a minimum point.

Largest and smallest value of a function on the interval(on a segment) are found using the same procedure, only taking into account the fact that, perhaps, not all critical points will lie within the specified interval. Those critical points that are outside the interval must be excluded from consideration. If there is only one critical point inside the interval, it will have either a maximum or a minimum. In this case, to determine the largest and smallest values ​​of the function, we also take into account the values ​​of the function at the ends of the interval.

For example, let's find the largest and smallest values ​​of the function

y(x) = 3sin(x) - 0.5x

at intervals:

So, the derivative of the function is

y?(x) = 3cos(x) - 0.5

We solve the equation 3cos(x) - 0.5 = 0

cos(x) = 0.5/3 = 0.16667

x = ±arccos(0.16667) + 2πk.

We find critical points on the interval [-9; 9]:

x = arccos(0.16667) - 2π*2 = -11.163 (not included in the interval)

x = -arccos(0.16667) - 2π*1 = -7.687

x = arccos(0.16667) - 2π*1 = -4.88

x = -arccos(0.16667) + 2π*0 = -1.403

x = arccos(0.16667) + 2π*0 = 1.403

x = -arccos(0.16667) + 2π*1 = 4.88

x = arccos(0.16667) + 2π*1 = 7.687

x = -arccos(0.16667) + 2π*2 = 11.163 (not included in the interval)

We find the function values ​​at critical values ​​of the argument:

y(-7.687) = 3cos(-7.687) - 0.5 = 0.885

y(-4.88) = 3cos(-4.88) - 0.5 = 5.398

y(-1.403) = 3cos(-1.403) - 0.5 = -2.256

y(1.403) = 3cos(1.403) - 0.5 = 2.256

y(4.88) = 3cos(4.88) - 0.5 = -5.398

y(7.687) = 3cos(7.687) - 0.5 = -0.885

It can be seen that on the interval [-9; 9] the function has the greatest value at x = -4.88:

x = -4.88, y = 5.398,

and the smallest - at x = 4.88:

x = 4.88, y = -5.398.

On the interval [-6; -3] we have only one critical point: x = -4.88. The value of the function at x = -4.88 is equal to y = 5.398.

Find the value of the function at the ends of the interval:

y(-6) = 3cos(-6) - 0.5 = 3.838

y(-3) = 3cos(-3) - 0.5 = 1.077

On the interval [-6; -3] we have the greatest value of the function

y = 5.398 at x = -4.88

smallest value -

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the convex and concave sides?

To find all the inflection points of the line y = f(x), you need to find the second derivative, equate it to zero (solve the equation) and test all those values ​​of x for which the second derivative is zero, infinite or does not exist. If, when passing through one of these values, the second derivative changes sign, then the graph of the function has an inflection at this point. If it doesn’t change, then there is no bend.

The roots of the equation f? (x) = 0, as well as possible points of discontinuity of the function and the second derivative, divide the domain of definition of the function into a number of intervals. The convexity on each of their intervals is determined by the sign of the second derivative. If the second derivative at a point on the interval under study is positive, then the line y = f(x) is concave upward, and if negative, then downward.

How to find the extrema of a function of two variables?

To find the extrema of the function f(x,y), differentiable in the domain of its specification, you need:

1) find the critical points, and for this - solve the system of equations

fх? (x,y) = 0, fу? (x,y) = 0

2) for each critical point P0(a;b) investigate whether the sign of the difference remains unchanged

for all points (x;y) sufficiently close to P0. If the difference remains positive, then at point P0 we have a minimum, if negative, then we have a maximum. If the difference does not retain its sign, then there is no extremum at point P0.

The extrema of the function are determined similarly for more arguments.

Dear friends! The group of tasks related to the derivative includes tasks - the condition gives a graph of a function, several points on this graph and the question is:

At what point is the derivative greatest (smallest)?

Let's briefly repeat:

The derivative at a point is equal to the slope of the tangent passing throughthis point on the graph.

UThe global coefficient of the tangent, in turn, is equal to the tangent of the angle of inclination of this tangent.

*This refers to the angle between the tangent and the x-axis.

1. At intervals of increasing function, the derivative has a positive value.

2. At intervals of its decrease, the derivative has a negative value.

Consider the following sketch:

At points 1,2,4, the derivative of the function has a negative value, since these points belong to decreasing intervals.

At points 3,5,6, the derivative of the function has a positive value, since these points belong to increasing intervals.

As you can see, everything is clear with the meaning of the derivative, that is, it is not at all difficult to determine what sign it has (positive or negative) at a certain point in the graph.

Moreover, if we mentally construct tangents at these points, we will see that straight lines passing through points 3, 5 and 6 form angles with the oX axis ranging from 0 to 90 o, and straight lines passing through points 1, 2 and 4 form with the oX axis the angles range from 90 o to 180 o.

*The relationship is clear: tangents passing through points belonging to intervals of increasing functions form acute angles with the oX axis, tangents passing through points belonging to intervals of decreasing functions form obtuse angles with the oX axis.

Now the important question!

How does the value of the derivative change? After all, the tangent at different points on the graph of a continuous function forms different angles, depending on which point on the graph it passes through.

