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 smallest value The function is usually searched for on some interval X , which is either the entire scope 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 on 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 x=1 abscissa, 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] .
Solution.
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): 
Solution.
Let's start with the scope of the function. The square trinomial in the denominator of a fraction must not vanish: 
It is easy to check that all intervals from the condition of the problem belong to the domain of the function.
Let's differentiate the function: 
Obviously, the derivative exists on the entire domain of the function.
Let's find stationary points. The derivative vanishes at . This stationary point falls within the intervals (-3;1] and (-3;2) .
And now you can compare the results obtained at each point with the graph of the function. The blue dotted lines indicate the asymptotes.
This can end with finding the largest and smallest value of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest value of the function on any interval. This gives a clearer picture and a rigorous justification of the results.
In many problems, it is required to calculate the maximum or minimum value of a quadratic function. The maximum or minimum can be found if the original function is written in standard form: or through the coordinates of the parabola vertex: f (x) = a (x − h) 2 + k (\displaystyle f(x)=a(x-h)^(2)+k). Moreover, the maximum or minimum of any quadratic function can be calculated using mathematical operations.
Steps
The quadratic function is written in standard form
- In function coefficients a = 1 (\displaystyle a=1) and b = 10 (\displaystyle b=10)
- As a second example, consider the function . Here a = − 3 (\displaystyle a=-3) and b = 6 (\displaystyle b=6). Therefore, calculate the x-coordinate of the top of the parabola as follows:
-
Find the corresponding value of f(x). Substitute the found value of "x" into the original function to find the corresponding value of f(x). This is how you find the minimum or maximum of the function.
- In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) you calculated that the x-coordinate of the top of the parabola is x = − 5 (\displaystyle x=-5). In the original function, instead of x (\displaystyle x) substitute − 5 (\displaystyle -5)
- In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) you found that the x-coordinate of the vertex of the parabola is x = 1 (\displaystyle x=1). In the original function, instead of x (\displaystyle x) substitute 1 (\displaystyle 1) to find its maximum value:
-
Write down the answer. Reread the condition of the problem. If you need to find the coordinates of the vertex of the parabola, write down both values in your answer x (\displaystyle x) and y (\displaystyle y)(or f (x) (\displaystyle f(x))). If you need to calculate the maximum or minimum of a function, write down only the value in your answer y (\displaystyle y)(or f (x) (\displaystyle f(x))). Look again at the sign of the coefficient a (\displaystyle a) to check if you calculated the maximum or minimum.
The quadratic function is written in terms of the coordinates of the vertex of the parabola
-
Write the quadratic function in terms of the coordinates of the vertex of the parabola. Such an equation has the following form:
Determine the direction of the parabola. To do this, look at the sign of the coefficient a (\displaystyle a). If the coefficient a (\displaystyle a) positive, the parabola is directed upwards. If the coefficient a (\displaystyle a) negative, the parabola is pointing down. For example:
Find the minimum or maximum value of the function. If the function is written in terms of the coordinates of the parabola vertex, the minimum or maximum is equal to the value of the coefficient k (\displaystyle k). In the examples above:
Find the coordinates of the vertex of the parabola. If in the problem it is required to find the vertex of the parabola, its coordinates are (h , k) (\displaystyle (h,k)). Note that when a quadratic function is written in terms of the coordinates of the parabola vertex, the subtraction operation must be enclosed in brackets (x − h) (\displaystyle (x-h)), so the value h (\displaystyle h) taken with the opposite sign.
How to calculate the minimum or maximum using mathematical operations
Let us first consider the standard form of the equation. Write the quadratic function in standard form: f (x) = a x 2 + b x + c (\displaystyle f(x)=ax^(2)+bx+c). If necessary, bring like terms and rearrange them to get the standard equation.
Find the first derivative. The first derivative of a quadratic function, which is written in standard form, is equal to f ′ (x) = 2 a x + b (\displaystyle f^(\prime )(x)=2ax+b).
