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 optimal value any 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 material we will tell you how to calculate the largest and smallest values of an explicitly defined function with one variable y=f(x) y = f (x) .

Basic 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 value 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 a function argument 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 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 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 largest value of the function will be achieved at the point with the abscissa on the right boundary of the interval, and the smallest at 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 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.

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.
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.
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.
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. 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 you study this method, we advise you to review how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and/or smallest value of a function on an open or infinite interval, perform the following steps sequentially.

First you need to check whether the given interval is a subset of the domain of definition of this function.
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 fractional rational indicator. If these points are missing, then you can proceed to the next step.
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 given interval, then we immediately go 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 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) .

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 quadratic trinomial, which should not go 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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Petite and pretty simple task from the category of those that serve 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 started playing sunny bunny theory in order to soon focus on practice, which, despite its claimed ease, contains glass shards in the sand. In this regard, I recommend that you conscientiously consider the few examples of this page. To solve practical tasks 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 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 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-sided limit equal to the value at this point:

Imagine that green dots- these are the nails on which the magic elastic band is attached:

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 segmentfunction reaches its accurate top edge 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 greatest value is where the most high point graphics, and the smallest 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 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 a maximum and, by the will of fate, the same number is the largest value of the function on the segment. But, of course, such a coincidence does not always occur.

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 most large number, write down the answer.

We sit down on the shore 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 instance for independent decision:

Example 6

Find the maximum and minimum value functions on an interval

The standard algorithm for solving such problems involves, after finding the zeros of the function, determining the signs of the derivative on the intervals. Then the calculation of values at the found maximum (or minimum) points and at the boundary of the interval, depending on what question is in the condition.

I advise you to do things a little differently. Why? I wrote about this.

I propose to solve such problems as follows:

1. Find the derivative.
2. Find the zeros of the derivative.
3. Determine which of them belong to this interval.
4. We calculate the values of the function at the boundaries of the interval and points of step 3.
5. We draw a conclusion (answer the question posed).

While solving the presented examples, solving quadratic equations is not discussed in detail; you must be able to do this. They should also know.

Let's look at examples:

77422. Find the largest value of the function y=x 3 –3x+4 on the segment [–2;0].

Let's find the zeros of the derivative:

The point x = –1 belongs to the interval specified in the condition.

We calculate the values of the function at points –2, –1 and 0:

The largest value of the function is 6.

Answer: 6

77425. Find the smallest value of the function y = x 3 – 3x 2 + 2 on the segment.

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

The point x = 2 belongs to the interval specified in the condition.

We calculate the values of the function at points 1, 2 and 4:

The smallest value of the function is –2.

Answer: –2

77426. Find the largest value of the function y = x 3 – 6x 2 on the segment [–3;3].

Let's find the derivative of the given function:

Let's find the zeros of the derivative:

The interval specified in the condition contains the point x = 0.

We calculate the values of the function at points –3, 0 and 3:

The smallest value of the function is 0.

Answer: 0

77429. Find the smallest value of the function y = x 3 – 2x 2 + x +3 on the segment.

Let's find the derivative of the given function:

3x 2 – 4x + 1 = 0

We get the roots: x 1 = 1 x 1 = 1/3.

The interval specified in the condition contains only x = 1.

Let's find the values of the function at points 1 and 4:

We found that the smallest value of the function is 3.

Answer: 3

77430. Find the largest value of the function y = x 3 + 2x 2 + x + 3 on the segment [– 4; –1].

Let's find the derivative of the given function:

Let's find the zeros of the derivative, solve quadratic equation:

3x 2 + 4x + 1 = 0

Let's get the roots:

The interval specified in the condition contains the root x = –1.

We find the values of the function at points –4, –1, –1/3 and 1:

We found that the largest value of the function is 3.

Answer: 3

77433. Find the smallest value of the function y = x 3 – x 2 – 40x +3 on the segment.

Let's find the derivative of the given function:

Let's find the zeros of the derivative and solve the quadratic equation:

3x 2 – 2x – 40 = 0

Let's get the roots:

The interval specified in the condition contains the root x = 4.

Find the function values at points 0 and 4:

We found that the smallest value of the function is –109.

Answer: –109

Let's consider a way to determine the largest and smallest values of functions without a derivative. This approach can be used if you have big problems. The principle is simple - we substitute all integer values from the interval into the function (the fact is that in all such prototypes the answer is an integer).

77437. Find the smallest value of the function y=7+12x–x 3 on the segment [–2;2].

Substitute points from –2 to 2: View solution

77434. Find the largest value of the function y=x 3 + 2x 2 – 4x + 4 on the segment [–2;0].

That's all. Good luck to you!

Sincerely, Alexander Krutitskikh.

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

Largest and smallest value of a function

The greatest value of a function is the greatest, the least value is the smallest of all its values.

A function can have only one largest and only one smallest value, or it may have none at all. Finding the largest and lowest values continuous functions based on the following properties these functions:

1) If in a certain interval (finite or infinite) the function y=f(x) is continuous and has only one extremum and if this is a maximum (minimum), then it will be the largest (smallest) value of the function in this interval.

2) If the function f(x) is continuous on a certain segment, then it necessarily has the largest and smallest values on this segment. These values are reached either at extremum points lying inside the segment, or at the boundaries of this segment.

To find the largest and smallest values on a segment, it is recommended to use the following scheme:

1. Find the derivative.

2. Find critical points of the function at which =0 or does not exist.

3. Find the values of the function at critical points and at the ends of the segment and select from them the largest f max and the smallest f max.

