Option 1. at
1. Graph of a function y=f(x) shown in the figure.
Specify the largest value for 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) given 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 the function y= -2x2+8x -7.
1) -2; 2) 7; 3) 1;
6. Find the smallest value of the 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 the 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 the function y=2sin-1.
1) -1 ; 2) -3 ; 3) -2 ; 4) - .
Test 14. Extremes. 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 for 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 the 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=2x2+24x -25 takes 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) ,
the graph of which is shown in the figure. 1 x
0 1
1) -1,5; 2) -1; 3) -3;
10. Find the largest value of the function y=log11 (121 – x2 ).
1) 11;; 3) 1;
11. Find the largest value of the function y=2cos+3.
1) 5 ; 2) 3 ; 3) 2 ; 4) .
Answers :
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, for 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 a closed domain $D$.
- Find critical points of the function $z=f(x,y)$ belonging to the domain $D$. Calculate the function values at critical points.
- Investigate the behavior of the function $z=f(x,y)$ on the boundary of the region $D$, 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, select the largest and smallest.
What are critical points? show\hide
Under critical points imply points at which 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 first-order partial derivatives are equal to zero are called stationary points. Thus, stationary points are a subset of critical points.
Example No. 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 a given area, which we will denote by the letter $D$. We are given the equations of three straight lines that 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 (Ox axis). Well, to construct the line $y=x+1$, we will find two points through which we will draw this 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 at which the straight line $y=x+1$ intersects the lines $x=3$ and $y=0$. Why is this better? Because we will kill a couple of birds with one stone: we will get two points to construct the straight line $y=x+1$ and at the same time find out at what points this straight line intersects other lines limiting the given area. The line $y=x+1$ intersects the line $x=3$ at the point $(3;4)$, and the line $y=0$ intersects at the point $(-1;0)$. In order not to clutter up the progress 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 intersection point 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. $$
The solution to such a system is trivial: substituting $x=3$ into the first equation we will have: $y=3+1=4$. The point $(3;4)$ is the desired intersection point of the lines $y=x+1$ and $x=3$.
Now let's find the intersection point of the lines $y=x+1$ and $y=0$. Let us again compose and solve the system of equations:
$$ \left \( \begin(aligned) & y=x+1;\\ & y=0. \end(aligned) \right. $$
Substituting $y=0$ into the first equation, we get: $0=x+1$, $x=-1$. The point $(-1;0)$ is the desired intersection point of the lines $y=x+1$ and $y=0$ (x-axis).
Everything is ready to build a drawing that will look like this:
The question of the note seems obvious, because everything is visible in the picture. However, it is worth remembering that a drawing cannot serve as evidence. The drawing is for illustrative purposes only.
Our area was defined using the equations of lines that bound it. Obviously, these lines define a triangle, right? Or is it not entirely obvious? Or maybe we are given a different area, bounded by the same lines:
Of course, the condition says that the area is closed, so the picture shown is incorrect. But to avoid such ambiguities, it is better to define regions by inequalities. Are we interested in the part of the plane located under the straight line $y=x+1$? Ok, so $y ≤ x+1$. Should our area be located above the line $y=0$? Great, that means $y ≥ 0$. By the way, the last two inequalities can easily be combined into one: $0 ≤ y ≤ x+1$.
$$ \left \( \begin(aligned) & 0 ≤ y ≤ x+1;\\ & x ≤ 3. \end(aligned) \right. $$
These inequalities define the region $D$, and they define it unambiguously, without allowing any ambiguity. But how does this help us with the question stated at the beginning of the note? It will also help :) We need to check whether the point $M_1(1;1)$ belongs to the region $D$. Let us substitute $x=1$ and $y=1$ into the system of inequalities that define this region. If both inequalities are satisfied, then the point lies inside the region. If at least one of the inequalities is not satisfied, then the point does not belong to the region. So:
$$ \left \( \begin(aligned) & 0 ≤ 1 ≤ 1+1;\\ & 1 ≤ 3. \end(aligned) \right. \;\; \left \( \begin(aligned) & 0 ≤ 1 ≤ 2;\\ & 1 ≤ 3. \end(aligned) \right $$.
Both inequalities are valid. Point $M_1(1;1)$ belongs to region $D$.
Now it’s the turn to study the behavior of the function at the boundary of the region, i.e. let's go to . Let's start with the straight line $y=0$.
