Peeling skin can be caused by a variety of factors. To deal with ...
The standard algorithm for solving such problems assumes, after finding the zeros of the function, the determination of the signs of the derivative on the intervals. Then the calculation of the values at the found points of the maximum (or minimum) and at the border of the interval, depending on what question is in the condition.
I advise you to do it a little differently. Why? I wrote about it.
I propose to solve such tasks as follows:
1. Find the derivative.
2. Find the zeros of the derivative.
3. Determine which of them belong to the given interval.
4. Calculate the values of the function at the boundaries of the interval and points of item 3.
5. We draw a conclusion (we answer the question posed).
In the course of solving the presented examples, the solution of quadratic equations was not considered in detail, you should be able to do this. They should also know.
Let's look at some examples:
77422. Find the largest value of the function y = x 3 –3х + 4 on the segment [–2; 0].
Find the zeros of the derivative:
The point x = –1 belongs to the interval indicated 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:
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:
Find the zeros of the derivative:
The point x = 0 belongs to the interval specified in the condition.
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.
Only x = 1 belongs to the interval specified in the condition.
Let's find the values of the function at points 1 and 4:
We got 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:
Find the zeros of the derivative, solve the quadratic equation:
3x 2 + 4x + 1 = 0
We get the roots:
The interval indicated in the condition belongs to the root x = –1.
Find the values of the function at points –4, –1, –1/3 and 1:
We got 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:
Find the zeros of the derivative, solve the quadratic equation:
3x 2 - 2x - 40 = 0
We get the roots:
The interval indicated in the condition belongs to the root x = 4.
We find the values of the function at points 0 and 4:
We got that the smallest value of the function is –109.
Answer: –109
Consider a method for determining the largest and smallest values of functions without a derivative. This approach can be used if you have big problems with the definition of the derivative. 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. Success to you!
Best regards, Alexander Krutitskikh.
P.S: I would be grateful if you could tell us about the site on social networks.
Let's see how to explore a function using a graph. It turns out, looking at the graph, you can find out everything that interests us, namely:
- function domain
- function range
- function zeros
- intervals of increasing and decreasing
- maximum and minimum points
- the largest and smallest value of the function on the segment.
Let's clarify the terminology:
Abscissa is the horizontal coordinate of the point.
Ordinate is the vertical coordinate.
Abscissa axis- a horizontal axis, most often called an axis.
Y-axis- vertical axis, or axis.
Argument is the independent variable on which the values of the function depend. Most often indicated.
In other words, we ourselves choose, substitute functions into the formula and get.
Domain functions - the set of those (and only those) values of the argument for which the function exists.
It is indicated by: or.
In our figure, the domain of the function is a segment. It is on this segment that the graph of the function is drawn. Only here this function exists.
Function range is the set of values that a variable takes. In our picture, this is a segment - from the lowest to the highest value.
Function zeros- points where the value of the function is equal to zero, that is. In our figure, these are points and.
Function values are positive where . In our figure, these are gaps and.
Function values are negative where . We have this interval (or interval) from to.
The most important concepts are increasing and decreasing function on some set. As a set, you can take a segment, an interval, a union of intervals, or the entire number line.
Function is increasing
In other words, the more, the more, that is, the chart goes to the right and up.
Function decreases on the set, if for any and belonging to the set, the inequality follows from the inequality.
For a decreasing function, a larger value corresponds to a smaller value. The graph goes to the right and down.
In our figure, the function increases in the interval and decreases in the intervals and.
Let's define what is maximum and minimum points of the function.
Maximum point- this is an internal point of the domain of definition, such that the value of the function in it is greater than at all points sufficiently close to it.
In other words, the maximum point is such a point, the value of the function at which more than in the neighboring ones. This is a local "mound" on the chart.
In our figure - the maximum point.
Minimum point- an internal point of the domain of definition, such that the value of the function in it is less than at all points sufficiently close to it.
That is, the minimum point is such that the value of the function in it is less than in the neighboring ones. This is a local “hole” on the chart.
In our picture - the minimum point.
The point is the boundary. It is not an internal point of the domain of definition and therefore does not fit the definition of a maximum point. After all, she has no neighbors on the left. In the same way, it cannot be a minimum point on our chart.
The maximum and minimum points are collectively called extremum points of the function... In our case, this is and.
And what to do if you need to find, for example, minimum function on the segment? In this case, the answer is. because minimum function is its value at the minimum point.
Likewise, the maximum of our function is. It is reached at a point.
We can say that the extrema of the function are equal to and.
Sometimes in tasks you need to find largest and smallest function values on a given segment. They do not necessarily coincide with extremes.
