With this service, you can find the largest and smallest value of a function one variable f(x) with the design of the solution in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables . You can also find the intervals of increase and decrease of the function.
Function entry rules:
A necessary condition for an extremum of a function of one variable
The equation f "0 (x *) \u003d 0 is a necessary condition for the extremum of a function of one variable, i.e. at the point x * the first derivative of the function must vanish. It selects stationary points x c at which the function does not increase and does not decrease .A sufficient condition for an extremum of a function of one variable
Let f 0 (x) be twice differentiable with respect to x belonging to the set D . If at the point x * the condition is met:F" 0 (x *) = 0
f"" 0 (x *) > 0
Then the point x * is the point of the local (global) minimum of the function.
If at the point x * the condition is met:
F" 0 (x *) = 0
f"" 0 (x *)< 0
That point x * is a local (global) maximum.
Example #1. Find the largest and smallest values of the function: on the segment .
Decision. 
The critical point is one x 1 = 2 (f'(x)=0). This point belongs to the segment . (The point x=0 is not critical, since 0∉).
We calculate the values of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 for x=2; f max =9 at x=1
Example #2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Decision.
Find the derivative of the function: y’=1-2cos(x) . Let us find the critical points: 1-cos(x)=2, cos(x)=1, x=± π / 3 +2πk, k∈Z. We find y''=2sin(x), calculate , so x= π / 3 +2πk, k∈Z are the minimum points of the function; 
, so x=- π / 3 +2πk, k∈Z are the maximum points of the function.
Example #3. Investigate the extremum function in the neighborhood of the point x=0.
Decision. Here it is necessary to find the extrema of the function. If the extremum x=0 , then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides, the derivative changes sign. At these points, one has to apply other methods to study functions to an extremum.
From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest value of a function. What is it connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life, one has to solve the problem of optimizing some parameters. And this is the problem of finding the largest and smallest values of the function.
It should be noted that the largest and smallest value of a function is usually sought on some interval X , which is either the entire domain of the function or part of the domain. The interval X itself can be a line segment, an open interval 
, an infinite interval .
In this article, we will talk about finding the largest and smallest values of an explicitly given function of one variable y=f(x) .
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The largest and smallest value of a function - definitions, illustrations.
Let us briefly dwell on the main definitions.
The largest value of the function 
, which for any 
the inequality is true.
The smallest value of the function y=f(x) on the interval X is called such a value 
, which for any the inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) value accepted in the interval under consideration with the abscissa.
Stationary points are the values of the argument at which the derivative of the function vanishes.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its maximum (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take on the largest and smallest values at points where the first derivative of this function does not exist, and the function itself is defined.
Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.
For clarity, we give a graphic illustration. Look at the pictures - and much will become clear.
On the segment
In the first figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points inside the segment [-6;6] .
Consider the case shown in the second figure. Change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.
In figure No. 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.
In the open range
In the fourth figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points within the open interval (-6;6) .
On the interval , no conclusions can be drawn about the largest value.
At infinity
In the example shown in the seventh figure, the function takes the largest value (max y ) at a stationary point with the abscissa x=1 , and the smallest value (min y ) is reached at the right boundary of the interval. At minus infinity, the values of the function asymptotically approach y=3 .
On the interval, the function does not reach either the smallest or the largest value. As x=2 tends to the right, the function values tend to minus infinity (the straight line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values asymptotically approach y=3 . A graphic illustration of this example is shown in Figure 8.
Algorithm for finding the largest and smallest values of a continuous function on the segment .
We write an algorithm that allows us to find the largest and smallest value of a function on a segment.
- We find the domain of the function and check if it contains the entire segment .
- We find all points at which the first derivative does not exist and which are contained in the segment (usually such points occur in functions with an argument under the module sign and in power functions with a fractional-rational exponent). If there are no such points, then go to the next point.
- We determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them fall into the segment, then go to the next step.
- We calculate the values of the function at the selected stationary points (if any), at points where the first derivative does not exist (if any), and also at x=a and x=b .
- From the obtained values of the function, we select the largest and smallest - they will be the desired maximum and smallest values of the function, respectively.
Let's analyze the algorithm when solving an example for finding the largest and smallest values of a function on a segment.
Example.
Find the largest and smallest value of a function
- on the segment;
- on the interval [-4;-1] .
Decision.
The domain of the function is the entire set of real numbers, except for zero, that is, . Both segments fall within the domain of definition.
We find the derivative of the function with respect to: 
Obviously, the derivative of the function exists at all points of the segments and [-4;-1] .
Stationary points are determined from the equation . The only real root is x=2 . This stationary point falls into the first segment.
For the first case, we calculate the values of the function at the ends of the segment and at a stationary point, that is, for x=1 , x=2 and x=4 : 
Therefore, the largest value of the function 
is reached at x=1 , and the smallest value 
– at x=2 .
