How to find the largest value of a function using its derivative. Largest and smallest value of a function

The process of searching for the smallest and largest values ​​of a function on a segment is reminiscent of a fascinating flight around an object (graph of a function) in a helicopter, firing at certain points from a long-range cannon and selecting very special points from these points for control shots. Points are selected in a certain way and according to certain rules. By what rules? We will talk about this further.

If the function y = f(x) is continuous on the interval [ a, b] , then it reaches on this segment least And highest values . This can happen either in extremum points, or at the ends of the segment. Therefore, to find least And the largest values ​​of the function , continuous on the interval [ a, b] , you need to calculate its values ​​in all critical points and at the ends of the segment, and then choose the smallest and largest from them.

Let, for example, you need to determine highest value functions f(x) on the segment [ a, b] . To do this, you need to find all its critical points lying on [ a, b] .

Critical point called the point at which function defined, and her derivative either equal to zero or does not exist. Then you should calculate the values ​​of the function at the critical points. And finally, one should compare the values ​​of the function at critical points and at the ends of the segment ( f(a) And f(b)). The largest of these numbers will be the largest value of the function on the segment [a, b] .

Problems of finding smallest function values .

We look for the smallest and largest values ​​of the function together

Example 1. Find the smallest and largest values ​​of a function on the segment [-1, 2] .

Solution. Find the derivative of this function. Let's equate the derivative to zero () and get two critical points: and . To find the smallest and largest values ​​of a function on a given segment, it is enough to calculate its values ​​at the ends of the segment and at the point, since the point does not belong to the segment [-1, 2]. These function values ​​are as follows: , , . It follows from this that smallest value functions(indicated in red on the graph below), equal to -7, is achieved at the right end of the segment - at point , and greatest(also red on the graph), equals 9, - at the critical point.

If a function is continuous in a certain interval and this interval is not a segment (but is, for example, an interval; the difference between an interval and a segment: the boundary points of the interval are not included in the interval, but the boundary points of the segment are included in the segment), then among the values ​​of the function there may not be to be the smallest and the greatest. So, for example, the function shown in the figure below is continuous on ]-∞, +∞[ and does not have the greatest value.

However, for any interval (closed, open or infinite), the following property of continuous functions is true.

Example 4. Find the smallest and largest values ​​of a function on the segment [-1, 3] .

Solution. We find the derivative of this function as the derivative of the quotient:

.

We equate the derivative to zero, which gives us one critical point: . It belongs to the segment [-1, 3] . To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Let's compare these values. Conclusion: equal to -5/13, at point and highest value equal to 1 at point .

We continue to look for the smallest and largest values ​​of the function together

There are teachers who, on the topic of finding the smallest and largest values ​​of a function, do not give students examples to solve that are more complex than those just discussed, that is, those in which the function is a polynomial or a fraction, the numerator and denominator of which are polynomials. But we will not limit ourselves to such examples, since among teachers there are those who like to force students to think in full (the table of derivatives). Therefore, the logarithm and trigonometric function will be used.

Example 6. Find the smallest and largest values ​​of a function on the segment .

Solution. We find the derivative of this function as derivative of the product :

We equate the derivative to zero, which gives one critical point: . It belongs to the segment. To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Result of all actions: the function reaches its minimum value, equal to 0, at the point and at the point and highest value, equal e², at the point.

Example 7. Find the smallest and largest values ​​of a function on the segment .

Solution. Find the derivative of this function:

We equate the derivative to zero:

The only critical point belongs to the segment. To find the smallest and largest values ​​of a function on a given segment, we find its values ​​at the ends of the segment and at the found critical point:

Conclusion: the function reaches its minimum value, equal to , at the point and highest value, equal , at the point .

In applied extremal problems, finding the smallest (maximum) values ​​of a function, as a rule, comes down to finding the minimum (maximum). But bigger practical interest have not the minimums or maximums themselves, but those values ​​of the argument at which they are achieved. When solving applied problems, an additional difficulty arises - composing functions that describe the phenomenon or process under consideration.

Example 8. A tank with a capacity of 4, having the shape of a parallelepiped with a square base and open at the top, must be tinned. What size should the tank be so that the least amount of material is used to cover it?

Solution. Let x- base side, h- tank height, S- its surface area without cover, V- its volume. The surface area of ​​the tank is expressed by the formula, i.e. is a function of two variables. To express S as a function of one variable, we use the fact that , from where . Substituting the found expression h into the formula for S:

Let's examine this function to its extremum. It is defined and differentiable everywhere in ]0, +∞[ , and

.