In our case smallest function value on the segment is equal to and coincides with the minimum of the function. But its greatest value on this segment is equal to. It is reached at the left end of the line.
In any case, the largest and smallest values of a continuous function on a segment are achieved either at the extremum points or at the ends of the segment.
In practice, it is quite common to use a 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 any parameter. To solve such problems correctly, you need to understand well what the largest and smallest value of a function is.
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We usually define these values within a certain interval x, which may in turn correspond to the entire domain of the function or its part. It can be like a segment [a; b] and 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
Let's start, as always, with the formulation of basic definitions.
Definition 1
The largest value of the function y = f (x) on some interval x is the value maxy = f (x 0) x ∈ X, which for any value xx ∈ X, x ≠ x 0 makes the inequality f (x) ≤ f (x 0).
Definition 2
The smallest value of the function y = f (x) on some interval x is the value minx ∈ 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 fairly obvious. It is even simpler to say this: the largest value of a function is its largest value in a known interval at x 0, and the smallest is the smallest accepted value in the same interval at x 0.
Definition 3
Stationary points are those values of the argument of a function at which its derivative vanishes.
Why do we need to know what stationary points are? To answer this question, one must recall Fermat's theorem. It follows from it that a stationary point is a point at which the extremum of the differentiable function is located (i.e., its local minimum or maximum). Consequently, the function will take the smallest or largest value over a certain interval exactly at one of the stationary points.
A function can also take the largest or smallest value at those points at which the function itself is definite, and its first derivative does not exist.
The first question that arises when studying this topic: in all cases, can we determine the largest or smallest value of a function on a given segment? No, we cannot do this when the boundaries of a given interval coincide with the boundaries of the domain of definition, or if we are dealing with an infinite interval. It also happens that a function in a given segment or at infinity will take infinitely small or infinitely large values. In these cases, it is not possible to determine the highest and / or lowest value.
These points will become clearer after being shown 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 obtain that the largest value of the function will be achieved at a point with an abscissa in the right boundary of the interval, and the smallest - at a stationary point.
In the third figure, the abscissas of the points represent the boundary points of the segment [- 3; 2]. They correspond to the highest and lowest values of the given function.
Now let's look at the fourth figure. In it, the function takes m a x y (the largest value) and m i n y (the smallest value) at stationary points on the open interval (- 6; 6).
If we take the interval [1; 6), then we can say that the smallest value of the function on it will be achieved at a stationary point. The greatest value will be unknown to us. The function could take its largest value at x equal to 6 if x = 6 belonged to the interval. It is this case that is depicted in graph 5.
On graph 6, this function acquires the smallest value at the right border of the interval (- 3; 2], and we cannot draw definite conclusions about the largest value.
In Figure 7, we see that the function will have m a x y at a stationary point with an abscissa equal to 1. The function will reach its smallest value at the border of the interval on the right side. At minus infinity, the values of the function 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 the vertical asymptote. If the abscissa tends to plus infinity, then the values of the function will asymptotically approach y = 3. It is this case that is depicted in Figure 8.
In this section, we will give a sequence of actions that must be performed to find the largest or smallest value of a function on a certain segment.
- First, let's find the domain of the function. Let us check whether the segment specified in the condition is included in it.
- Now let's calculate the points contained in this segment, where the first derivative does not exist. Most often they can be found in functions, the argument of which 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 the given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then choose the appropriate roots. If we do not get any stationary points or they do not fall into the given segment, then we proceed to the next step.
- We 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 got a series of function values, from which we now need to select the largest and the smallest. These will be the largest and smallest values 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 value on the segments [1; 4] and [- 4; - 1 ] .
Solution:
Let's start by finding the domain of this function. In this case, it will be the set of all real numbers except 0. In other words, D (y): x ∈ (- ∞; 0) ∪ 0; + ∞. Both segments specified in the condition will be inside the definition area.
Now we calculate the derivative of the function according to the rule for differentiating the fraction:
y "= 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 xx 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; - 1 ] .
Now we need to define the stationary points of the function. We do this using the equation x 3 - 8 x 3 = 0. It only has one valid root, which is 2. It will be a stationary point of the function and will fall into the first segment [1; 4 ] .
We calculate the values of the function at the ends of the first segment and at a 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 - for x = 2.
The second segment does not include any stationary points, so we need to calculate the values of the function 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 studying this method, we advise you to repeat how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and / or the smallest value of a function on an open or infinite interval, perform the following steps in sequence.
- First, you need to check whether the specified interval will be a subset of the scope of this function.