The straight line $y=0$ (abscissa axis) limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Let's substitute $y=0$ into the given function $z(x,y)=x^2+2xy-y^2-4x$. We denote the function of one variable $x$ obtained as a result of substitution as $f_1(x)$:
$$ f_1(x)=z(x,0)=x^2+2x\cdot 0-0^2-4x=x^2-4x. $$
Now for the function $f_1(x)$ we need to find the largest and smallest values on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of this function and equate it to zero:
$$ f_(1)^(")(x)=2x-4;\\ 2x-4=0; \; x=2. $$
The value $x=2$ belongs to the segment $-1 ≤ x ≤ 3$, so we will also add $M_2(2;0)$ to the list of points. In addition, let us calculate the values of the function $z$ at the ends of the segment $-1 ≤ x ≤ 3$, i.e. at points $M_3(-1;0)$ and $M_4(3;0)$. By the way, if the point $M_2$ did not belong to the segment under consideration, then, of course, there would be no need to calculate the value of the function $z$ in it.
So, let's calculate the values of the function $z$ at points $M_2$, $M_3$, $M_4$. You can, of course, substitute the coordinates of these points into the original expression $z=x^2+2xy-y^2-4x$. For example, for point $M_2$ we get:
$$z_2=z(M_2)=2^2+2\cdot 2\cdot 0-0^2-4\cdot 2=-4.$$
However, the calculations can be simplified a little. To do this, it is worth remembering that on the segment $M_3M_4$ we have $z(x,y)=f_1(x)$. I'll write it down in detail:
\begin(aligned) & z_2=z(M_2)=z(2,0)=f_1(2)=2^2-4\cdot 2=-4;\\ & z_3=z(M_3)=z(- 1,0)=f_1(-1)=(-1)^2-4\cdot (-1)=5;\\ & z_4=z(M_4)=z(3,0)=f_1(3)= 3^2-4\cdot 3=-3. \end(aligned)
Of course, there is usually no need for such detailed records, and in the future we will write down all calculations briefly:
$$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.$$
Now let's turn to the straight line $x=3$. This straight line limits the region $D$ under the condition $0 ≤ y ≤ 4$. Let's substitute $x=3$ into the given function $z$. As a result of this substitution we get the function $f_2(y)$:
$$ f_2(y)=z(3,y)=3^2+2\cdot 3\cdot y-y^2-4\cdot 3=-y^2+6y-3. $$
For the function $f_2(y)$ we need to find the largest and smallest values on the interval $0 ≤ y ≤ 4$. Let's find the derivative of this function and equate it to zero:
$$ f_(2)^(")(y)=-2y+6;\\ -2y+6=0; \; y=3. $$
The value $y=3$ belongs to the segment $0 ≤ y ≤ 4$, so we will also add $M_5(3;3)$ to the previously found points. In addition, it is necessary to calculate the value of the function $z$ at the points at the ends of the segment $0 ≤ y ≤ 4$, i.e. at points $M_4(3;0)$ and $M_6(3;4)$. At point $M_4(3;0)$ we have already calculated the value of $z$. Let us calculate the value of the function $z$ at points $M_5$ and $M_6$. Let me remind you that on the segment $M_4M_6$ we have $z(x,y)=f_2(y)$, therefore:
\begin(aligned) & z_5=f_2(3)=-3^2+6\cdot 3-3=6; & z_6=f_2(4)=-4^2+6\cdot 4-3=5. \end(aligned)
And finally, consider the last boundary of the region $D$, i.e. straight line $y=x+1$. This straight line limits the region $D$ under the condition $-1 ≤ x ≤ 3$. Substituting $y=x+1$ into the function $z$, we will have:
$$ f_3(x)=z(x,x+1)=x^2+2x\cdot (x+1)-(x+1)^2-4x=2x^2-4x-1. $$
Once again we have a function of one variable $x$. And again we need to find the largest and smallest values of this function on the interval $-1 ≤ x ≤ 3$. Let's find the derivative of the function $f_(3)(x)$ and equate it to zero:
$$ f_(3)^(")(x)=4x-4;\\ 4x-4=0; \; x=1. $$
The value $x=1$ belongs to the interval $-1 ≤ x ≤ 3$. If $x=1$, then $y=x+1=2$. Let's add $M_7(1;2)$ to the list of points and find out what the value of the function $z$ is at this point. Points at the ends of the segment $-1 ≤ x ≤ 3$, i.e. points $M_3(-1;0)$ and $M_6(3;4)$ were considered earlier, we already found the value of the function in them.