We equate the derivative to zero () and find the critical point. In addition, when the derivative does not exist, but this value is not included in the domain of definition and therefore cannot be an extremum point. So, this is the only critical point. Let's check it for the presence of an extremum using the second sufficient sign. Let's find the second derivative. When the second derivative is greater than zero (). This means that when the function reaches a minimum . Since this minimum is the only extremum of this function, it is its smallest value. So, the side of the base of the tank should be 2 m, and its height should be .

Example 9. From point A located on the railway line, to the point WITH, located at a distance from it l, cargo must be transported. The cost of transporting a weight unit per unit distance by rail is equal to , and by highway it is equal to . To what point M lines railway a highway should be built to transport cargo from A V WITH was the most economical (section AB railway is assumed to be straight)?

Let the function y =f(X) is continuous on the interval [ a, b]. As is known, such a function reaches its maximum and minimum values ​​on this segment. The function can take these values ​​either at the internal 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 segment [ 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, when x=A and x = b;

4) from all calculated values ​​of the function, select 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.

Study of a function for convexity and inflection point.

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

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

Algorithm for examining convexity and inflection point:

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

2. Plot critical points on the number line, dividing it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upward, if, then the function is convex downward.

3. If, when passing through a critical point of the second kind, the sign changes 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. Study of a function for asymptotes.

Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point on the graph to this line tends to zero as the point on the graph moves indefinitely from the origin.

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

Definition. The straight line is called vertical asymptote function graphics 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 – break point.

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

Example.

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

General scheme for studying functions and constructing graphs.

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 (if x= 0 and at y = 0).

3. Examine the evenness and oddness of the function ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).

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

5. Find the 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 function graph.

8. Based on the research conducted, construct a graph of the function.

Example. Explore the function and build its graph.

1) D (y) =

x= 4 – break point.

2) When x = 0,

(0; ‒ 5) – point of intersection with oh.

At y = 0,

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

4) We examine for asymptotes.

a) vertical

b) horizontal

c) find the oblique asymptotes where

‒oblique asymptote equation

5) B given equation there is no need to find intervals of monotonicity of the function.

6)

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

What is an extremum of a function and what is the necessary condition for an extremum?

The extremum of a function is the maximum and minimum of the function.

Prerequisite The maximum and minimum (extremum) of a function are as follows: if the function f(x) has an extremum at the point x = a, then at this point the derivative is either zero, or infinite, or does not exist.

This condition is necessary, but not sufficient. The derivative at the point x = a can go to zero, infinity, or not exist without the function having an extremum at this point.

What is the sufficient condition for the extremum of a function (maximum or minimum)?

First condition:

If, in sufficient proximity to the point x = a, the derivative f?(x) is positive to the left of a and negative to the right of a, then at the point x = a the function f(x) has maximum

If, in sufficient proximity to the point x = a, the derivative f?(x) is negative to the left of a and positive to the right of a, then at the point x = a the function f(x) has minimum provided that the function f(x) here is continuous.

Instead, you can use the second sufficient condition for the extremum of a function:

Let at the point x = a the first derivative f?(x) vanish; if the second derivative f??(a) is negative, then the function f(x) has a maximum at the point x = a, if it is positive, then it has a minimum.

What is the critical point of a function and how to find it?

This is the value of the function argument at which the function has an extremum (i.e. maximum or minimum). To find it you need find the derivative function f?(x) and, equating it to zero, solve the equation f?(x) = 0. The roots of this equation, as well as those points at which the derivative of this function does not exist, are critical points, i.e., values ​​of the argument at which there can be an extremum. They can be easily identified by looking at derivative graph: we are interested in those values ​​of the argument at which the graph of the function intersects the abscissa axis (Ox axis) and those at which the graph suffers discontinuities.

For example, let's find extremum of a parabola.

Function y(x) = 3x2 + 2x - 50.

Derivative of the function: y?(x) = 6x + 2

Solve the equation: y?(x) = 0

6x + 2 = 0, 6x = -2, x = -2/6 = -1/3

In this case, the critical point is x0=-1/3. It is with this argument value that the function has extremum. To him find, substitute the found number in the expression for the function instead of “x”:

y0 = 3*(-1/3)2 + 2*(-1/3) - 50 = 3*1/9 - 2/3 - 50 = 1/3 - 2/3 - 50 = -1/3 - 50 = -50.333.

How to determine the maximum and minimum of a function, i.e. its largest and smallest values?

If the sign of the derivative when passing through the critical point x0 changes from “plus” to “minus”, then x0 is maximum point; if the sign of the derivative changes from minus to plus, then x0 is minimum point; if the sign does not change, then at point x0 there is neither a maximum nor a minimum.