- Let us determine all the points that are contained in the required interval and in which the first derivative does not exist. Usually they are in functions where the argument is enclosed in the modulus sign, and in power functions with fractionally rational exponent. If these points are missing, then you can proceed to the next step.
- Now we will determine which stationary points fall into the given interval. First, we equate the derivative to 0, solve the equation, and find suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.
- If the interval is of the form [a; b), then we need to calculate the value of the function at the point x = a and the one-sided limit lim x → b - 0 f (x).
- If the interval has the form (a; b], then we need to calculate the value of the function at the point x = b and the one-sided limit lim x → a + 0 f (x).
- If the interval has the form (a; b), then we need to calculate the one-sided limits lim x → b - 0 f (x), lim x → a + 0 f (x).
- If the interval is of the form [a; + ∞), then it is necessary 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], calculate the value at the point x = b and the limit at minus infinity lim x → - ∞ f (x).
- If - ∞; b, then we assume the one-sided limit lim x → b - 0 f (x) and the limit at minus infinity lim x → - ∞ f (x)
- If - ∞; + ∞, then we consider the limits at minus and plus infinity lim x → + ∞ f (x), lim x → - ∞ f (x).
- In the end, you need to draw a conclusion based on the obtained function values and limits. There are many possibilities here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest value of the function. Below we will analyze one typical example. Detailed descriptions will help you understand what's what. If necessary, you can return to Figures 4 - 8 in the first part of the material.
Condition: given the function y = 3 e 1 x 2 + x - 6 - 4. Calculate its highest and lowest values in the intervals - ∞; - 4, - ∞; - 3, (- 3; 1], (- 3; 2), [1; 2), 2; + ∞, [4; + ∞).
Solution
The first step is to find the domain of the function. The denominator of the fraction contains a square trinomial, which should not vanish:
x 2 + x - 6 = 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 got the domain of the function, to which all the intervals specified in the condition belong.
Now let's differentiate the function and get:
y "= 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, the derivatives of the function exist over the entire domain of its definition.
Let's move on to finding stationary points. The derivative of the function vanishes at x = - 1 2. This is a stationary point located in the intervals (- 3; 1] and (- 3; 2).
We calculate the value of the function at x = - 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 maxyx ∈ (- ∞; - 4] = y (- 4) = 3 e 1 6 - 4. This does not allow us to unambiguously determine the smallest value of the function. We can only to conclude that there is a limitation - 1 at the bottom, since it is to this value that the function approaches asymptotically at minus infinity.
The peculiarity of the second interval is that it does not contain a single stationary point and not a single strict boundary. Therefore, we cannot calculate either the largest or the smallest value of the function. Having determined the limit at minus infinity and as the argument tends to - 3 from the left side, we will 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 values of the function will be located in the interval - 1; + ∞
To find the largest value of the function in the third interval, we determine its value at the stationary point x = - 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
We have found that the function will take the greatest value at the stationary point maxyx ∈ (3; 1] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. , Is the presence of a restriction from below to - 4.
For the interval (- 3; 2), we 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
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 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, and it is impossible to find the smallest one.
On the interval (2; + ∞), the function will reach neither the largest nor the smallest value, i.e. it will take values 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 for 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 with a dotted line.
That's all we wanted to tell you about finding the largest and smallest function value. The sequences of actions that we have given will help you make the necessary calculations as quickly and easily as possible. But remember that it is often useful to first find out at what intervals the function will decrease and at what intervals it will increase, after which you can draw further conclusions. This way you can more accurately determine the largest and smallest value of the function and justify the results.
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In this article I will talk about algorithm for finding the highest and lowest value functions, minimum and maximum points.
From the theory, it will definitely come in handy derivatives table and differentiation rules... All this is in this plate:
Algorithm for finding the highest and lowest value.
It is more convenient for me to explain with a specific example. Consider:
Example: Find the largest value of the function y = x ^ 5 + 20x ^ 3–65x on the segment [–4; 0].
Step 1. We take the derivative.
Y "= (x ^ 5 + 20x ^ 3–65x)" = 5x ^ 4 + 20 * 3x ^ 2 - 65 = 5x ^ 4 + 60x ^ 2 - 65
Step 2. Finding extremum points.
Extremum point we call such points at which a function reaches its highest or lowest value.
To find the extremum points, you need to equate the derivative of the function to zero (y "= 0)
5x ^ 4 + 60x ^ 2 - 65 = 0
Now we solve this biquadratic equation and the found roots are our extremum points.
I solve such equations by replacing t = x ^ 2, then 5t ^ 2 + 60t - 65 = 0.