$$z_7=f_3(1)=2\cdot 1^2-4\cdot 1-1=-3.$$
The second step of the solution is completed. We received seven values:
$$z_1=-2;\;z_2=-4;\;z_3=5;\;z_4=-3;\;z_5=6;\;z_6=5;\;z_7=-3.$$
Let's turn to. Choosing the largest and smallest values from the numbers obtained in the third paragraph, we will have:
$$z_(min)=-4; \; z_(max)=6.$$
The problem is solved, all that remains is to write down the answer.
Answer: $z_(min)=-4; \; z_(max)=6$.
Example No. 2
Find the largest and smallest values of the function $z=x^2+y^2-12x+16y$ in the region $x^2+y^2 ≤ 25$.
First, let's build a drawing. The equation $x^2+y^2=25$ (this is the boundary line of a given area) defines a circle with a center at the origin (i.e. at the point $(0;0)$) and a radius of 5. The inequality $x^2 +y^2 ≤ $25 satisfy all points inside and on the mentioned circle.
We will act according to. Let's find partial derivatives and find out the critical points.
$$ \frac(\partial z)(\partial x)=2x-12; \frac(\partial z)(\partial y)=2y+16. $$
There are no points at which the found partial derivatives do not exist. Let us find out at what points both partial derivatives are simultaneously equal to zero, i.e. let's find stationary points.
$$ \left \( \begin(aligned) & 2x-12=0;\\ & 2y+16=0. \end(aligned) \right. \;\; \left \( \begin(aligned) & x =6;\\ & y=-8. \end(aligned) \right $$.
We have obtained a stationary point $(6;-8)$. However, the found point does not belong to the region $D$. This is easy to show without even resorting to drawing. Let's check whether the inequality $x^2+y^2 ≤ 25$ holds, which defines our region $D$. If $x=6$, $y=-8$, then $x^2+y^2=36+64=100$, i.e. the inequality $x^2+y^2 ≤ 25$ does not hold. Conclusion: point $(6;-8)$ does not belong to area $D$.
So, there are no critical points inside the region $D$. Let's move on to... We need to study the behavior of a function on the boundary of a given region, i.e. on the circle $x^2+y^2=25$. We can, of course, express $y$ in terms of $x$, and then substitute the resulting expression into our function $z$. From the equation of a circle we get: $y=\sqrt(25-x^2)$ or $y=-\sqrt(25-x^2)$. Substituting, for example, $y=\sqrt(25-x^2)$ into the given function, we will have:
$$ 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. $$
The further solution will be completely identical to the study of the behavior of the function at the boundary of the region in the previous example No. 1. However, it seems to me more reasonable to apply the Lagrange method in this situation. We will be interested only in the first part of this method. After applying the first part of the Lagrange method, we will obtain points at which we will examine the function $z$ for minimum and maximum values.
We compose the Lagrange function:
$$ 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. right. \;\; \left \( \begin(aligned) & x+\lambda x=6;\\ & y+\lambda y=-8;\\ & x^2+y^2=25. \end( aligned)\right.$$
To solve this system, let's immediately point out that $\lambda\neq -1$. Why $\lambda\neq -1$? Let's try to substitute $\lambda=-1$ into the first equation:
$$ x+(-1)\cdot x=6; \; x-x=6; \; 0=6. $$
The resulting contradiction $0=6$ indicates that the value $\lambda=-1$ is unacceptable. Output: $\lambda\neq -1$. Let's express $x$ and $y$ in terms of $\lambda$:
\begin(aligned) & x+\lambda x=6;\; x(1+\lambda)=6;\; x=\frac(6)(1+\lambda). \\ & y+\lambda y=-8;\; y(1+\lambda)=-8;\; y=\frac(-8)(1+\lambda). \end(aligned)
I believe that it becomes obvious here why we specifically stipulated the condition $\lambda\neq -1$. This was done to fit the expression $1+\lambda$ into the denominators without interference. That is, to be sure that the denominator $1+\lambda\neq 0$.