For the second case, we calculate the values of the function only at the ends of the segment [-4;-1] (since it does not contain a single stationary point): 
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 those cases when it is necessary to determine the optimal value of a parameter. To solve such problems correctly, one must have a good understanding of what the largest and smallest value of a function are.
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Usually we define these values within some interval x , which in turn can correspond to the entire scope of the function or part of it. It can be either 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 describe how the largest and smallest value of an explicitly given function with one variable y=f(x) y = f (x) is calculated.
Basic definitions
We begin, as always, with the formulation of the main definitions.
Definition 1
The largest value of the function y = f (x) on some 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 0) .
Definition 2
The smallest value of the function y = f (x) on some 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(x0) .
These definitions are fairly obvious. It can be even simpler to say this: the largest value of a function is its largest value on a known interval at the abscissa x 0, and the smallest is the smallest accepted value on the same interval at x 0.
Definition 3
Stationary points are such values of the 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 a 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 exactly at one of the stationary points.
Another function can take on the largest or smallest value at those points at which the function itself is definite, and its first derivative does not exist.
The first question that arises when studying this topic is: in all cases, can we determine the maximum or minimum value of a function on a given interval? No, we cannot do this when the boundaries of the given interval will coincide with the boundaries of the domain of definition, or if we are dealing with an infinite interval. It also happens that a function in a given interval or at infinity will take on infinitely small or infinitely large values. In these cases, it is not possible to determine the largest and/or smallest value.
These moments will become more understandable after the image on the graphs:
The first figure shows us a function that takes on the largest and smallest values (m a x y and m i n y) at stationary points located on the interval [ - 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 get that the largest value of the function will be achieved at the point with the abscissa in the right boundary of the interval, and the smallest - 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 the 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 in 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 reached at a stationary point. We will not know the maximum value. The function could take the largest value at x equal to 6 if x = 6 belonged to the interval. It is this case that is shown in Figure 5.
On graph 6, this function acquires the smallest value in the right border of the interval (- 3 ; 2 ] , and we cannot draw definite conclusions about the largest value.
In figure 7, we see that the function will have m a x y at the stationary point, having an abscissa equal to 1 . The function reaches its minimum value at the interval boundary on the right side. At minus infinity, the values of the function will asymptotically approach y = 3 .
If we take an interval x ∈ 2 ; + ∞ , then we will see that the given function will not take on it either the smallest or the largest value. 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 values of the function will asymptotically approach y = 3. This is the case shown in Figure 8.
In this paragraph, we will give a sequence of actions that must be performed to find the largest or smallest value of a function on a certain interval.
- First, let's find the domain 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 find out which stationary points fall into a given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then choose the appropriate roots. If we do not get a single stationary point or they do not fall into a given segment, then we proceed to the next step.
- Let us determine what values the function will take at the given stationary points (if any), or at those points where the first derivative does not exist (if any), or we calculate the values for x = a and x = b .
- 5. We have a series of function values, from which we now need to choose the largest and smallest. This will be the largest and smallest values of the function that we need to find.
Let's see how to apply this algorithm correctly when solving problems.
Example 1
Condition: the function y = x 3 + 4 x 2 is given. Determine its largest and smallest value on the segments [ 1 ; 4 ] and [ - 4 ; - one ] .
Decision:
Let's start by finding the domain of this function. In this case, it will be the set of all real numbers except 0 . In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; +∞ . Both segments specified in the condition will be inside the definition area.
Now we calculate the derivative of the function according to the rule of differentiation of a fraction:
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 the function will exist at all points of the segments [ 1 ; 4 ] and [ - 4 ; - one ] .
Now we need to determine the stationary points of the function. Let's do this with 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 the given 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 have obtained 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 any stationary points, so we need to calculate the function values only at the ends of the given segment:
y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3
Hence, 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 learning 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, we perform the following steps in sequence.
- First you need to check whether the given interval will be a subset of the domain of the given function.
- Let us determine all the points that are contained in the required interval and at which the first derivative does not exist. Usually they occur in functions where the argument is enclosed in the sign of the module, and in power functions with a fractionally rational exponent. If these points are missing, then you can proceed to the next step.
- Now we determine which stationary points fall into a given interval. First, we equate the derivative to 0, solve the equation and find suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.
- If the interval looks like [ 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 looks like [ a ; + ∞) , then it is necessary to calculate the value at the point x = a and the limit to 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 to 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 values of the function and limits. There are many options here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest value of the function. Below we will consider 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 a 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 ; +∞) .
Decision
First of all, we find the domain of the function. The denominator of the fraction is a square 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 scope 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 on the entire domain of its 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 is 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) \u003d 3 e 1 (- 4) 2 + (- 4) - 6 - 4 \u003d 3 e 1 6 - 4 ≈ - 0. 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1
Since 3 e 1 6 - 4 > - 1 , then 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 to conclude that there is a limit below - 1 , since it is to this value that the function approaches asymptotically at minus infinity.
