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How to find the smallest value of a function from an equation. The largest and smallest values of a function of two variables in a closed region

A miniature and rather simple task of the kind that serves as a lifeline for a floating student. In nature, the sleepy realm of mid-July, so it's time to settle down with a laptop on the beach. Played early in the morning sunbeam theory in order to soon focus on practice, which, despite its claimed lightness, contains glass fragments in the sand. In this regard, I recommend conscientiously consider a few examples of this page. To solve practical tasks, you need to be able to find derivatives and understand the material of the article Intervals of monotonicity and extrema of a function.

First, briefly about the main thing. In a lesson about function continuity I gave the definition of continuity at a point and continuity on an interval. The exemplary behavior of a function on a segment is formulated similarly. A function is continuous on a segment if:

1) it is continuous on the interval ;
2) continuous at a point on right and at the point left.

The second paragraph deals with the so-called unilateral continuity functions at a point. There are several approaches to its definition, but I will stick to the line started earlier:

The function is continuous at a 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 that point:

Imagine that the green dots are the nails on which the magic rubber band is attached:

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 hedge above, a hedge below, and our product grazes in a paddock. Thus, a function continuous on a segment is bounded on it. In the course of mathematical analysis, this seemingly simple fact is stated and rigorously proved Weierstrass' first theorem.... Many people are annoyed that elementary statements are tediously substantiated in mathematics, but there is important meaning. Suppose a certain inhabitant of the terry Middle Ages pulled the 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! Indeed, how do you know what awaits us beyond the horizon? After all, once the Earth was considered flat, so today even ordinary teleportation requires proof =)

According to second Weierstrass theorem, continuous on the segmentfunction reaches its exact top edge and his exact bottom edge .

The number is also called the maximum value of the function on the segment and denoted by , and the number - the minimum value of the function on the segment marked .

In our case:

Note : in theory, records are common .

Roughly speaking, highest value is located where the high point graphics, and the smallest - where is the lowest point.

Important! As already pointed out in the article on extrema of the function, the largest value of the function And smallest function value – NOT THE SAME, What function maximum And function minimum. So, in this example, the number is the minimum of the function, but not the minimum value.

By the way, what happens outside the segment? Yes, even the flood, in the context of the problem under consideration, this does not interest us at all. The task involves only finding two numbers and that's it!

Moreover, the solution is purely analytical, therefore, no need to draw!

The algorithm lies on the surface and suggests itself from the above figure:

1) Find the function values in critical points, that belong to this segment.

Catch one more goodie: there is no need to check a sufficient condition for an extremum, since, as just shown, the presence of a minimum or maximum not yet guaranteed what is the minimum or maximum value. The demonstration function reaches its maximum and, by the will of fate, the same number is the largest value of the function on the interval . But, of course, such a coincidence does not always take place.

So, at 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 they have extrema or not.

2) We calculate the values of the function at the ends of the segment.

3) Among the values of the function found in the 1st and 2nd paragraphs, we select the smallest and most big number, write down the answer.

We sit on the shore of the blue sea and hit the heels in shallow water:

Example 1

Find the largest and smallest value functions on the segment

Solution:
1) Calculate the values of the function at critical points belonging to this segment:

Let us calculate the value of the function at the second critical point:

2) Calculate the values of the function at the ends of the segment:

3) "Bold" results were obtained with exponentials and logarithms, which significantly complicates their comparison. For this reason, we will arm ourselves with a calculator or Excel and calculate the approximate values, not forgetting that:

Now everything is clear.

Answer:

Fractional-rational instance for independent solution:

Example 6

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

The largest (smallest) value of the function is the largest (smallest) accepted value of the ordinate in the considered interval.

To find the largest or smallest value of a function, you need to:

Check which stationary points are included in the given segment.
Calculate the value of the function at the ends of the segment and at stationary points from step 3
Choose from the results obtained the largest or smallest value.

To find the maximum or minimum points, you need to:

Find the derivative of the function $f"(x)$
Find stationary points by solving the equation $f"(x)=0$
Factorize the derivative of a function.
Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the obtained intervals, using the notation of clause 3.
Find the maximum or minimum points according to the rule: if at a point the derivative changes sign from plus to minus, then this will be the maximum point (if from minus to plus, then this will be the minimum point). In practice, it is convenient to use the image of arrows on the intervals: on the interval where the derivative is positive, the arrow is drawn upwards and vice versa.