For the example considered:

We take an arbitrary value of the argument to the left of the critical point: x = -1

At x = -1, the value of the derivative will be y?(-1) = 6*(-1) + 2 = -6 + 2 = -4 (i.e. the sign is “minus”).

Now we take an arbitrary value of the argument to the right of the critical point: x = 1

At x = 1, the value of the derivative will be y(1) = 6*1 + 2 = 6 + 2 = 8 (i.e. the sign is “plus”).

As you can see, the derivative changed sign from minus to plus when passing through the critical point. This means that at the critical value x0 we have a minimum point.

Largest and smallest value of a function on the interval(on a segment) are found using the same procedure, only taking into account the fact that, perhaps, not all critical points will lie within the specified interval. Those critical points that are outside the interval must be excluded from consideration. If there is only one critical point within the interval, it will have either a maximum or a minimum. In this case, to determine the largest and smallest values ​​of the function, we also take into account the values ​​of the function at the ends of the interval.

For example, let's find the largest and smallest values ​​of the function

y(x) = 3sin(x) - 0.5x

at intervals:

So, the derivative of the function is

y?(x) = 3cos(x) - 0.5

We solve the equation 3cos(x) - 0.5 = 0

cos(x) = 0.5/3 = 0.16667

x = ±arccos(0.16667) + 2πk.

We find critical points on the interval [-9; 9]:

x = arccos(0.16667) - 2π*2 = -11.163 (not included in the interval)

x = -arccos(0.16667) - 2π*1 = -7.687

x = arccos(0.16667) - 2π*1 = -4.88

x = -arccos(0.16667) + 2π*0 = -1.403

x = arccos(0.16667) + 2π*0 = 1.403

x = -arccos(0.16667) + 2π*1 = 4.88

x = arccos(0.16667) + 2π*1 = 7.687

x = -arccos(0.16667) + 2π*2 = 11.163 (not included in the interval)

We find the function values ​​at critical values ​​of the argument:

y(-7.687) = 3cos(-7.687) - 0.5 = 0.885

y(-4.88) = 3cos(-4.88) - 0.5 = 5.398

y(-1.403) = 3cos(-1.403) - 0.5 = -2.256

y(1.403) = 3cos(1.403) - 0.5 = 2.256

y(4.88) = 3cos(4.88) - 0.5 = -5.398

y(7.687) = 3cos(7.687) - 0.5 = -0.885

It can be seen that on the interval [-9; 9] the function has the greatest value at x = -4.88:

x = -4.88, y = 5.398,

and the smallest - at x = 4.88:

x = 4.88, y = -5.398.

On the interval [-6; -3] we have only one critical point: x = -4.88. The value of the function at x = -4.88 is equal to y = 5.398.

Find the value of the function at the ends of the interval:

y(-6) = 3cos(-6) - 0.5 = 3.838

y(-3) = 3cos(-3) - 0.5 = 1.077

On the interval [-6; -3] we have the greatest value of the function

y = 5.398 at x = -4.88

smallest value -

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the convex and concave sides?

To find all the inflection points of the line y = f(x), you need to find the second derivative, equate it to zero (solve the equation) and test all those values ​​of x for which the second derivative is zero, infinite or does not exist. If, when passing through one of these values, the second derivative changes sign, then the graph of the function has an inflection at this point. If it doesn’t change, then there is no bend.

The roots of the equation f? (x) = 0, as well as possible points of discontinuity of the function and the second derivative, divide the domain of definition of the function into a number of intervals. The convexity on each of their intervals is determined by the sign of the second derivative. If the second derivative at a point on the interval under study is positive, then the line y = f(x) is concave upward, and if negative, then downward.

How to find the extrema of a function of two variables?

To find the extrema of the function f(x,y), differentiable in the domain of its specification, you need:

1) find the critical points, and for this - solve the system of equations

fх? (x,y) = 0, fу? (x,y) = 0

2) for each critical point P0(a;b) investigate whether the sign of the difference remains unchanged

for all points (x;y) sufficiently close to P0. If the difference remains positive, then at point P0 we have a minimum, if negative, then we have a maximum. If the difference does not retain its sign, then there is no extremum at point P0.

The extrema of the function are determined similarly for more arguments.

Petite and pretty simple task from the category of those that serve 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 problems 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 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 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 doesn't 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 extremes 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 the 1st and 2nd paragraphs, select the smallest and most large number, write down the answer.

We sit down on the shore 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 instance for independent decision:

Example 6

Find the maximum and minimum values ​​of a function on a segment

Sometimes in problems B15 there are “bad” functions for which it is difficult to find a derivative. Previously, this only happened during sample tests, but now these tasks are so common that they can no longer be ignored when preparing for the real Unified State Exam.