Let us substitute the resulting expressions for $x$ and $y$ into the third equation of the system, i.e. in $x^2+y^2=25$:
$$ \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. $$
From the resulting equality it follows that $1+\lambda=2$ or $1+\lambda=-2$. Hence we have two values of the parameter $\lambda$, namely: $\lambda_1=1$, $\lambda_2=-3$. Accordingly, we get two pairs of values $x$ and $y$:
\begin(aligned) & x_1=\frac(6)(1+\lambda_1)=\frac(6)(2)=3; \; y_1=\frac(-8)(1+\lambda_1)=\frac(-8)(2)=-4. \\ & x_2=\frac(6)(1+\lambda_2)=\frac(6)(-2)=-3; \; y_2=\frac(-8)(1+\lambda_2)=\frac(-8)(-2)=4. \end(aligned)
So, we have obtained two points of a possible conditional extremum, i.e. $M_1(3;-4)$ and $M_2(-3;4)$. Let's find the values of the function $z$ at points $M_1$ and $M_2$:
\begin(aligned) & z_1=z(M_1)=3^2+(-4)^2-12\cdot 3+16\cdot (-4)=-75; \\ & z_2=z(M_2)=(-3)^2+4^2-12\cdot(-3)+16\cdot 4=125. \end(aligned)
We should select the largest and smallest values from those we obtained in the first and second steps. But in this case the choice is small :) We have:
$$ z_(min)=-75; \; z_(max)=125. $$
Answer: $z_(min)=-75; \; z_(max)=$125.
A miniature and fairly simple task of the kind that serves as a life preserver for a floating student. It's mid-July in nature, so it's time to settle down with your laptop on the beach. Early in the morning, the sunbeam of theory began to play, in order to soon focus on practice, which, despite the declared ease, contains shards of glass in the sand. In this regard, I recommend that you conscientiously consider the few examples of this page. To solve practical tasks you must be able to find derivatives and understand the material of the article Monotonicity intervals and extrema of the function.
First, briefly about the main thing. In the lesson about continuity of function I gave the definition of continuity at a point and continuity at an interval. The exemplary behavior of a function on a segment is formulated in a similar way. A function is continuous on an interval if:
1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.
In the second paragraph we talked about the so-called one-sided continuity functions at a point. There are several approaches to defining it, but I will stick to the line I started earlier:
The function is continuous at the point on right, if it is defined at a given point and its right-hand limit coincides with the value of the function at a given point: 
. It is continuous at the point left, if defined at a given point and its left-hand limit is equal to the value at this point: 
Imagine that the green dots are nails with a magic elastic band attached to them:
Mentally take the red line in your hands. Obviously, no matter how far we stretch the graph up and down (along the axis), the function will still remain limited– a fence at the top, a fence at the bottom, and our product grazes in the paddock. Thus, a function continuous on an interval is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and strictly proven. Weierstrass's first theorem....Many people are annoyed that elementary statements are tediously substantiated in mathematics, but this has an important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled a graph into the sky beyond the limits of visibility, this was inserted. Before the invention of the telescope, the limited function in space was not at all obvious! Really, how do you know what awaits us over the horizon? After all, the Earth was once considered flat, so today even ordinary teleportation requires proof =)
According to Weierstrass's second theorem, continuous on a segmentfunction reaches its exact upper bound and yours exact bottom edge .
The number is also called the maximum value of the function on the segment and are denoted by , and the number is the minimum value of the function on the segment marked .
In our case: 
Note
: in theory, recordings are common 
.
Roughly speaking, the largest value is where the highest point on the graph is, and the smallest value is where the lowest point is.
Important! As already emphasized in the article about extrema of the function, greatest function value And smallest function value – NOT THE SAME, What maximum function And minimum function. So, in the example under consideration, the number is the minimum of the function, but not the minimum value.
By the way, what happens outside the segment? Yes, even a flood, in the context of the problem under consideration, this does not interest us at all. The task only involves finding two numbers 
and that's it!
Moreover, the solution is purely analytical, therefore no need to make a drawing!
The algorithm lies on the surface and suggests itself from the above figure:
1) Find the values of the function in critical points, which belong to this segment.
Catch another bonus: here there is no need to check the sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum does not guarantee yet, what is the minimum or maximum value. The demonstration function reaches a maximum and, by the will of fate, the same number is the largest value of the function on the segment. But, of course, such a coincidence does not always occur.