Set the derivative to zero. Recall that the derivative of a function is equal to the slope of the function at a certain point. At the minimum or maximum, the slope is zero. Therefore, to find the minimum or maximum value of a function, the derivative must be equated to zero. In our example:
-
Write the function in standard form. A quadratic function is a function whose equation includes a variable x 2 (\displaystyle x^(2)). The equation may or may not include a variable x (\displaystyle x). If an equation includes a variable with an exponent greater than 2, it does not describe a quadratic function. If necessary, bring like terms and rearrange them to write the function in standard form.
The graph of a quadratic function is a parabola. The branches of a parabola point up or down. If the coefficient a (\displaystyle a) with a variable x 2 (\displaystyle x^(2)) a (\displaystyle a)
Calculate -b/2a. Meaning − b 2 a (\displaystyle -(\frac (b)(2a))) is the coordinate x (\displaystyle x) top of the parabola. If a quadratic function written in standard form a x 2 + b x + c (\displaystyle ax^(2)+bx+c), use the coefficients for x (\displaystyle x) and x 2 (\displaystyle x^(2)) in the following way:
Sometimes in problems B15 there are "bad" functions for which it is difficult to find the derivative. Previously, this was only on probes, but now these tasks are so common that they can no longer be ignored when preparing for this exam.
In this case, other tricks work, one of which is - monotone.
The function f (x) is called monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following is true:
x 1< x 2 ⇒ f (x 1) < f (x2).
The function f (x) is called monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following is true:
x 1< x 2 ⇒ f (x 1) > f( x2).
In other words, for an increasing function, the larger x is, the larger f(x) is. For a decreasing function, the opposite is true: the more x , the less f(x).
For example, the logarithm increases monotonically if the base a > 1 and decreases monotonically if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.
f (x) = log a x (a > 0; a ≠ 1; x > 0)
The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:
The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, показательная функция определена для всех чисел, а не только для x > 0:
f (x) = a x (a > 0)
Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where monotony is broken.
All these functions are never found in their pure form. Polynomials, fractions and other nonsense are added to them, because of which it becomes difficult to calculate the derivative. What happens in this case - now we will analyze.
Parabola vertex coordinates
Most often, the function argument is replaced with square trinomial of the form y = ax 2 + bx + c . Its graph is a standard parabola, in which we are interested in:
- Parabola branches - can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
- The vertex of the parabola is the extremum point of a quadratic function, at which this function takes its smallest (for a > 0) or largest (a< 0) значение.
Of greatest interest is top of a parabola, the abscissa of which is calculated by the formula:
So, we have found the extremum point of the quadratic function. But if the original function is monotonic, for it the point x 0 will also be an extremum point. Thus, we formulate the key rule:
The extremum points of the square trinomial and the complex function it enters into coincide. Therefore, you can look for x 0 for a square trinomial, and forget about the function.
From the above reasoning, it remains unclear what kind of point we get: a maximum or a minimum. However, the tasks are specifically designed so that it does not matter. Judge for yourself:
- There is no segment in the condition of the problem. Therefore, it is not required to calculate f(a) and f(b). It remains to consider only the extremum points;
- But there is only one such point - this is the top of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.
Thus, the solution of the problem is greatly simplified and reduced to just two steps:
- Write out the parabola equation y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a;
- Find the value of the original function at this point: f (x 0). If there are no additional conditions, this will be the answer.
At first glance, this algorithm and its justification may seem complicated. I deliberately do not post a "bare" solution scheme, since the thoughtless application of such rules is fraught with errors.
Consider the real tasks from the trial exam in mathematics - this is where this technique is most common. At the same time, we will make sure that in this way many problems of B15 become almost verbal.
Under the root is a quadratic function y \u003d x 2 + 6x + 13. The graph of this function is a parabola with branches up, since the coefficient a \u003d 1\u003e 0.
Top of the parabola:
x 0 \u003d -b / (2a) \u003d -6 / (2 1) \u003d -6 / 2 \u003d -3
Since the branches of the parabola are directed upwards, at the point x 0 \u003d −3, the function y \u003d x 2 + 6x + 13 takes on the smallest value.