In this case, other techniques work, one of which is monotone.

A function f (x) is said to be monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following holds:

x 1< x 2 ⇒ f (x 1) < f (x 2).

A function f (x) is said to be monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following holds:

x 1< x 2 ⇒ f (x 1) > f ( x 2).

In other words, for an increasing function, the larger x, the larger f(x). For a decreasing function the opposite is true: the larger x, the less f(x).

For example, the logarithm increases monotonically if the base a > 1, and monotonically decreases if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.

f (x) = log a x (a > 0; a ≠ 1; x > 0)

The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:

The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, показательная функция определена для всех чисел, а не только для x > 0:

f (x) = a x (a > 0)

Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where the monotony is broken.

All these functions are never found in pure form. They add polynomials, fractions and other nonsense, which makes it difficult to calculate the derivative. Let's look at what happens in this case.

Parabola vertex coordinates

Most often the function argument is replaced with quadratic trinomial of the form y = ax 2 + bx + c. Its graph is a standard parabola in which we are interested in:

1. The branches of a parabola can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
2. The vertex of a parabola is the extremum point of a quadratic function at which this function takes its minimum (for a > 0) or maximum (a< 0) значение.

Of greatest interest is vertex of 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, let us formulate the key rule:

Extremum points of a quadratic trinomial and complex function, which it is included in, coincide. Therefore, you can look for x 0 for a quadratic trinomial, and forget about the function.

From the above reasoning, it remains unclear which point we get: maximum or minimum. However, the tasks are specifically designed so that this does not matter. Judge for yourself:

1. There is no segment in the problem statement. Therefore, there is no need to calculate f(a) and f(b). It remains to consider only the extremum points;
2. But there is only one such point - this is the vertex of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.

Thus, solving the problem is greatly simplified and comes down to just two steps:

1. Write out the equation of the parabola y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a ;
2. Find the value of the original function at this point: f (x 0). If no additional conditions no, that will be the answer.

At first glance, this algorithm and its rationale may seem complex. I deliberately do not post a “bare” solution diagram, since thoughtless application of such rules is fraught with errors.

Let's look at real problems from trial Unified State Exam in mathematics - this is where this technique is found most often. At the same time, we will make sure that in this way many B15 problems become almost oral.

Under the root stands quadratic function y = x 2 + 6x + 13. The graph of this function is a parabola with branches upward, since the coefficient a = 1 > 0.

Vertex of the parabola:

x 0 = −b /(2a ) = −6/(2 1) = −6/2 = −3

Since the branches of the parabola are directed upward, at the point x 0 = −3 the function y = x 2 + 6x + 13 takes on its minimum value.

The root increases monotonically, which means 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 there is again a quadratic function: y = x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.

Vertex of the parabola:

x 0 = −b /(2a ) = −2/(2 1) = −2/2 = −1

So, at the point x 0 = −1 the quadratic function takes on its minimum value. But the function y = log 2 x is monotonic, so:

y min = y (−1) = log 2 ((−1) 2 + 2 · (−1) + 9) = ... = log 2 8 = 3

The exponent contains the 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 the greatest value will be at the found point x 0 = −2:

An attentive reader will probably notice that we did not write out the range of permissible values ​​of the root and logarithm. But this was not required: inside there are functions whose values ​​are always positive.

Corollaries from the domain of a function

Sometimes simply finding the vertex of the parabola is not enough to solve Problem B15. The value you are looking for may lie at the end of the segment, and not at all at the extremum point. If the problem does not indicate a segment at all, look at range of acceptable values original function. Namely:

Please note again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how this works with specific examples:

Task. Find the largest value of the function:

Under the root is again a quadratic function: y = 3 − 2x − x 2 . Its graph is a parabola, but branches down because a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический square root of a negative number does not exist.

We write out the range of permissible values ​​(APV):

3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; 1]

Now let's find the vertex of the parabola:

x 0 = −b /(2a ) = −(−2)/(2 · (−1)) = 2/(−2) = −1

The point x 0 = −1 belongs to the segment ODZ - and this is good. Now we calculate 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 there cannot be negative numbers in the logarithm, so we write out the ODZ:

6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)

Please note: the inequality is strict, so the ends do not belong to the ODZ. This differs the logarithm from the root, where the ends of the segment suit us quite well.

We are looking for the vertex of the parabola:

x 0 = −b /(2a ) = −6/(2 · (−1)) = −6/(−2) = 3

The vertex of the parabola fits according to the ODZ: x 0 = 3 ∈ (1; 5). But since we are not interested in the ends of the segment, we calculate 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