A feature of the second interval is that it does not have a single stationary point and not a single strict boundary. Therefore, we cannot calculate either the largest or the smallest value of the function. By defining the limit at minus infinity and as the argument tends to - 3 on the left side, we get only the range 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 maximum value of the function in the third interval, we determine its value at the stationary point x = - 1 2 if x = 1 . We also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:
y - 1 2 = 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 largest 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. All that we know , is the presence of a lower limit to - 4 .
For the interval (- 3 ; 2), let's take the results of the previous calculation and once again calculate what the one-sided limit is equal to when tending to 2 from 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
Hence, 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 bounded from below by the number - 4 .
Based on what we did in the two previous calculations, we can assert that on the interval [ 1 ; 2) the function will take the largest value at x = 1, and 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 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 talk about finding the largest and smallest value of a function. Those 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 on what intervals the function will decrease and on what intervals it will increase, after which further conclusions can be drawn. So you can more accurately determine the largest and smallest value of the function and justify the results.
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Often in physics and mathematics it is required to find the smallest value of a function. How to do this, we will now tell.
How to find the smallest value of a function: instruction
- To calculate the smallest value of a continuous function on a given interval, you need to follow this algorithm:
- Find the derivative of a function.
- Find on a given segment the points at which the derivative is equal to zero, as well as all the 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 of the values is the smallest.
- Find out what value the function has at the endpoints. Determine the smallest value of the function at these points.
- Compare the received data with the smallest value. The smaller of the received numbers will be the smallest value of the function.
Note that in the event that a function on a segment does not have the smallest points, this means that it increases or decreases on this segment. 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 step of the algorithm, you will need to solve a simple linear equation with one root. Solve the equation using the drawing 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 as follows. At the endpoints of the function value, compute the one-sided limit of the function. In other words, solve an equation in which the tending points are given by the value a+0 and b+0, where a and b are the names of the critical points.
Now you know how to find the smallest value of a function. The main thing is to do all the calculations correctly, accurately and without errors.
Let the function y=f(X) continuous on the segment [ a, b]. As is known, such a function reaches its maximum and minimum values on this interval. The function can take these values either at an interior point of the segment [ a, b], or on the boundary of the segment.
To find the largest and smallest values of a function on the interval [ a, b] necessary:
1) find the critical points of the function in the interval ( a, b);
2) calculate the values of the function at the found critical points;
3) calculate the values of the function at the ends of the segment, that is, for x=a and x = b;
4) from all the calculated values of the function, choose the largest and smallest.
Example. Find the largest and smallest values of a function
on the segment.
Finding critical points:
These points lie inside the segment ; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;
at the point x= 3 and at the point x= 0.
Investigation of a function for convexity and an inflection point.
Function y = f (x) called convexup in between (a, b) , if its graph lies under a tangent drawn at any point of this interval, and is called convex down (concave) if its graph lies above the tangent.
The point at the transition through which the convexity is replaced by concavity or vice versa is called inflection point.
Algorithm for studying for convexity and inflection point:
1. Find the critical points of the second kind, that is, the points at which the second derivative is equal to zero or does not exist.
2. Put critical points on the number line, breaking it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upwards, if, then the function is convex downwards.
3. If, when passing through a critical point of the second kind, it changes sign and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find its ordinate.
Asymptotes of the graph of a function. Investigation of a function into asymptotes.
Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point of the graph to this line tends to zero with an unlimited removal of the graph point from the origin.
There are three types of asymptotes: vertical, horizontal and inclined.
Definition. Direct called vertical asymptote function graph y = f(x), if at least one of the one-sided limits of the function at this point is equal to infinity, that is
where is the discontinuity point of the function, that is, it does not belong to the domain of definition.
Example.
D( y) = (‒ ∞; 2) (2; + ∞)
x= 2 - breaking point.
Definition. Straight y=A called horizontal asymptote function graph y = f(x) at , if
Example.
|
x | |||
|
y |
Definition. Straight y=kx +b (k≠ 0) is called oblique asymptote function graph y = f(x) at , where
General scheme for the study of functions and plotting.
Function research algorithmy = f(x) :
1. Find the domain of the function D (y).
2. Find (if possible) the points of intersection of the graph with the coordinate axes (with x= 0 and at y = 0).
3. Investigate for even and odd functions ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).
4. Find the asymptotes of the graph of the function.
5. Find intervals of monotonicity of the function.
6. Find the extrema of the function.
7. Find the intervals of convexity (concavity) and inflection points of the graph of the function.
8. On the basis of the conducted research, construct a graph of the function.
Example. Investigate the function and plot its graph.
1) D (y) =
x= 4 - breaking point.
2) When x = 0,
(0; – 5) – point of intersection with oy.
At y = 0,
3)
y(‒
x)=
general function (neither even nor odd).
4) We investigate for asymptotes.
a) vertical
b) horizontal
c) find oblique asymptotes where
‒oblique asymptote equation
5) In this equation, it is not required to find intervals of monotonicity of the function.
6)
These critical points partition the entire domain of the function on the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the obtained results in the form of the following table.