Reducing the equation by 5, we get: t ^ 2 + 12t - 13 = 0
D = 12 ^ 2 - 4 * 1 * (- 13) = 196
T_ (1) = (-12 + sqrt (196)) / 2 = (-12 + 14) / 2 = 1
T_ (2) = (-12 - sqrt (196)) / 2 = (-12 - 14) / 2 = -13
We make the reverse change x ^ 2 = t:
X_ (1 and 2) = ± sqrt (1) = ± 1
x_ (3 and 4) = ± sqrt (-13) (exclude, there can be no negative numbers under the root, unless of course we are talking about complex numbers)
Total: x_ (1) = 1 and x_ (2) = -1 - these are our extremum points.
Step 3. Determine the highest and lowest value.
Substitution method.
In the condition we were given the segment [b] [- 4; 0]. The point x = 1 is not included in this segment. So we are not considering it. But besides the point x = -1, we also need to consider the left and right boundaries of our segment, that is, points -4 and 0. To do this, we substitute all these three points into the original function. Notice the original one - this is the one given in the condition (y = x ^ 5 + 20x ^ 3-65x), some begin to substitute in the derivative ...
Y (-1) = (-1) ^ 5 + 20 * (- 1) ^ 3 - 65 * (- 1) = -1 - 20 + 65 = [b] 44
y (0) = (0) ^ 5 + 20 * (0) ^ 3 - 65 * (0) = 0
y (-4) = (-4) ^ 5 + 20 * (- 4) ^ 3 - 65 * (- 4) = -1024 - 1280 + 260 = -2044
This means that the maximum value of the function is [b] 44 and it is reached at the point [b] -1, which is called the maximum point of the function on the segment [-4; 0].
We decided and received an answer, we are great, you can relax. But stop! Don't you think it is somehow too difficult to count y (-4)? In a limited time environment, it is better to use another method, I call it like this:
Through intervals of constancy.
These intervals are found for the derivative of the function, that is, for our biquadratic equation.
I do it in the following way. I draw a directional line. I place the points: -4, -1, 0, 1. Despite the fact that 1 is not included in the given segment, it should still be marked in order to correctly determine the intervals of constancy. Let's take some number many times greater than 1, let's say 100, mentally substitute it into our biquadratic equation 5 (100) ^ 4 + 60 (100) ^ 2 - 65. Even without counting anything, it becomes obvious that at point 100 the function has plus sign. This means that it has a plus sign for the intervals from 1 to 100. When passing through 1 (we go from right to left), the function will change its sign to minus. When passing through point 0, the function will retain its sign, since this is only the boundary of the segment, and not the root of the equation. When passing through -1, the function will again change its sign to plus.
From theory, we know that where the derivative of the function is (and we just drew it for it) changes sign from plus to minus (point -1 in our case) function reaches its local maximum (y (-1) = 44 as calculated earlier) on this interval (this is logically very clear, the function stopped increasing, since it reached its maximum and began to decrease).
Accordingly, where the derivative of the function changes sign from minus to plus, achieved local minimum of the function... Yes, yes, we also found the point of the local minimum, this is 1, and y (1) is the minimum value of the function on the segment, let's say from -1 to + ∞. Pay great attention that this is only a LOCAL MINIMUM, that is, a minimum on a certain segment. Since the real (global) minimum the function will reach somewhere there, at -∞.
In my opinion, the first method is simpler theoretically, and the second is simpler from the point of view of arithmetic operations, but much more complicated from the point of view of theory. Indeed, sometimes there are cases when the function does not change sign when passing through the root of the equation, and in general you can get confused with these local, global maxima and minima, although you will have to master it well if you plan to enter a technical university (and for why else take the profile exam and solve this task). But practice and only practice will once and for all teach you how to solve such problems. And you can train on our website. Here .
If you have any questions, or something is not clear, be sure to ask. I will gladly answer you and make changes, additions to the article. Remember we are making this site together!
Let the function $ z = f (x, y) $ be defined and continuous in some bounded closed domain $ D $. Let the given function in this region have finite partial derivatives of the first order (except, perhaps, a finite number of points). To find the largest and smallest values of a function of two variables in a given closed area, three steps of a simple algorithm are required.
Algorithm for finding the largest and smallest values of the function $ z = f (x, y) $ in the closed domain $ D $.
- Find the critical points of the function $ z = f (x, y) $ that belong to the domain $ D $. Calculate the values of the function at critical points.
- Investigate the behavior of the function $ z = f (x, y) $ on the boundary of the domain $ D $, finding the points of possible maximum and minimum values. Calculate the values of the function at the obtained points.
- From the values of the function obtained in the previous two paragraphs, choose the largest and the smallest.