The root is monotonically increasing, so x 0 is the minimum point of the entire function. We have:
A task. Find the smallest value of the function:
y = log 2 (x 2 + 2x + 9)
Under the logarithm is again a quadratic function: y \u003d x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.
Top of the parabola:
x 0 \u003d -b / (2a) \u003d -2 / (2 1) \u003d -2/2 \u003d -1
So, at the point x 0 = −1, the quadratic function takes on the smallest value. But the function y = log 2 x is monotone, so:
y min = y (−1) = log 2 ((−1) 2 + 2 (−1) + 9) = ... = log 2 8 = 3
The exponent is a quadratic function y = 1 − 4x − x 2 . Let's rewrite it in normal form: y = −x 2 − 4x + 1.
Obviously, the graph of this function is a parabola, branches down (a = −1< 0). Поэтому вершина будет точкой максимума:
x 0 = −b /(2a ) = −(−4)/(2 (−1)) = 4/(−2) = −2
The original function is exponential, it is monotone, so the largest value will be at the found point x 0 = −2:
An attentive reader will surely notice that we did not write out the area of \u200b\u200bpermissible values of the root and logarithm. But this was not required: inside there are functions whose values are always positive.
Consequences from the scope of a function
Sometimes, to solve problem B15, it is not enough just to find the vertex of the parabola. The desired value may lie at the end of the segment, but not at the extremum point. If the task does not specify a segment at all, look at tolerance range original function. Namely:
Pay attention again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how it works with specific examples:
A task. Find the largest value of the function:
Under the root is again a quadratic function: y \u003d 3 - 2x - x 2. Its graph is a parabola, but branches down since a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический квадратный корень из отрицательного числа не существует.
We write out the area of permissible values (ODZ):
3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; one]
Now find the vertex of the parabola:
x 0 = −b /(2a ) = −(−2)/(2 (−1)) = 2/(−2) = −1
The point x 0 = −1 belongs to the ODZ segment - and this is good. Now we consider the value of the function at the point x 0, as well as at the ends of the ODZ:
y(−3) = y(1) = 0
So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.
A task. Find the smallest value of the function:
y = log 0.5 (6x - x 2 - 5)
Inside the logarithm there is a quadratic function y \u003d 6x - x 2 - 5. This is a parabola with branches down, but there cannot be negative numbers in the logarithm, so we write out the ODZ:
6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)
Please note: the inequality is strict, so the ends do not belong to the ODZ. In this way, the logarithm differs from the root, where the ends of the segment suit us quite well.
Looking for the vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 (−1)) = −6/(−2) = 3
The top of the parabola fits along the ODZ: x 0 = 3 ∈ (1; 5). But since the ends of the segment do not interest us, we consider the value of the function only at the point x 0:
y min = y (3) = log 0.5 (6 3 − 3 2 − 5) = log 0.5 (18 − 9 − 5) = log 0.5 4 = −2
Let the function $z=f(x,y)$ be defined and continuous in some bounded closed domain $D$. Let the given function have finite partial derivatives of the first order in this region (with the possible exception of a finite number of points). To find the largest and smallest values of a function of two variables in a given closed region, 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 region $D$. Compute function values at critical points.
- Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$ by finding the points of possible maximum and minimum values. Calculate the function values at the obtained points.
- From the function values obtained in the previous two paragraphs, choose the largest and smallest.
What are critical points? show/hide
Under critical points imply points where both first-order partial derivatives are equal to 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 first-order partial derivatives 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 area, bounded by lines$x=3$, $y=0$ and $y=x+1$.