So, in the first step, it is faster and easier to calculate the values of the function at critical points belonging to the segment, without bothering whether there are extrema in them or not.
2) We calculate the values of the function at the ends of the segment.
3) Among the function values found in points 1 and 2, select the smallest and largest number and write down the answer.
We sit down on the shore of the blue sea and hit the shallow water with our heels:
Example 1
Find the largest and smallest values of a function on a segment
Solution:
1) Let's calculate the values of the function at critical points belonging to this segment:
Let's calculate the value of the function at the second critical point:
2) Let’s calculate the values of the function at the ends of the segment: 
3) “Bold” results were obtained with exponents and logarithms, which significantly complicates their comparison. For this reason, let’s arm ourselves with a calculator or Excel and calculate approximate values, not forgetting that: 
Now everything is clear.
Answer:
Fractional-rational example for independent solution:
Example 6
Find the maximum and minimum values of a function on a segment
\(\blacktriangleright\) In order to find the largest/smallest value of a function on the segment \(\) , it is necessary to schematically depict the graph of the function on this segment.
In problems from this subtopic, this can be done using the derivative: find the intervals of increasing (\(f">0\) ) and decreasing (\(f"<0\)
) функции, критические точки (где \(f"=0\)
или \(f"\)
не существует).
\(\blacktriangleright\) Do not forget that the function can take the largest/smallest value not only at the internal points of the segment \(\), but also at its ends.
\(\blacktriangleright\) The largest/smallest value of the function is the coordinate value \(y=f(x)\) .
\(\blacktriangleright\) The derivative of a complex function \(f(t(x))\) is found according to the rule: \[(\Large(f"(x)=f"(t)\cdot t"(x)))\]
\[\begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x)\\ \hline \textbf(1) & c & 0\\&&\\ \textbf(2) & x^a & a\cdot x^(a-1)\\&&\\ \textbf(3) & \ln x & \dfrac1x\\&&\\ \ textbf(4) & \log_ax & \dfrac1(x\cdot \ln a)\\&&\\ \textbf(5) & e^x & e^x\\&&\\ \textbf(6) & a^x & a^x\cdot \ln a\\&&\\ \textbf(7) & \sin x & \cos x\\&&\\ \textbf(8) & \cos x & -\sin x\\ \hline \end(array) \quad \quad \quad \quad \begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x) \\ \hline \textbf(9) & \mathrm(tg)\, x & \dfrac1(\cos^2 x)\\&&\\ \textbf(10) & \mathrm(ctg)\, x & -\ ,\dfrac1(\sin^2 x)\\&&\\ \textbf(11) & \arcsin x & \dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(12) & \ arccos x & -\,\dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(13) & \mathrm(arctg)\, x & \dfrac1(1+x^2)\\ &&\\ \textbf(14) & \mathrm(arcctg)\, x & -\,\dfrac1(1+x^2)\\ \hline \end(array)\]
Task 1 #2357
Task level: Equal to the Unified State Exam
Find the smallest value of the function \(y = e^(x^2 - 4)\) on the segment \([-10; -2]\) .
ODZ: \(x\) – arbitrary.
1) \
\ Thus, \(y" = 0\) for \(x = 0\) .
3) Let’s find intervals of constant sign \(y"\) on the segment under consideration \([-10; -2]\) :
4) Sketch of a graph on the segment \([-10; -2]\) :
Thus, the function reaches its smallest value at \([-10; -2]\) at \(x = -2\) .
\ Total: \(1\) – the smallest value of the function \(y\) on \([-10; -2]\) .
Answer: 1
Task 2 #2355
Task level: Equal to the Unified State Exam
\(y = \sqrt(2)\cdot\sqrt(x^2 + 1)\) on the segment \([-1; 1]\) .
ODZ: \(x\) – arbitrary.
1) \
Let's find critical points (that is, internal points of the function's domain of definition at which its derivative is equal to \(0\) or does not exist): \[\sqrt(2)\cdot\dfrac(x)(\sqrt(x^2 + 1)) = 0\qquad\Leftrightarrow\qquad x = 0\,.\] The derivative exists for any \(x\) .
2) Let’s find intervals of constant sign \(y"\) :
3) Let’s find intervals of constant sign \(y"\) on the segment under consideration \([-1; 1]\) :
4) Sketch of a graph on the segment \([-1; 1]\) :
Thus, the function reaches its greatest value at \([-1; 1]\) at \(x = -1\) or at \(x = 1\) . Let's compare the function values at these points.