What are tipping points? show \ hide
Under critical points means points where both first-order partial derivatives are zero (i.e. $ \ frac (\ partial z) (\ partial x) = 0 $ and $ \ frac (\ partial z) (\ partial y) = 0 $) or at least one partial derivative does not exist.
Often the points at which the partial derivatives of the first order are equal to zero are called stationary points... Thus, stationary points are a subset of critical points.
Example # 1
Find the largest and smallest values of the function $ z = x ^ 2 + 2xy-y ^ 2-4x $ in a closed region bounded by the lines $ x = 3 $, $ y = 0 $ and $ y = x + 1 $.
We will follow the above, but first we will deal with the drawing of the given area, which we will denote by the letter $ D $. We are given the equations of three straight lines, which limit this area. The straight line $ x = 3 $ passes through the point $ (3; 0) $ parallel to the ordinate axis (Oy axis). The straight line $ y = 0 $ is the equation of the abscissa axis (the Ox axis). Well, to construct a straight line $ y = x + 1 $, we find two points through which we draw this straight line. You can, of course, substitute a couple of arbitrary values instead of $ x $. For example, substituting $ x = 10 $, we get: $ y = x + 1 = 10 + 1 = 11 $. We found a point $ (10; 11) $ lying on the line $ y = x + 1 $. However, it is better to find those points at which the straight line $ y = x + 1 $ intersects the lines $ x = 3 $ and $ y = 0 $. Why is it better? Because we will lay down a couple of birds with one stone: we will get two points for constructing the line $ y = x + 1 $ and at the same time find out at what points this line intersects other lines that limit the given area. The straight line $ y = x + 1 $ intersects the straight line $ x = 3 $ at the point $ (3; 4) $, and the straight line $ y = 0 $ - at the point $ (- 1; 0) $. In order not to clutter the course of the decision with auxiliary explanations, I will put the question of obtaining these two points in a note.
How were the points $ (3; 4) $ and $ (- 1; 0) $ obtained? show \ hide
Let's start from the point of intersection of the lines $ y = x + 1 $ and $ x = 3 $. The coordinates of the desired point belong to both the first and second straight lines, therefore, to find the unknown coordinates, you need to solve the system of equations:
$$ \ left \ (\ begin (aligned) & y = x + 1; \\ & x = 3. \ end (aligned) \ right. $$
The solution to such a system is trivial: substituting $ x = 3 $ into the first equation, we will have: $ y = 3 + 1 = 4 $. The point $ (3; 4) $ is the desired intersection point of the lines $ y = x + 1 $ and $ x = 3 $.
Now let's find the point of intersection of the lines $ y = x + 1 $ and $ y = 0 $. Let's again compose and solve the system of equations:
$$ \ left \ (\ begin (aligned) & y = x + 1; \\ & y = 0. \ end (aligned) \ right. $$
Substituting $ y = 0 $ in the first equation, we get: $ 0 = x + 1 $, $ x = -1 $. The point $ (- 1; 0) $ is the desired intersection point of the lines $ y = x + 1 $ and $ y = 0 $ (abscissa axis).
Everything is ready for building a drawing, which will look like this:
The question of a note seems obvious, because everything can be seen from the picture. However, it is worth remembering that a drawing cannot serve as evidence. The figure is just an illustration for clarity.
Our area was defined using the equations of the lines that bound it. Obviously these lines define a triangle, right? Or is it not entirely obvious? Or maybe we are given a different area, bounded by the same straight lines:
Of course, the condition says the region is closed, so the figure shown is incorrect. But to avoid such ambiguities, it is better to define regions with inequalities. We are interested in the part of the plane located under the line $ y = x + 1 $? Ok, so $ y ≤ x + 1 $. Our area should be located above the line $ y = 0 $? Great, so $ y ≥ 0 $. By the way, the last two inequalities can be easily combined into one: $ 0 ≤ y ≤ x + 1 $.
$$ \ left \ (\ begin (aligned) & 0 ≤ y ≤ x + 1; \\ & x ≤ 3. \ end (aligned) \ right. $$
These inequalities define the domain $ D $, and they define it unambiguously, without allowing any ambiguity. But how does this help us in the question indicated at the beginning of the note? It will also help :) We need to check if the point $ M_1 (1; 1) $ belongs to the area $ D $. Substitute $ x = 1 $ and $ y = 1 $ into the system of inequalities that define this domain. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities fails, then the point of the region does not belong. So:
$$ \ left \ (\ begin (aligned) & 0 ≤ 1 ≤ 1 + 1; \\ & 1 ≤ 3. \ end (aligned) \ right. \; \; \ left \ (\ begin (aligned) & 0 ≤ 1 ≤ 2; \\ & 1 ≤ 3. \ end (aligned) \ right. $$
Both inequalities are valid. Point $ M_1 (1; 1) $ belongs to area $ D $.