We will follow the above, but first we will deal with the drawing of a 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 y-axis (axis Oy). The straight line $y=0$ is the equation of the abscissa axis (Ox axis). Well, to construct a straight line $y=x+1$ let's 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 have found the point $(10;11)$ lying on the line $y=x+1$. However, it is better to find those points where the line $y=x+1$ intersects with 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 straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines that bound the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ - at the point $(-1;0)$. In order not to clutter up the course of the solution 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 lines, so to find unknown coordinates, you need to solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & x=3. \end(aligned) \right. $$
The solution of such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the point of intersection of the lines $y=x+1$ and $y=0$. Again, we compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (abscissa axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything can be seen from the figure. However, it is worth remembering that the drawing cannot serve as evidence. The figure is just an illustration for clarity.
Our area was set using the equations of lines that limit it. It's obvious that these lines define a triangle, don't they? Or not quite obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is wrong. But to avoid such ambiguities, it is better to define regions by inequalities. We are interested in the part of the plane located under the line $y=x+1$? Ok, so $y ≤ x+1$. Our area should be located above the line $y=0$? Great, so $y ≥ 0$. By the way, the last two inequalities are easily combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the domain $D$, and define it uniquely, without any ambiguities. But how does this help us in the question at the beginning of the footnote? It will also help :) We need to check if the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right.$$
Both inequalities are true. The point $M_1(1;1)$ belongs to the region $D$.
Now it is the turn to investigate the behavior of the function on the boundary of the domain, i.e. go to. Let's start with the straight line $y=0$.
The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. The resulting substitution function of one variable $x$ will be denoted as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we also add $M_2(2;0)$ to the list of points. In addition, we calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at the points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at the points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points in the original expression $z=x^2+2xy-y^2-4x$. For example, for the point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a bit. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll spell it out in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, there is usually no need for such detailed entries, and in the future we will begin to write down all calculations in a shorter way:
$$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.$$
Now let's turn to the straight line $x=3$. This line bounds the domain $D$ under the condition $0 ≤ y ≤ 4$. Substitute $x=3$ into the given function $z$. As a result of such a substitution, we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$, you need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we add $M_5(3;3)$ to the points found earlier. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at the points $M_4(3;0)$ and $M_6(3;4)$. At the point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at the points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; &z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And, finally, consider the last boundary of $D$, i.e. line $y=x+1$. This line bounds the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again, you need to find the largest and smallest values of this function on the segment $-1 ≤ x ≤ 3$. Find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. The points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we have already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We got seven values:
$$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.$$
The problem is solved, it remains only to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example #2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
Let's build a drawing first. The equation $x^2+y^2=25$ (this is the boundary line of the given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ 25$ satisfy all points inside and on the mentioned circle.
We will act on. Let's find partial derivatives and find out the critical points.
$$ \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.$$
We got a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check if the inequality $x^2+y^2 ≤ 25$, which defines our domain $D$, holds. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ is not satisfied. Conclusion: the point $(6;-8)$ does not belong to the region $D$.
Thus, there are no critical points inside $D$. Let's move on, to. We need to investigate the behavior of the function on the boundary of the given area, i.e. on the circle $x^2+y^2=25$. You can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the circle equation we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ 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. $$
The further solution will be completely identical to the study of the behavior of the function on the boundary of the region in the previous example No. 1. However, it seems to me more reasonable in this situation to apply the Lagrange method. We are only interested in the first part of this method. After applying the first part of the Lagrange method, we will get points at which and examine the function $z$ for the minimum and maximum values.
We compose the Lagrange function:
$$ 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.$$
To solve this system, let's immediately indicate that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ says that the value $\lambda=-1$ is invalid. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the $\lambda\neq -1$ condition. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator is $1+\lambda\neq 0$.
Let us substitute the obtained expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \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. $$
It follows from the resulting equality that $1+\lambda=2$ or $1+\lambda=-2$. Hence, we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we got two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Find the values of the function $z$ at the points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should choose the largest and smallest values from those that we obtained in the first and second steps. But in this case, the choice is small :) We have:
$$z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=125$.
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.
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 in a known interval at the abscissa x 0, and the smallest is the smallest accepted value in 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 ] .
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 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 ; four ] .
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 ; +∞) .
Solution
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 which intervals the function will decrease and on which 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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