*Or, speaking in simple language, the tangent is located as if “horizontally” or “vertically”. Look:

Straight lines form angles with the oX axis ranging from 0 to 90 o

Straight lines form angles with the oX axis ranging from 90° to 180°

Therefore, if you have any questions:

- at which of the given points on the graph does the derivative have the smallest value?

- at which of the given points on the graph does the derivative have the greatest value?

then to answer it is necessary to understand how the value of the tangent of the tangent angle changes in the range from 0 to 180 o.

*As already mentioned, the value of the derivative of the function at a point is equal to the tangent of the angle of inclination of the tangent to the oX axis.

The tangent value changes as follows:

When the angle of inclination of the straight line changes from 0° to 90°, the value of the tangent, and therefore the derivative, changes accordingly from 0 to +∞;

When the angle of inclination of the straight line changes from 90° to 180°, the value of the tangent, and therefore the derivative, changes accordingly –∞ to 0.

This can be clearly seen from the graph of the tangent function:

In simple terms:

At a tangent inclination angle from 0° to 90°

The closer it is to 0 o, the greater the value of the derivative will be close to zero (on the positive side).

The closer the angle is to 90°, the more the derivative value will increase towards +∞.

With a tangent inclination angle from 90° to 180°

The closer it is to 90 o, the more the derivative value will decrease towards –∞.

The closer the angle is to 180°, the greater the value of the derivative will be close to zero (on the negative side).

317543. The figure shows the graph of the function y = f(x) and the points are marked–2, –1, 1, 2. At which of these points is the derivative greatest? Please indicate this point in your answer.

We have four points: two of them belong to the intervals on which the function decreases (these are points –1 and 1) and two to the intervals on which the function increases (these are points –2 and 2).

We can immediately conclude that at points –1 and 1 the derivative has a negative value, and at points –2 and 2 it has a positive value. Therefore, in this case, it is necessary to analyze points –2 and 2 and determine which of them will have the largest value. Let's construct tangents passing through the indicated points:

The value of the tangent of the angle between straight line a and the abscissa axis will be greater than the value of the tangent of the angle between straight line b and this axis. This means that the value of the derivative at point –2 will be greatest.

Let's answer the following question: at which point –2, –1, 1 or 2 is the value of the derivative most negative? Please indicate this point in your answer.

The derivative will have a negative value at points belonging to the decreasing intervals, so let’s consider points –2 and 1. Let’s construct tangents passing through them:

We see that the obtuse angle between straight line b and the oX axis is “closer” to 180 O , therefore its tangent will be greater than the tangent of the angle formed by the straight line a and the oX axis.

Thus, at the point x = 1, the value of the derivative will be greatest negative.

317544. The figure shows the graph of the function y = f(x) and the points are marked–2, –1, 1, 4. At which of these points is the derivative the smallest? Please indicate this point in your answer.

We have four points: two of them belong to the intervals at which the function decreases (these are points –1 and 4) and two to the intervals at which the function increases (these are points –2 and 1).

We can immediately conclude that at points –1 and 4 the derivative has a negative value, and at points –2 and 1 it has a positive value. Therefore, in this case, it is necessary to analyze points –1 and 4 and determine which of them will have the smallest value. Let's construct tangents passing through the indicated points:

The value of the tangent of the angle between straight line a and the abscissa axis will be greater than the value of the tangent of the angle between straight line b and this axis. This means that the value of the derivative at the point x = 4 will be the smallest.

Answer: 4

I hope I haven’t “overloaded” you with the amount of writing. In fact, everything is very simple, you just need to understand the properties of the derivative, its geometric meaning and how the value of the tangent of the angle changes from 0 to 180 o.

1. First, determine the signs of the derivative at these points (+ or -) and select the necessary points (depending on the question posed).

2. Construct tangents at these points.

3. Using the tangesoid graph, schematically mark the angles and displayAlexander.

P.S: I would be grateful if you tell me about the site on social networks.

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.

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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.

In graph 6, this function acquires its smallest value at the right boundary of the interval (- 3; 2 ], and we cannot draw definite conclusions about the largest value.

In Figure 7 we see that the function will have m a x y at a stationary point having an abscissa equal to 1. The function will reach its minimum value at the boundary of the interval c 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 whose exponent 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 is a subset of the domain of definition of this 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 in functions where the argument is enclosed in the modulus sign, and in 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.

Condition: given function y = 3 e 1 x 2 + x - 6 - 4 . Calculate its largest and smallest value in the intervals - ∞ ; - 4, - ∞; - 3 , (- 3 ; 1 ] , (- 3 ; 2) , [ 1 ; 2) , 2 ; + ∞ , [ 4 ; + ∞) .

Solution

First of all, we find the domain of definition of the function. The denominator of the fraction contains a quadratic trinomial, which should not turn to 0:

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

We have obtained the domain of definition of the function to which all the intervals specified in the condition belong.