Now the turn has come to investigate the behavior of the function on the boundary of the region, i.e. go to. Let's start with the line $ y = 0 $.
The straight line $ y = 0 $ (abscissa axis) limits the area $ D $ under the condition $ -1 ≤ x ≤ 3 $. Substitute $ y = 0 $ into the given function $ z (x, y) = x ^ 2 + 2xy-y ^ 2-4x $. The function of one variable $ x $ obtained as a result of substitution is denoted as $ f_1 (x) $:
$$ f_1 (x) = z (x, 0) = x ^ 2 + 2x \ cdot 0-0 ^ 2-4x = x ^ 2-4x. $$
Now for the function $ f_1 (x) $ you need to find the largest and smallest values on the segment $ -1 ≤ x ≤ 3 $. Let's find the derivative of this function and equate it to zero:
$$ f_ (1) ^ (") (x) = 2x-4; \\ 2x-4 = 0; \; x = 2. $$
The value $ x = 2 $ belongs to the segment $ -1 ≤ x ≤ 3 $, so add $ M_2 (2; 0) $ to the list of points. In addition, we calculate the values of the function $ z $ at the ends of the segment $ -1 ≤ x ≤ 3 $, i.e. at the points $ M_3 (-1; 0) $ and $ M_4 (3; 0) $. By the way, if the point $ M_2 $ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $ z $ in it.
So, let's calculate the values of the function $ z $ at the points $ M_2 $, $ M_3 $, $ M_4 $. You can, of course, substitute the coordinates of these points into the original expression $ z = x ^ 2 + 2xy-y ^ 2-4x $. For example, for the point $ M_2 $ we get:
$$ z_2 = z (M_2) = 2 ^ 2 + 2 \ cdot 2 \ cdot 0-0 ^ 2-4 \ cdot 2 = -4. $$
However, the calculations can be simplified a little. To do this, it is worth remembering that on the segment $ M_3M_4 $ we have $ z (x, y) = f_1 (x) $. I'll write it down in detail:
\ begin (aligned) & z_2 = z (M_2) = z (2,0) = f_1 (2) = 2 ^ 2-4 \ cdot 2 = -4; \\ & z_3 = z (M_3) = z (- 1,0) = f_1 (-1) = (- 1) ^ 2-4 \ cdot (-1) = 5; \\ & z_4 = z (M_4) = z (3,0) = f_1 (3) = 3 ^ 2-4 \ cdot 3 = -3. \ end (aligned)
Of course, there is usually no need for such detailed notes, and in the future we will write down all calculations in a shorter manner:
$$ z_2 = f_1 (2) = 2 ^ 2-4 \ cdot 2 = -4; \; z_3 = f_1 (-1) = (- 1) ^ 2-4 \ cdot (-1) = 5; \; z_4 = f_1 (3) = 3 ^ 2-4 \ cdot 3 = -3. $$
Now let's turn to the line $ x = 3 $. This line limits the region $ D $ under the condition $ 0 ≤ y ≤ 4 $. Substitute $ x = 3 $ into the given function $ z $. As a result of this substitution, we get the function $ f_2 (y) $:
$$ f_2 (y) = z (3, y) = 3 ^ 2 + 2 \ cdot 3 \ cdot y-y ^ 2-4 \ cdot 3 = -y ^ 2 + 6y-3. $$
For the function $ f_2 (y) $, you need to find the largest and smallest values on the segment $ 0 ≤ y ≤ 4 $. Let's find the derivative of this function and equate it to zero:
$$ f_ (2) ^ (") (y) = - 2y + 6; \\ -2y + 6 = 0; \; y = 3. $$
The value $ y = 3 $ belongs to the segment $ 0 ≤ y ≤ 4 $, so add $ M_5 (3; 3) $ to the previously found points. In addition, it is necessary to calculate the value of the function $ z $ at the points at the ends of the segment $ 0 ≤ y ≤ 4 $, i.e. at the points $ M_4 (3; 0) $ and $ M_6 (3; 4) $. At the point $ M_4 (3; 0) $ we have already calculated the value of $ z $. Let's calculate the value of the function $ z $ at the points $ M_5 $ and $ M_6 $. Let me remind you that on the segment $ M_4M_6 $ we have $ z (x, y) = f_2 (y) $, therefore:
\ begin (aligned) & z_5 = f_2 (3) = - 3 ^ 2 + 6 \ cdot 3-3 = 6; & z_6 = f_2 (4) = - 4 ^ 2 + 6 \ cdot 4-3 = 5. \ end (aligned)
And finally, consider the last boundary of the domain $ D $, i.e. straight line $ y = x + 1 $. This line limits the area $ D $ under the condition $ -1 ≤ x ≤ 3 $. Substituting $ y = x + 1 $ into the function $ z $, we will have:
$$ f_3 (x) = z (x, x + 1) = x ^ 2 + 2x \ cdot (x + 1) - (x + 1) ^ 2-4x = 2x ^ 2-4x-1. $$
