Finding the smallest value of a function on a segment. How to find the largest and smallest values of a function in a bounded closed region? Largest and smallest value of a function
Often in physics and mathematics you need to find smallest value functions. We will now tell you how to do this.
How to find the smallest value of a function: instructions
Find on a given segment the points at which the derivative is equal to zero, as well as all critical points. Then find out the values of the function at these points, that is, solve the equation where x is equal to zero. Find out which value is the smallest.
Determine what value the function has on end points. Determine the smallest value of the function at these points.
Compare the obtained data with the lowest value. The smaller of the resulting numbers will be the smallest value of the function.
Note that if a function on a segment does not have smallest points, this means that in a given segment it increases or decreases. Therefore, the smallest value should be calculated on the finite segments of the function.
In all other cases, the value of the function is calculated according to the specified algorithm. At each point of the algorithm you will need to solve a simple linear equation with one root. Solve the equation using a picture to avoid mistakes.
How to find the smallest value of a function on a half-open segment? On a half-open or open period of the function, the smallest value should be found in the following way. At the end points of the function value, calculate the one-sided limit of the function. In other words, solve an equation in which the tending points are given by the values a+0 and b+0, where a and b are the names critical points.
Now you know how to find the smallest value of a function. The main thing is to do all calculations correctly, accurately and without errors.
In this article I will talk about how to apply the skill of finding to the study of a function: to find its largest or smallest value. And then we will solve several problems from Task B15 from Open Bank tasks for .
As usual, let's first remember the theory.
At the beginning of any study of a function, we find it
To find the largest or smallest value of a function, you need to examine on which intervals the function increases and on which it decreases.
To do this, we need to find the derivative of the function and examine its intervals of constant sign, that is, the intervals over which the derivative retains its sign.
Intervals over which the derivative of a function is positive are intervals of increasing function.
Intervals on which the derivative of a function is negative are intervals of decreasing function.
1 . Let's solve task B15 (No. 245184)
To solve it, we will follow the following algorithm:
a) Find the domain of definition of the function
b) Let's find the derivative of the function.
c) Let's equate it to zero.
d) Let us find the intervals of constant sign of the function.
e) Find the point at which the function takes highest value.
f) Find the value of the function at this point.
I explain the detailed solution to this task in the VIDEO TUTORIAL:
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2. Let's solve task B15 (No. 282862)
Find the largest value of the function on the segment
It is obvious that the function takes the greatest value on the segment at the maximum point, at x=2. Let's find the value of the function at this point:
Answer: 5
3. Let's solve task B15 (No. 245180):
Find the largest value of the function
1. title="ln5>0">, , т.к. title="5>1">, поэтому это число не влияет на знак неравенства.!}
2. Because according to the domain of definition of the original function title="4-2x-x^2>0">, следовательно знаменатель дроби всегда больще нуля и дробь меняет знак только в нуле числителя.!}
3. Numerator equal to zero at . Let's check if it belongs ODZ functions. To do this, let’s check whether the condition title="4-2x-x^2>0"> при .!}
Title="4-2(-1)-((-1))^2>0">,
this means that the point belongs to the ODZ function
Let's examine the sign of the derivative to the right and left of the point:
We see that the function takes on its greatest value at point . Now let's find the value of the function at:
Remark 1. Note that in this problem we did not find the domain of definition of the function: we only fixed the restrictions and checked whether the point at which the derivative is equal to zero belongs to the domain of definition of the function. This turned out to be sufficient for this task. However, this is not always the case. It depends on the task.
Note 2. When studying behavior complex function you can use this rule:
if the external function of a complex function is increasing, then the function takes its greatest value at the same point at which internal function takes the greatest value. This follows from the definition of an increasing function: a function increases on interval I if a larger value of the argument from this interval corresponds to a larger value of the function.
if the outer function of a complex function is decreasing, then the function takes on its largest value at the same point at which the inner function takes on its smallest value . This follows from the definition of a decreasing function: a function decreases on interval I if a larger value of the argument from this interval corresponds to a smaller value of the function
In our example, the external function increases throughout the entire domain of definition. Under the sign of the logarithm there is an expression - quadratic trinomial, which, with a negative leading coefficient, takes the greatest value at the point . Next, we substitute this x value into the function equation and find its greatest value.
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 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, the solution was not considered in detail quadratic equations, 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.
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 and solve the 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 the 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.
Problem statement 2:
Given a function that is defined and continuous on a certain interval. You need to find the largest (smallest) value of the function on this interval.
If a function is defined and continuous in a closed interval, then it reaches its maximum and minimum values in this interval.
The function can reach its largest and smallest values either by internal points gap or at its boundaries. Let's illustrate all the possible options.
Explanation: 1) The function reaches its greatest value on the left boundary of the interval at point , and its minimum value on the right boundary of the interval at point . 2) The function reaches its greatest value at the point (this is the maximum point), and its minimum value at the right boundary of the interval at the point. 3) The function reaches its maximum value on the left boundary of the interval at point , and its minimum value at point (this is the minimum point). 4) The function is constant on the interval, i.e. it reaches its minimum and maximum values at any point in the interval, and the minimum and maximum values are equal to each other. 5) The function reaches its maximum value at point , and its minimum value at point (despite the fact that the function has both a maximum and a minimum on this interval). 6) The function reaches its greatest value at a point (this is the maximum point), and its minimum value at a point (this is the minimum point). Comment:
“Maximum” and “maximum value” are different things. This follows from the definition of maximum and the intuitive understanding of the phrase “maximum value”.
Algorithm for solving problem 2.
4) Select the largest (smallest) from the obtained values and write down the answer.
Example 4:
Determine the largest and smallest value of a function on the segment. Solution: 1) Find the derivative of the function. 2) Find stationary points (and points suspected of extremum) by solving the equation. Pay attention to the points at which there is no two-sided finite derivative. 3) Calculate the values of the function at stationary points and at the boundaries of the interval.
4) Select the largest (smallest) from the obtained values and write down the answer.
The function on this segment reaches its greatest value at the point with coordinates .
The function on this segment reaches its minimum value at the point with coordinates .
You can verify the correctness of the calculations by looking at the graph of the function under study.
Comment: The function reaches its greatest value at the maximum point, and its minimum at the boundary of the segment.
A special case.
Suppose we need to find the maximum and minimum value some function on an interval. After completing the first point of the algorithm, i.e. derivative calculation, it becomes clear that, for example, it only takes negative values over the entire considered segment. Remember that if the derivative is negative, then the function decreases. We found that the function decreases over the entire segment. This situation is shown in graph No. 1 at the beginning of the article.
The function decreases on the segment, i.e. it has no extrema points. From the picture you can see that the function will take the smallest value on the right boundary of the segment, and the largest value on the left. if the derivative on the segment is positive everywhere, then the function increases. The smallest value is on the left border of the segment, the largest is on the right.
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 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 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 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 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, let’s find out which stationary points fall into 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, 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 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. He only has one real root, equal to 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:
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 will be a subset of the domain of the given 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 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 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 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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on the segment
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