Now let's differentiate the function and get:

y" = 3 e 1 x 2 + x - 6 - 4 " = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " = = 3 · e 1 x 2 + x - 6 · 1 " · x 2 + x - 6 - 1 · x 2 + x - 6 " (x 2 + x - 6) 2 = - 3 · (2 ​​x + 1) · e 1 x 2 + x - 6 x 2 + x - 6 2

Consequently, derivatives of a function exist throughout its entire domain of definition.

Let's move on to finding stationary points. The derivative of the function becomes 0 at x = - 1 2 . This is a stationary point that lies in the intervals (- 3 ; 1 ] and (- 3 ; 2) .

Let's calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ], as well as the limit at minus infinity:

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

Since 3 e 1 6 - 4 > - 1, it means that m a x y x ∈ (- ∞ ; - 4 ] = y (- 4) = 3 e 1 6 - 4. This does not allow us to uniquely determine the smallest value of the function. We can only conclude that there is a constraint below - 1, since it is to this value that the function approaches asymptotically at minus infinity.

The peculiarity of the second interval is that there is not a single stationary point and not a single strict boundary in it. Consequently, we will not be able to calculate either the largest or smallest value of the function. Having defined the limit at minus infinity and as the argument tends to - 3 on the left side, we get only an interval of values:

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

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

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

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

It turned out that the function will take the greatest value at a stationary point m a x y x ∈ (3; 1 ] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. Everything we know , is the presence of a lower limit to - 4 .

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

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

This means that m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4, and the smallest value cannot be determined, and the values ​​of the function are limited from below by the number - 4.

Based on what we got in the two previous calculations, we can say that on the interval [ 1 ; 2) the function will take the largest value at x = 1, but it is impossible to find the smallest.

On the interval (2 ; + ∞) the function will not reach either the largest or the smallest value, i.e. it will take values ​​from the interval - 1 ; + ∞ .

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

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

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

That's all we wanted to tell you about finding the largest and smallest values ​​of a function. The sequences of actions that we have given will help you make the necessary calculations as quickly and simply as possible. But remember that it is often useful to first find out at which intervals the function will decrease and at which it will increase, after which you can draw further conclusions. This way you can more accurately determine the largest and smallest values ​​of the function and justify the results obtained.

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A miniature and fairly simple problem of the kind that serves as a life preserver for a floating student. It's mid-July in nature, so it's time to settle down with your laptop on the beach. Early in the morning, the sunbeam of theory began to play, in order to soon focus on practice, which, despite the declared ease, contains shards of glass in the sand. In this regard, I recommend that you conscientiously consider the few examples of this page. To solve practical problems you must be able to find derivatives and understand the material of the article Monotonicity intervals and extrema of the function.

First, briefly about the main thing. In the lesson about continuity of function I gave the definition of continuity at a point and continuity at an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on an interval if:

1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.

In the second paragraph we talked about the so-called one-sided continuity functions at a point. There are several approaches to defining it, but I will stick to the line I started earlier:

The function is continuous at the point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: . It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at this point:

Imagine that the green dots are nails with a magic elastic band attached to them:

Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited– a fence at the top, a fence at the bottom, and our product grazes in the paddock. Thus, a function continuous on an interval is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and strictly proven. Weierstrass's first theorem....Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled a graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Really, how do you know what awaits us over the horizon? After all, the Earth was once considered flat, so today even ordinary teleportation requires proof =)

According to Weierstrass's second theorem, continuous on a segmentthe function reaches its exact upper bound and yours exact bottom edge .

The number is also called the maximum value of the function on the segment and are denoted by , and the number is the minimum value of the function on the segment marked .

In our case:

Note : in theory, recordings are common .

Roughly speaking, the largest value is where the highest point on the graph is, and the smallest value is where the lowest point is.

Important! As already emphasized in the article about extrema of the function, greatest function value And smallest function value – NOT THE SAME, What maximum function And minimum function. So, in the example under consideration, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even a flood, in the context of the problem under consideration, this does not interest us at all. The task only involves finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore no need to make a drawing!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the values ​​of the function in critical points, which belong to this segment.

Catch another bonus: here there is no need to check the sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum doesn't guarantee yet, what is the minimum or maximum value. The demonstration function reaches its maximum and, by the will of fate, the same number is highest value functions on the interval. But, of course, such a coincidence does not always take place.

So, in the first step, it is faster and easier to calculate the values ​​of the function at critical points belonging to the segment, without bothering whether there are extrema in them or not.

2) We calculate the values ​​of the function at the ends of the segment.

3) Among the function values ​​found in the 1st and 2nd paragraphs, select the smallest and largest number and write down the answer.

We sit down on the shore of the blue sea and hit the shallow water with our heels:

Example 1

Find the largest and smallest values ​​of a function on a segment

Solution:
1) Let us calculate the values ​​of the function at critical points belonging to this segment:

Let's calculate the value of the function at the second critical point:

2) Let’s calculate the values ​​of the function at the ends of the segment:

3) “Bold” results were obtained with exponents and logarithms, which significantly complicates their comparison. For this reason, let’s arm ourselves with a calculator or Excel and calculate approximate values, not forgetting that:

Now everything is clear.

Answer:

Fractional-rational example for independent solution:

Example 6

Find the maximum and minimum values ​​of a function on a segment