Again we got a function of one variable $ x $. And again you need to find the largest and smallest values of this function on the segment $ -1 ≤ x ≤ 3 $. Find the derivative of the function $ f_ (3) (x) $ and equate it to zero:
$$ f_ (3) ^ (") (x) = 4x-4; \\ 4x-4 = 0; \; x = 1. $$
The value $ x = 1 $ belongs to the segment $ -1 ≤ x ≤ 3 $. If $ x = 1 $, then $ y = x + 1 = 2 $. Let's add $ M_7 (1; 2) $ to the list of points and find out what the value of the function $ z $ is equal to at this point. Points at the ends of the segment $ -1 ≤ x ≤ 3 $, i.e. points $ M_3 (-1; 0) $ and $ M_6 (3; 4) $, were considered earlier, we already found the value of the function in them.
$$ z_7 = f_3 (1) = 2 \ cdot 1 ^ 2-4 \ cdot 1-1 = -3. $$
The second step of the solution is over. We got seven values:
$$ z_1 = -2; \; z_2 = -4; \; z_3 = 5; \; z_4 = -3; \; z_5 = 6; \; z_6 = 5; \; z_7 = -3. $$
Let's turn to. Choosing the largest and smallest values from those numbers that were obtained in the third paragraph, we will have:
$$ z_ (min) = - 4; \; z_ (max) = 6. $$
The problem is solved, it remains only to write down the answer.
Answer: $ z_ (min) = - 4; \; z_ (max) = 6 $.
Example No. 2
Find the largest and smallest values of the function $ z = x ^ 2 + y ^ 2-12x + 16y $ in the region $ x ^ 2 + y ^ 2 ≤ 25 $.
Let's build a blueprint first. The equation $ x ^ 2 + y ^ 2 = 25 $ (this is the boundary line of a given area) defines a circle centered at the origin (that is, at the point $ (0; 0) $) and radius 5. To the inequality $ x ^ 2 + y ^ 2 ≤ 25 $ satisfy all points inside and on the mentioned circle.
We will act according to. Let's find the partial derivatives and find out the critical points.
$$ \ frac (\ partial z) (\ partial x) = 2x-12; \ frac (\ partial z) (\ partial y) = 2y + 16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. find stationary points.
$$ \ left \ (\ begin (aligned) & 2x-12 = 0; \\ & 2y + 16 = 0. \ end (aligned) \ right. \; \; \ left \ (\ begin (aligned) & x = 6; \\ & y = -8. \ End (aligned) \ right. $$
We got the stationary point $ (6; -8) $. However, the found point does not belong to the region $ D $. This is easy to show without even drawing a drawing. Let us check if the inequality $ x ^ 2 + y ^ 2 ≤ 25 $, which defines our domain $ D $, holds. If $ x = 6 $, $ y = -8 $, then $ x ^ 2 + y ^ 2 = 36 + 64 = 100 $, i.e. the inequality $ x ^ 2 + y ^ 2 ≤ 25 $ is not satisfied. Conclusion: the point $ (6; -8) $ does not belong to the area $ D $.
So, there are no critical points inside the region $ D $. Moving on, to. We need to investigate the behavior of the function at the boundary of a given area, i.e. on the circle $ x ^ 2 + y ^ 2 = 25 $. You can, of course, express $ y $ in terms of $ x $, and then substitute the resulting expression into our function $ z $. From the equation of the circle we get: $ y = \ sqrt (25-x ^ 2) $ or $ y = - \ sqrt (25-x ^ 2) $. Substituting, for example, $ y = \ sqrt (25-x ^ 2) $ into the given function, we will have:
$$ z = x ^ 2 + y ^ 2-12x + 16y = x ^ 2 + 25-x ^ 2-12x + 16 \ sqrt (25-x ^ 2) = 25-12x + 16 \ sqrt (25-x ^ 2); \; \; -5≤ x ≤ 5. $$
The further solution will be completely identical to the study of the behavior of the function on the boundary of the region in the previous example # 1. However, it seems to me more reasonable in this situation to apply the Lagrange method. We will only be interested in the first part of this method. After applying the first part of the Lagrange method, we get the points at which we examine the function $ z $ for minimum and maximum values.