\ Total: \(2\) – the largest value of the function \(y\) on \([-1; 1]\) .
Answer: 2
Task 3 #2356
Task level: Equal to the Unified State Exam
Find the smallest value of the function \(y = \cos 2x\) on the segment \(\) .
ODZ: \(x\) – arbitrary.
1) \
Let's find critical points (that is, internal points of the function's domain of definition at which its derivative is equal to \(0\) or does not exist): \[-2\cdot \sin 2x = 0\qquad\Leftrightarrow\qquad 2x = \pi n, n\in\mathbb(Z)\qquad\Leftrightarrow\qquad x = \dfrac(\pi n)(2), n\in\mathbb(Z)\,.\] The derivative exists for any \(x\) .
2) Let’s find intervals of constant sign \(y"\) :
(here there is an infinite number of intervals in which the signs of the derivative alternate).
3) Let’s find intervals of constant sign \(y"\) on the segment under consideration \(\):
4) Sketch of a graph on the segment \(\) :
Thus, the function reaches its smallest value on \(\) at \(x = \dfrac(\pi)(2)\) .
\ Total: \(-1\) – the smallest value of the function \(y\) on \(\) .
Answer: -1
Task 4 #915
Task level: Equal to the Unified State Exam
Find the largest value of the function
\(y = -\log_(17)(2x^2 - 2\sqrt(2)x + 2)\).
ODZ: \(2x^2 - 2\sqrt(2)x + 2 > 0\) . Let's decide on ODZ:
1) Let us denote \(2x^2-2\sqrt(2)x+2=t(x)\) , then \(y(t)=-\log_(17)t\) .
Let's find critical points (that is, internal points of the function's domain of definition at which its derivative is equal to \(0\) or does not exist): \[-\dfrac(1)(\ln 17)\cdot\dfrac(4x-2\sqrt(2))(2x^2-2\sqrt(2)x+2) = 0\qquad\Leftrightarrow\qquad 4x-2\sqrt(2) = 0\]– on the ODZ, from where we find the root \(x = \dfrac(\sqrt(2))(2)\) . The derivative of the function \(y\) does not exist when \(2x^2-2\sqrt(2)x+2 = 0\), but this equation has a negative discriminant, therefore, it has no solutions. In order to find the largest/smallest value of a function, you need to understand how its graph looks schematically.
2) Let’s find intervals of constant sign \(y"\) :
3) Sketch of the graph:
Thus, the function reaches its greatest value at \(x = \dfrac(\sqrt(2))(2)\) :
\(y\left(\dfrac(\sqrt(2))(2)\right) = -\log_(17)1 = 0\),
Total: \(0\) – the largest value of the function \(y\) .
Answer: 0
Task 5 #2344
Task level: Equal to the Unified State Exam
Find the smallest value of the function
\(y = \log_(3)(x^2 + 8x + 19)\).
ODZ: \(x^2 + 8x + 19 > 0\) . Let's decide on ODZ:
1) Let us denote \(x^2 + 8x + 19=t(x)\) , then \(y(t)=\log_(3)t\) .
Let's find critical points (that is, internal points of the function's domain of definition at which its derivative is equal to \(0\) or does not exist): \[\dfrac(1)(\ln 3)\cdot\dfrac(2x+8)(x^2 + 8x + 19) = 0\qquad\Leftrightarrow\qquad 2x+8 = 0\]– on the ODZ, from where we find the root \(x = -4\) . The derivative of the function \(y\) does not exist when \(x^2 + 8x + 19 = 0\), but this equation has a negative discriminant, therefore, it has no solutions. In order to find the largest/smallest value of a function, you need to understand how its graph looks schematically.
2) Let’s find intervals of constant sign \(y"\) :
3) Sketch of the graph:
Thus, \(x = -4\) is the minimum point of the function \(y\) and the smallest value is achieved at it:
\(y(-4) = \log_(3)3 = 1\) .
Total: \(1\) – the smallest value of the function \(y\) .
Answer: 1
Task 6 #917
Task level: More difficult than the Unified State Exam
Find the largest value of the function
\(y = -e^((x^2 - 12x + 36 + 2\ln 2))\).