When deciding applied problems, in particular optimization, important have tasks of finding the largest and smallest values (global maximum and global minimum) of a function on the interval X. To solve such problems, one should, based on the condition, select an independent variable and express the value under study through this variable. Then find the desired largest or smallest value of the resulting function. In this case, the interval of change of the independent variable, which can be finite or infinite, is also determined from the conditions of the problem.

Example. Reservoir shaped like an open top rectangular parallelepiped with a square bottom, you need to tin the inside. What should be the dimensions of the tank if its capacity is 108 liters? water so that the cost of tinning it is minimal?

Solution. The cost of coating a tank with tin will be minimal if, for a given capacity, its surface area is minimal. Let us denote by a dm the side of the base, b dm the height of the tank. Then the area S of its surface is equal to

AND

The resulting relationship establishes the relationship between the surface area of the reservoir S (function) and the side of the base a (argument). Let us examine the function S for an extremum. Let's find the first derivative, equate it to zero and solve the resulting equation:

Hence a = 6. (a) > 0 for a > 6, (a)< 0 при а < 6. Следовательно, при а = 6 функция S имеет минимум. Если а = 6, то b = 3. Таким образом, затраты на лужение резервуара емкостью 108 литров будут наименьшими, если он имеет размеры 6дм х 6дм х 3дм.

Example. Find the largest and smallest values of a function on the interval.

Solution: Specified function continuous throughout number axis. Derivative of a function

Derivative for and for . Let's calculate the function values at these points:

The values of the function at the ends of the given interval are equal. Therefore, the largest value of the function is equal to at , the smallest value of the function is equal to at .

Self-test questions

1. Formulate L'Hopital's rule for revealing uncertainties of the form. List various types uncertainties for which L'Hopital's rule can be used.

2. Formulate the signs of increasing and decreasing functions.

3. Define the maximum and minimum of a function.

4. Formulate necessary condition existence of an extremum.

5. What values of the argument (which points) are called critical? How to find these points?

6. What are sufficient signs of the existence of an extremum of a function? Outline a scheme for studying a function at an extremum using the first derivative.

7. Outline a scheme for studying a function at an extremum using the second derivative.

8. Define convexity and concavity of a curve.

9. What is called the inflection point of the graph of a function? Indicate a method for finding these points.

10. Formulate the necessary and sufficient signs convexity and concavity of a curve on a given segment.

11. Define the asymptote of a curve. How to find vertical, horizontal and oblique asymptotes function graphics?

12. Outline general scheme researching a function and plotting its graph.

13. Formulate a rule for finding the largest and smallest values of a function on a given segment.

In this article we will analyze all types of problems to find

Let's remember geometric meaning derivative: if a tangent is drawn to the graph of a function at a point, then the slope coefficient of the tangent ( equal to tangent angle between the tangent and the positive direction of the axis) is equal to the derivative of the function at the point.

Let's take it on a tangent arbitrary point with coordinates:

And consider a right triangle:

In this triangle

From here

This is the equation of the tangent drawn to the graph of the function at the point.

To write the tangent equation, we only need to know the equation of the function and the point at which the tangent is drawn. Then we can find and .

There are three main types of tangent equation problems.

1. Given a point of contact

2. The tangent slope coefficient is given, that is, the value of the derivative of the function at the point.

3. Given are the coordinates of the point through which the tangent is drawn, but which is not the point of tangency.

Let's look at each type of task.

1. Write the equation of the tangent to the graph of the function at the point .

b) Find the value of the derivative at point . First let's find the derivative of the function

Let's substitute the found values into the tangent equation:

Let's open the brackets on the right side of the equation. We get:

Answer: .

2. Find the abscissa of the points at which the functions are tangent to the graph parallel to the x-axis.

If the tangent is parallel to the x-axis, therefore the angle between the tangent and the positive direction of the axis equal to zero, therefore the tangent of the tangent angle is zero. This means that the value of the derivative of the function at the points of contact is zero.

a) Find the derivative of the function .

b) Let's equate the derivative to zero and find the values in which the tangent is parallel to the axis:

Equating each factor to zero, we get:

Answer: 0;3;5

3. Write equations for tangents to the graph of a function , parallel direct .

A tangent is parallel to a line. The slope of this line is -1. Since the tangent is parallel to this line, therefore, the slope of the tangent is also -1. That is we know the slope of the tangent, and, thereby, derivative value at the point of tangency.

This is the second type of problem to find the tangent equation.

So, we are given the function and the value of the derivative at the point of tangency.

a) Find the points at which the derivative of the function is equal to -1.

First, let's find the derivative equation.

Let's equate the derivative to the number -1.

Let's find the value of the function at the point.

(by condition)

b) Find the equation of the tangent to the graph of the function at point .

Let's find the value of the function at the point.

(by condition).

Let's substitute these values into the tangent equation:

Answer:

4. Write the equation of the tangent to the curve , passing through a point

First, let's check whether the point is a tangent point. If a point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Let's substitute the coordinates of the point into the equation of the function.

Title="1sqrt(8-3^2)">. Мы получили под корнем !} negative number, the equality is not true, and the point does not belong to the graph of the function and is not a point of contact.

This is the last type of problem to find the tangent equation. First of all we need to find the abscissa of the tangent point.

Let's find the value.

Let be the point of contact. The point belongs to the tangent to the graph of the function. If we substitute the coordinates of this point into the tangent equation, we get the correct equality: