Find the value of the function on the interval. How to solve B15 problems without derivatives
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, exponential function defined for all numbers, not just 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.
Let's consider the real problems from trial exam in mathematics - this is where this technique occurs most often. At the same time, we will make sure that in this way many problems of B15 become almost verbal.
Under the root is 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:
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 monotonic, so highest 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:
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 Square root from a negative number does not exist.
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.
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 = 6x − x 2 − 5. This is a parabola with branches down, but in the logarithm it cannot be negative numbers, 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's see how to explore a function using a graph. It turns out that looking at the graph, you can find out everything that interests us, namely:
- function scope
- function range
- function zeros
- periods of increase and decrease
- high and low points
- the largest and smallest value of the function on the segment.
Let's clarify the terminology:
Abscissa is the horizontal coordinate of the point.
Ordinate- vertical coordinate.
abscissa- the horizontal axis, most often called the axis.
Y-axis- vertical axis, or axis.
Argument is an independent variable on which the values of the function depend. Most often indicated.
In other words, we ourselves choose , substitute in the function formula and get .
Domain functions - the set of those (and only those) values of the argument for which the function exists.
Denoted: or .
In our figure, the domain of the function is a segment. It is on this segment that the graph of the function is drawn. Only here this function exists.
Function range is the set of values that the variable takes. In our figure, this is a segment - from the lowest to the highest value.
Function zeros- points where the value of the function is equal to zero, i.e. . In our figure, these are the points and .
Function values are positive where . In our figure, these are the intervals and .
Function values are negative where . We have this interval (or interval) from to.
The most important concepts - increasing and decreasing function on some set. As a set, you can take a segment, an interval, a union of intervals, or the entire number line.
Function increases
In other words, the more , the more , that is, the graph goes to the right and up.
Function decreases on the set if for any and belonging to the set the inequality implies the inequality .
For a decreasing function, a larger value corresponds to a smaller value. The graph goes right and down.
In our figure, the function increases on the interval and decreases on the intervals and .
Let's define what is maximum and minimum points of the function.
Maximum point- this is an internal point of the domain of definition, such that the value of the function in it is greater than in all points sufficiently close to it.
In other words, the maximum point is such a point, the value of the function at which more than in neighboring ones. This is a local "hill" on the chart.
In our figure - the maximum point.
Low point- an internal point of the domain of definition, such that the value of the function in it is less than in all points sufficiently close to it.
That is, the minimum point is such that the value of the function in it is less than in neighboring ones. On the graph, this is a local “hole”.
In our figure - the minimum point.
The point is the boundary. It is not an interior point of the domain of definition and therefore does not fit the definition of a maximum point. After all, she has no neighbors on the left. In the same way, there can be no minimum point on our chart.
The maximum and minimum points are collectively called extremum points of the function. In our case, this is and .
But what if you need to find, for example, function minimum on the cut? In this case, the answer is: because function minimum is its value at the minimum point.
Similarly, the maximum of our function is . It is reached at the point .
We can say that the extrema of the function are equal to and .
Sometimes in tasks you need to find the largest and smallest values of the function on a given segment. They do not necessarily coincide with extremes.
In our case smallest function value on the interval is equal to and coincides with the minimum of the function. But its largest value on this segment is equal to . It is reached at the left end of the segment.
In any case, the largest and smallest values of a continuous function on a segment are achieved either at the extremum points or at the ends of the segment.
And to solve it, you need minimal knowledge of the topic. The next academic year, everyone wants to go on vacation, and to bring this moment closer, I immediately get down to business:
Let's start with the area. The area referred to in the condition is limited closed set of points in the plane. For example, a set of points bounded by a triangle, including the ENTIRE triangle (if from borders“Poke out” at least one point, then the area will no longer be closed). In practice, there are also areas of rectangular, round and slightly more complex shapes. It should be noted that in theory mathematical analysis strict definitions are given limitations, isolation, boundaries, etc., but I think everyone is aware of these concepts on an intuitive level, and more is not needed now.
The flat area is standardly denoted by the letter , and, as a rule, is given analytically - by several equations (not necessarily linear); less often inequalities. A typical verbal turnover: "closed arealimited by lines".
An integral part of the task under consideration is the construction of the area on the drawing. How to do it? It is necessary to draw all the listed lines (in this case 3 straight) and analyze what happened. The desired area is usually lightly hatched, and its border is highlighted with a bold line: 
The same area can be set linear inequalities: , which for some reason are more often written as an enumeration list, and not system.
Since the boundary belongs to the region, then all inequalities, of course, non-strict.
And now the crux of the matter. Imagine that the axis goes straight to you from the origin of coordinates. Consider a function that continuous in each area point. The graph of this function is surface, and the small happiness is that in order to solve today's problem, we do not need to know what this surface looks like at all. It can be located above, below, cross the plane - all this is not important. And the following is important: according to Weierstrass theorems, continuous v limited closed area, the function reaches its maximum (of the "highest") and least (of the "lowest") values to be found. These values are achieved or v stationary points, belonging to the regionD , or at points that lie on the boundary of this region. From which follows a simple and transparent solution algorithm:
Example 1
In a limited enclosed area
Solution: First of all, you need to depict the area on the drawing. Unfortunately, it is technically difficult for me to make an interactive model of the problem, and therefore I will immediately give the final illustration, which shows all the "suspicious" points found during the study. Usually they are put down one after another as they are found: 
Based on the preamble, the decision can be conveniently divided into two points:
I) Let's find stationary points. This is a standard action that we have repeatedly performed in the lesson. about extrema of several variables:
Found stationary point belongs areas: (mark it on the drawing), which means that we should calculate the value of the function at a given point:
- as in the article The largest and smallest values of a function on a segment, I will highlight the important results in bold. In a notebook, it is convenient to circle them with a pencil.
Pay attention to our second happiness - there is no point in checking sufficient condition for an extremum. Why? Even if at the point the function reaches, for example, local minimum, then this DOES NOT MEAN that the resulting value will be minimal throughout the region (see the beginning of the lesson about unconditional extremes) .
What if the stationary point does NOT belong to the area? Almost nothing! It should be noted that and go to the next paragraph.
II) We investigate the border of the region.
Since the border consists of the sides of a triangle, it is convenient to divide the study into 3 subparagraphs. But it is better to do it not anyhow. From my point of view, at first it is more advantageous to consider segments parallel to the coordinate axes, and first of all, those lying on the axes themselves. To catch the whole sequence and logic of actions, try to study the ending "in one breath":
1) Let's deal with the lower side of the triangle. To do this, we substitute directly into the function:
Alternatively, you can do it like this: 
Geometrically, this means that coordinate plane (which is also given by the equation)"cut out" from surfaces"spatial" parabola, the top of which immediately falls under suspicion. Let's find out where is she:
- the resulting value "hit" in the area, and it may well be that at the point (mark on the drawing) the function reaches the largest or smallest value in the entire area. Anyway, let's do the calculations:
Other "candidates" are, of course, the ends of the segment. Calculate the values of the function at points 
(mark on the drawing):
Here, by the way, you can perform an oral mini-check on the "stripped down" version: 
2) To study the right side of the triangle, we substitute it into the function and “put things in order there”:
Here we immediately perform a rough check, “ringing” the already processed end of the segment:
, perfect.
The geometric situation is related to the previous point:
- the resulting value also “entered the scope of our interests”, which means that we need to calculate what the function is equal to at the point that has appeared:
Let's examine the second end of the segment:
Using the function 
, let's check: 
3) Everyone probably knows how to explore the remaining side. We substitute into the function and carry out simplifications:
Line ends 
have already been investigated, but on the draft we still check whether we found the function correctly 
:
– coincided with the result of the 1st subparagraph;
– coincided with the result of the 2nd subparagraph.
It remains to find out if there is something interesting inside the segment :
- there is! Substituting a straight line into the equation, we get the ordinate of this “interestingness”:
We mark a point on the drawing and find the corresponding value of the function:
Let's control the calculations according to the "budget" version 
:
, order.
And the final step: CAREFULLY look through all the "fat" numbers, I recommend even beginners to make a single list: 
from which we choose the largest and smallest values. Answer write in the style of the problem of finding the largest and smallest values of the function on the interval:
I'll comment again just in case. geometric sense result:
- here is the most high point surfaces in the area ;
- here is the lowest point of the surface in the area.
In the analyzed problem, we found 7 “suspicious” points, but their number varies from task to task. For a triangular region, the minimum "exploration set" consists of three points. This happens when the function, for example, sets plane- it is quite clear that there are no stationary points, and the function can reach the maximum / minimum values only at the vertices of the triangle. But there are no such examples once, twice - usually you have to deal with some kind of surface of the 2nd order.