We compose the Lagrange function:
$$ F = z (x, y) + \ lambda \ cdot (x ^ 2 + y ^ 2-25) = x ^ 2 + y ^ 2-12x + 16y + \ lambda \ cdot (x ^ 2 + y ^ 2 -25). $$
We find the partial derivatives of the Lagrange function and compose the corresponding system of equations:
$$ F_ (x) ^ (") = 2x-12 + 2 \ lambda x; \; \; F_ (y) ^ (") = 2y + 16 + 2 \ lambda y. \\ \ left \ (\ begin (aligned) & 2x-12 + 2 \ lambda x = 0; \\ & 2y + 16 + 2 \ lambda y = 0; \\ & x ^ 2 + y ^ 2-25 = 0. \ end (aligned) \ right. \; \; \ left \ (\ begin (aligned) & x + \ lambda x = 6; \\ & y + \ lambda y = -8; \\ & x ^ 2 + y ^ 2 = 25. \ end ( aligned) \ right. $$
To solve this system, let's immediately indicate that $ \ lambda \ neq -1 $. Why $ \ lambda \ neq -1 $? Let's try to substitute $ \ lambda = -1 $ in the first equation:
$$ x + (- 1) \ cdot x = 6; \; x-x = 6; \; 0 = 6. $$
The resulting contradiction $ 0 = 6 $ indicates that the value $ \ lambda = -1 $ is invalid. Output: $ \ lambda \ neq -1 $. Let us express $ x $ and $ y $ in terms of $ \ lambda $:
\ begin (aligned) & x + \ lambda x = 6; \; x (1+ \ lambda) = 6; \; x = \ frac (6) (1+ \ lambda). \\ & y + \ lambda y = -8; \; y (1+ \ lambda) = - 8; \; y = \ frac (-8) (1+ \ lambda). \ end (aligned)
I believe that it becomes obvious here why we specifically stipulated the condition $ \ lambda \ neq -1 $. This was done to smoothly place the expression $ 1 + \ lambda $ in the denominators. That is, to be sure that the denominator is $ 1 + \ lambda \ neq 0 $.
Substitute the obtained expressions for $ x $ and $ y $ into the third equation of the system, i.e. at $ x ^ 2 + y ^ 2 = 25 $:
$$ \ left (\ frac (6) (1+ \ lambda) \ right) ^ 2 + \ left (\ frac (-8) (1+ \ lambda) \ right) ^ 2 = 25; \\ \ frac ( 36) ((1+ \ lambda) ^ 2) + \ frac (64) ((1+ \ lambda) ^ 2) = 25; \\ \ frac (100) ((1+ \ lambda) ^ 2) = 25 ; \; (1+ \ lambda) ^ 2 = 4. $$
From the obtained equality it follows that $ 1 + \ lambda = 2 $ or $ 1 + \ lambda = -2 $. Hence, we have two values of the $ \ lambda $ parameter, namely: $ \ lambda_1 = 1 $, $ \ lambda_2 = -3 $. Accordingly, we get two pairs of values $ x $ and $ y $:
\ begin (aligned) & x_1 = \ frac (6) (1+ \ lambda_1) = \ frac (6) (2) = 3; \; y_1 = \ frac (-8) (1+ \ lambda_1) = \ frac (-8) (2) = - 4. \\ & x_2 = \ frac (6) (1+ \ lambda_2) = \ frac (6) (- 2) = - 3; \; y_2 = \ frac (-8) (1+ \ lambda_2) = \ frac (-8) (- 2) = 4. \ end (aligned)
So, we got two points of a possible conditional extremum, i.e. $ M_1 (3; -4) $ and $ M_2 (-3; 4) $. Let's find the values of the function $ z $ at the points $ M_1 $ and $ M_2 $:
\ begin (aligned) & z_1 = z (M_1) = 3 ^ 2 + (- 4) ^ 2-12 \ cdot 3 + 16 \ cdot (-4) = - 75; \\ & z_2 = z (M_2) = (- 3) ^ 2 + 4 ^ 2-12 \ cdot (-3) +16 \ cdot 4 = 125. \ end (aligned)
You should choose the largest and smallest values from those that we got in the first and second steps. But in this case, the choice is small :) We have:
$$ z_ (min) = - 75; \; z_ (max) = 125. $$
Answer: $ z_ (min) = - 75; \; z_ (max) = 125 $.
