The largest and smallest values ​​of a function on a segment. How to find the maximum or minimum of a quadratic function Find the smallest and largest integer values ​​of a function

Let the function y=f(X) continuous on the interval [ a, b]. As is known, such a function reaches its maximum and minimum values ​​on this interval. The function can take these values ​​either at an interior point of the segment [ a, b], or on the boundary of the segment.

To find the largest and smallest values ​​of a function on the interval [ a, b] necessary:

1) find the critical points of the function in the interval ( a, b);

2) calculate the values ​​of the function at the found critical points;

3) calculate the values ​​of the function at the ends of the segment, that is, for x=A and x = b;

4) from all the calculated values ​​of the function, choose the largest and smallest.

Example. Find the largest and smallest values ​​of a function

on the segment.

Finding critical points:

These points lie inside the segment ; y(1) = ‒ 3; y(2) = ‒ 4; y(0) = ‒ 8; y(3) = 1;

at the point x= 3 and at the point x= 0.

Investigation of a function for convexity and an inflection point.

Function y = f (x) called convexup in between (a, b) , if its graph lies under a tangent drawn at any point of this interval, and is called convex down (concave) if its graph lies above the tangent.

The point at the transition through which the convexity is replaced by concavity or vice versa is called inflection point.

Algorithm for studying for convexity and inflection point:

1. Find the critical points of the second kind, that is, the points at which the second derivative is equal to zero or does not exist.

2. Put critical points on the number line, breaking it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upwards, if, then the function is convex downwards.

3. If, when passing through a critical point of the second kind, it changes sign and at this point the second derivative is equal to zero, then this point is the abscissa of the inflection point. Find its ordinate.

Asymptotes of the graph of a function. Investigation of a function into asymptotes.

Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point of the graph to this line tends to zero with an unlimited removal of the graph point from the origin.

There are three types of asymptotes: vertical, horizontal and inclined.

Definition. Direct called vertical asymptote function graph y = f(x), if at least one of the one-sided limits of the function at this point is equal to infinity, that is

where is the discontinuity point of the function, that is, it does not belong to the domain of definition.

Example.

D( y) = (‒ ∞; 2) (2; + ∞)

x= 2 - breaking point.

Definition. Straight y=A called horizontal asymptote function graph y = f(x) at , if

Example.

x
y

Definition. Straight y=kx +b (k≠ 0) is called oblique asymptote function graph y = f(x) at , where

General scheme for the study of functions and plotting.

Function research algorithmy = f(x) :

1. Find the domain of the function D (y).

2. Find (if possible) the points of intersection of the graph with the coordinate axes (with x= 0 and at y = 0).

3. Investigate for even and odd functions ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).

4. Find the asymptotes of the graph of the function.

5. Find intervals of monotonicity of the function.

6. Find the extrema of the function.

7. Find the intervals of convexity (concavity) and inflection points of the graph of the function.

8. On the basis of the conducted research, construct a graph of the function.

Example. Investigate the function and plot its graph.

1) D (y) =

x= 4 - breaking point.

2) When x = 0,

(0; – 5) – point of intersection with oy.

At y = 0,

3) y(‒ x)= general function (neither even nor odd).

4) We investigate for asymptotes.

a) vertical

b) horizontal

c) find oblique asymptotes where

‒oblique asymptote equation

5) In this equation, it is not required to find intervals of monotonicity of the function.

6)

These critical points partition the entire domain of the function on the interval (˗∞; ˗2), (˗2; 4), (4; 10) and (10; +∞). It is convenient to present the results obtained in the form of the following table:

				no extra.

It can be seen from the table that the point X= ‒2‒maximum point, at the point X= 4‒ no extremum, X= 10 – minimum point.

Substitute the value (‒ 3) into the equation:

9 + 24 ‒ 20 > 0

25 ‒ 40 ‒ 20 < 0

121 ‒ 88 ‒ 20 > 0

The maximum of this function is

(– 2; – 4) – maximum extremum.

The minimum of this function is

(10; 20) is the minimum extremum.

7) examine the convexity and inflection point of the graph of the function

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 $D$.

1. Find the critical points of the function $z=f(x,y)$ that belong to the region $D$. Compute function values ​​at critical points.
2. 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.
3. 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 maximum and minimum values ​​of the function $z=x^2+2xy-y^2-4x$ in the 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 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 $D$ 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.

1. First, let's find the domain of the function. Let's check whether the segment specified in the condition is included in it.
2. 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.
3. 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.
4. 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. 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 ; - 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 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 ; - 1 ] .

Now we need to determine the stationary points of the function. Let's do this with the equation x 3 - 8 x 3 = 0. It has only one real root, which is 2. It will be a stationary point of the function and will fall into the first segment [ 1 ; 4 ] .

Let us calculate the values ​​of the function at the ends of the first segment and at the given point, i.e. for x = 1 , x = 2 and x = 4:

y(1) = 1 3 + 4 1 2 = 5 y(2) = 2 3 + 4 2 2 = 3 y(4) = 4 3 + 4 4 2 = 4 1 4

We have obtained that the largest value of the function m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 will be achieved at x = 1 , and the smallest m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – at x = 2 .

The second segment does not include any stationary points, so we need to calculate the function values ​​only at the ends of the given segment:

y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3

Hence, m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .

Answer: For the segment [ 1 ; 4 ] - m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 , for the segment [ - 4 ; - 1 ] - m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .

See picture:

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

1. First you need to check whether the given interval will be a subset of the domain of the given function.
2. 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.
3. 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) .

1. 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.

Example 2

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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With this service, you can find the largest and smallest value of a function one variable f(x) with the design of the solution in Word. If the function f(x,y) is given, therefore, it is necessary to find the extremum of the function of two variables . You can also find the intervals of increase and decrease of the function.

Function entry rules:

A necessary condition for an extremum of a function of one variable

The equation f "0 (x *) \u003d 0 is a necessary condition for the extremum of a function of one variable, i.e. at the point x * the first derivative of the function must vanish. It selects stationary points x c at which the function does not increase and does not decrease .

A sufficient condition for an extremum of a function of one variable

Let f 0 (x) be twice differentiable with respect to x belonging to the set D . If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then the point x * is the point of the local (global) minimum of the function.

If at the point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

That point x * is a local (global) maximum.

Example #1. Find the largest and smallest values ​​of the function: on the segment .
Solution.

The critical point is one x 1 = 2 (f'(x)=0). This point belongs to the segment . (The point x=0 is not critical, since 0∉).
We calculate the values ​​of the function at the ends of the segment and at the critical point.
f(1)=9, f(2)= 5 / 2 , f(3)=3 8 / 81
Answer: f min = 5 / 2 for x=2; f max =9 at x=1

Example #2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let us find the critical points: 1-cos(x)=2, cos(x)=1, x=± π / 3 +2πk, k∈Z. We find y''=2sin(x), calculate , so x= π / 3 +2πk, k∈Z are the minimum points of the function; , so x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example #3. Investigate the extremum function in the neighborhood of the point x=0.
Solution. Here it is necessary to find the extrema of the function. If the extremum x=0 , then find out its type (minimum or maximum). If among the found points there is no x = 0, then calculate the value of the function f(x=0).
It should be noted that when the derivative on each side of a given point does not change its sign, the possible situations are not exhausted even for differentiable functions: it may happen that for an arbitrarily small neighborhood on one side of the point x 0 or on both sides, the derivative changes sign. At these points, one has to apply other methods to study functions to an extremum.

Example #4. Divide the number 49 into two terms, the product of which will be the largest.
Solution. Let x be the first term. Then (49-x) is the second term.
The product will be maximal: x (49-x) → max

Option 1. at

1. Graph of a function y=f(x) shown in the figure.

Specify the largest value of this function 1

on the segment [ a; b]. A 0 1 b x

1) 2,5; 2) 3; 3) 4; 4) 2.

https://pandia.ru/text/78/524/images/image003_127.gif" width="242" height="133 src="> 1) -4; 2) -2; 3) 4; 4) 2.

4. Functions y=f(x) set on the segment [ a; b]. at

The figure shows a graph of its derivative

y=f ´(x). Explore for extremes 1 b

function y=f(x). Please indicate the quantity in your answer. a 0 1 x

minimum points.

1) 6; 2) 7; 3) 4;

5. Find the largest value of a function y \u003d -2x2 + 8x -7.

1) -2; 2) 7; 3) 1;

6. Find the smallest value of a function on the segment .

1) https://pandia.ru/text/78/524/images/image005_87.gif" width="17" height="48 src=">.

7. Find the smallest value of a function y=|2x+3| - .

1) - https://pandia.ru/text/78/524/images/image006_79.gif" width="17" height="47"> ; 4) - .

https://pandia.ru/text/78/524/images/image009_67.gif" width="144" height="33 src="> has a minimum at the point xo=1.5?

1) 5; 2) -6; 3) 4; 4) 6.at

9. Specify the largest value of the function y=f(x) ,

1 x

0 1

1) 2,5; 2) 3; 3) -3;

y=lg(100 – x2 ).

1) 10 ; 2) 100 ; 3) 2 ; 4) 1 .

11. Find the smallest value of a function y=2sin-1.

1) -1 ; 2) -3 ; 3) -2 ; 4) - .

Test 14 The largest (smallest) value of the function.

https://pandia.ru/text/78/524/images/image013_44.gif" width="130" height="115 src=">1. Graph of the function y=f(x) shown in the figure.

Specify the smallest value of this function 1

on the segment [ a; b]. A b

0 1 x

1) 0; 2) - 4 ,5; 3) -2; 4) - 3.

2. at The figure shows a graph of the function y=f(x).

How many maximum points does the function have?

1

0 1 x 1) 5; 2) 6; 3) 4; 4) 1.

3. At what point is the function y \u003d 2x2 + 24x -25 takes on the smallest value?

https://pandia.ru/text/78/524/images/image018_37.gif" width="76" height="48"> on the segment [-3;-1].

1) - https://pandia.ru/text/78/524/images/image020_37.gif" width="17" height="47 src=">; 2); 4) - 5.

https://pandia.ru/text/78/524/images/image022_35.gif" width="135" height="33 src="> has a minimum at the point xo = -2?

; 2) -6;; 4) 6.at

9. Specify the smallest value of the function y=f(x) ,

whose graph is shown in the figure. 1 x

0 1

1) -1,5; 2) -1; 3) -3;

10. Find the largest value of a function y=log11 (121 – x2 ).

1) 11;; 3) 1;

11. Find the largest value of a function y=2cos+3.

1) 5 ; 2) 3 ; 3) 2 ; 4) .

Answers :