If you solve such tasks a little, then triangles can make your head spin, and therefore I have prepared unusual examples for you to make it square :))
Example 2
Find the largest and smallest values of a function 
in a closed area bounded by lines
Example 3
Find the largest and smallest values of a function in a bounded closed area.
Special attention pay attention to the rational order and technique of studying the boundary of the area, as well as to the chain of intermediate checks, which will almost completely avoid computational errors. Generally speaking, you can solve it as you like, but in some problems, for example, in the same Example 2, there is every chance to significantly complicate your life. An approximate example of finishing assignments at the end of the lesson.
We systematize the solution algorithm, otherwise, with my diligence of a spider, it somehow got lost in a long thread of comments of the 1st example:
- At the first step, we build an area, it is desirable to shade it, and highlight the border with a bold line. During the solution, points will appear that need to be put on the drawing.
– Find stationary points and calculate the values of the function only in those, which belong to the area . The obtained values are highlighted in the text (for example, circled with a pencil). If the stationary point does NOT belong to the area, then we mark this fact with an icon or verbally. If there are no stationary points at all, then we draw a written conclusion that they are absent. In any case, this item cannot be skipped!
– Exploring the border area. First, it is advantageous to deal with straight lines that are parallel to the coordinate axes (if there are any). The function values calculated at "suspicious" points are also highlighted. A lot has been said about the solution technique above and something else will be said below - read, re-read, delve into!
- From the selected numbers, select the largest and smallest values \u200b\u200band give an answer. Sometimes it happens that the function reaches such values at several points at once - in this case, all these points should be reflected in the answer. Let, for example, 
and it turned out that this is the smallest value. Then we write that
The final examples are devoted to other useful ideas that will come in handy in practice:
Example 4
Find the largest and smallest values of a function in a closed area 
.
I have kept the author's formulation, in which the area is given as a double inequality. This condition can be written in an equivalent system or in a more traditional form for this problem: 
I remind you that with non-linear we encountered inequalities on , and if you do not understand the geometric meaning of the entry, then please do not delay and clarify the situation right now ;-)
Solution, as always, begins with the construction of the area, which is a kind of "sole": 
Hmm, sometimes you have to gnaw not only the granite of science ....
I) Find stationary points: 
Idiot's dream system :) 
The stationary point belongs to the region, namely, lies on its boundary.
And so, it’s nothing ... fun lesson went - that’s what it means to drink the right tea =)
II) We investigate the border of the region. Without further ado, let's start with the x-axis:
1) If , then
Find where the top of the parabola is:
- Appreciate such moments - "hit" right to the point, from which everything is already clear. But don't forget to check: 
Let's calculate the values of the function at the ends of the segment: 
2) We will deal with the lower part of the “sole” “in one sitting” - without any complexes we substitute it into the function, moreover, we will only be interested in the segment:
Control:
Now this is already bringing some revival to the monotonous ride on a knurled track. Let's find the critical points:
We decide quadratic equation do you remember this one? ... However, remember, of course, otherwise you would not read these lines =) If in the two previous examples calculations were convenient in decimal fractions(which, by the way, is rare), then here we are waiting for the usual common fractions. We find the “x” roots and, using the equation, determine the corresponding “game” coordinates of the “candidate” points: 
Let's calculate the values of the function at the found points: 
Check the function yourself.
Now we carefully study the won trophies and write down answer:
Here are the "candidates", so the "candidates"!
For a standalone solution:
Example 5
Find the smallest and largest values of a function 
in a closed area 
An entry with curly braces reads like this: “a set of points such that”.
Sometimes in such examples they use Lagrange multiplier method, but the real need to use it is unlikely to arise. So, for example, if a function with the same area "de" is given, then after substitution into it - with a derivative of no difficulties; moreover, everything is drawn up in a “one line” (with signs) without the need to consider the upper and lower semicircles separately. But, of course, there are more complicated cases, where without the Lagrange function (where , for example, is the same circle equation) it's hard to get by - how hard it is to get by without a good rest!
All the best to pass the session and see you soon next season!
Solutions and answers:
Example 2: Solution: draw the area on the drawing:
\(\blacktriangleright\) In order to find the maximum/smallest value of a function on the segment \(\) , it is necessary to schematically depict the graph of the function on this segment.
In the problems from this subtopic, this can be done using the derivative: find the intervals of increase (\(f">0\) ) and decrease (\(f"<0\)
) функции, критические точки (где \(f"=0\)
или \(f"\)
не существует).
\(\blacktriangleright\) Do not forget that the function can take the maximum/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 value of the coordinate \(y=f(x)\) .
\(\blacktriangleright\) The derivative of a complex function \(f(t(x))\) is searched 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 Examination
Find the smallest value of the function \(y = e^(x^2 - 4)\) on the interval \([-10; -2]\) .
ODZ: \(x\) - arbitrary.
1) \
\ So \(y" = 0\) when \(x = 0\) .
3) Let's find intervals of constant sign \(y"\) on the considered segment \([-10; -2]\) :
4) Sketch of the graph on the segment \([-10; -2]\) :
Thus, the function reaches its smallest value on \([-10; -2]\) at \(x = -2\) .
\ Total: \(1\) is the smallest value of the function \(y\) on \([-10; -2]\) .
Answer: 1
Task 2 #2355
Task level: Equal to the Unified State Examination
\(y = \sqrt(2)\cdot\sqrt(x^2 + 1)\) on the segment \([-1; 1]\) .
ODZ: \(x\) - arbitrary.
1) \
Let's find the critical points (that is, the internal points of the domain of the function, in 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) Find the intervals of constant sign \(y"\) :
3) Let's find intervals of constant sign \(y"\) on the considered segment \([-1; 1]\) :
4) Sketch of the graph on the segment \([-1; 1]\) :
Thus, the function reaches its maximum value on \([-1; 1]\) in \(x = -1\) or in \(x = 1\) . Let's compare the values of the function at these points.
\ Total: \(2\) is the largest value of the function \(y\) on \([-1; 1]\) .
Answer: 2
Task 3 #2356
Task level: Equal to the Unified State Examination
Find the smallest value of the function \(y = \cos 2x\) on the interval \(\) .
ODZ: \(x\) - arbitrary.
1) \
Let's find the critical points (that is, the internal points of the domain of the function, in 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) Find the 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 constancy \(y"\) on the considered segment \(\) :
4) Sketch of the graph on the segment \(\) :
Thus, the function reaches its smallest value on \(\) at \(x = \dfrac(\pi)(2)\) .
\ Total: \(-1\) is the smallest value of the function \(y\) on \(\) .
Answer: -1
Task 4 #915
Task level: Equal to the Unified State Examination
Find the largest value of a 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) Denote \(2x^2-2\sqrt(2)x+2=t(x)\) , then \(y(t)=-\log_(17)t\) .
Let's find the critical points (that is, the internal points of the domain of the function, in 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 for \(2x^2-2\sqrt(2)x+2 = 0\) , but this equation has a negative discriminant, so 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) Find the intervals of constant sign \(y"\) :
3) Graphic sketch:
Thus, the function reaches its maximum value at \(x = \dfrac(\sqrt(2))(2)\) :
\(y\left(\dfrac(\sqrt(2))(2)\right) = -\log_(17)1 = 0\),
Total: \(0\) is the largest value of the function \(y\) .
Answer: 0
Task 5 #2344
Task level: Equal to the Unified State Examination
Find the smallest value of a function
\(y = \log_(3)(x^2 + 8x + 19)\).
ODZ: \(x^2 + 8x + 19 > 0\) . Let's decide on ODZ:
1) Denote \(x^2 + 8x + 19=t(x)\) , then \(y(t)=\log_(3)t\) .
Let's find the critical points (that is, the internal points of the domain of the function, in 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 \u003d -4 \) . The derivative of the function \(y\) does not exist for \(x^2 + 8x + 19 = 0\) , 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) Find the intervals of constant sign \(y"\) :
3) Graphic sketch:
Thus, \(x = -4\) is the minimum point of the function \(y\) and the smallest value is reached in it:
\(y(-4) = \log_(3)3 = 1\) .
Total: \(1\) is the smallest value of the function \(y\) .
Answer: 1
Task 6 #917
Task level: More difficult than the exam
Find the largest value of a function
\(y = -e^((x^2 - 12x + 36 + 2\ln 2))\).
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 *) = 0 is necessary condition 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 or 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