Table of derivatives of some elementary functions:

Function	Derivative
$c$	$0$
$x$	$1$
$x^n, n∈N$	$nx^(n-1), n∈N$
$(1)/(x)$	$-(1)/(x^2)$
$(1)/x(^n), n∈N$	$-(n)/(x^(n+1)), n∈N$
$√^n(x), n∈N$	$(1)/(n√^n(x^(n-1)), n∈N$
$sinx$	$cosx$
$cosx$	$-sinx$
$tgx$	$(1)/(cos^2x)$
$ctgx$	$-(1)/(sin^2x)$
$cos^2x$	$-sin2x$
$sin^2x$	$sin2x$
$e^x$	$e^x$
$a^x$	$a^xlna$
$lnx$	$(1)/(x)$
$log_(a)x$	$(1)/(xlna)$

Basic rules of differentiation

1. The derivative of the sum and difference is equal to the derivative of each term

$(f(x) ± g(x))′= f′(x)± g′(x)$

Find the derivative of the function $f(x) = 3x^5 – cosx + (1)/(x)$

The derivative of the sum and difference is equal to the derivative of each term

$f′(x)=(3x^5)′–(cosx)′+((1)/(x))"=15x^4+sinx-(1)/(x^2)$

2. Derivative of a product.

$(f(x)∙g(x))′=f′(x)∙g(x)+f(x)∙g(x)′$

Find the derivative $f(x)=4x∙cosx$

$f′(x)=(4x)′∙cosx+4x∙(cosx)′=4∙cosx-4x∙sinx$

3. Derivative of the quotient

$((f(x))/(g(x)))"=(f^"(x)∙g(x)-f(x)∙g(x)")/(g^2(x) )$

Find the derivative $f(x)=(5x^5)/(e^x)$

$f"(x)=((5x^5)"∙e^x-5x^5∙(e^x)")/((e^x)^2)=(25x^4∙e^x- 5x^5∙e^x)/((e^x)^2)$

4. The derivative of a complex function is equal to the product of the derivative of the external function and the derivative of the internal function

$f(g(x))′=f′(g(x))∙g′(x)$

$f′(x)=cos′(5x)∙(5x)′= - sin(5x)∙5= -5sin(5x)$

Find the minimum point of the function $y=2x-ln⁡(x+11)+4$

1. Find the ODZ of the function: $x+11>0; x>-11$

2. Find the derivative of the function $y"=2-(1)/(x+11)=(2x+22-1)/(x+11)=(2x+21)/(x+11)$

3. Find stationary points by equating the derivative to zero

$(2x+21)/(x+11)=0$

A fraction is zero if the numerator is zero and the denominator is not zero

$2x+21=0; x≠-11$

4. Draw a coordinate line, place stationary points on it and determine the signs of the derivative in the obtained intervals. To do this, we substitute into the derivative any number from the extreme right region, for example, zero.

$y"(0)=(2∙0+21)/(0+11)=(21)/(11)>0$

5. At the minimum point, the derivative changes sign from minus to plus, therefore, the $-10.5$ point is the minimum point.

Answer: $-10.5$

Find the maximum value of the function $y=6x^5-90x^3-5$ on the segment $[-5;1]$

1. Find the derivative of the function $y′=30x^4-270x^2$

2. Equate the derivative to zero and find stationary points

$30x^4-270x^2=0$

Let's take the common factor $30x^2$ out of brackets

$30x^2(x^2-9)=0$

$30x^2(x-3)(x+3)=0$

Set each factor equal to zero

$x^2=0 ; x-3=0; x+3=0$

$x=0;x=3;x=-3$

3. Choose stationary points that belong to the given segment $[-5;1]$

Stationary points $x=0$ and $x=-3$ are suitable for us

4. Calculate the value of the function at the ends of the segment and at stationary points from item 3

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.

The necessary condition for the maximum and minimum (extremum) of the function is 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 vanish, go to infinity, or not exist without the function having an extremum at this point.

What is the sufficient condition for the extremum of the 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 itself, 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 itself, the function f(x) has minimum provided that the function f(x) is continuous here.

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

Let at the point x = and the first derivative f? (x) vanishes; if the second derivative f??(а) is negative, then the function f(x) has a maximum at the point x = a, if it is positive, then 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., the values of the argument at which there may 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 breaks.

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

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

Function derivative: y?(x) = 6x + 2

We 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 for this value of the argument that the function has extremum. To get it find, we 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 changes from “plus” to “minus” when passing through the critical point x0, 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 the point x0 there is neither a maximum nor a minimum.

For the considered example:

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

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

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

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

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

The largest and smallest value of the function on the interval(on the segment) are found by 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 inside the interval, it will either have 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) \u003d 3 sin (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 \u003d ± arccos (0.16667) + 2πk.

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

x \u003d arccos (0.16667) - 2π * 2 \u003d -11.163 (not included in the interval)

x \u003d -arccos (0.16667) - 2π * 1 \u003d -7.687

x \u003d arccos (0.16667) - 2π * 1 \u003d -4.88

x \u003d -arccos (0.16667) + 2π * 0 \u003d -1.403

x \u003d arccos (0.16667) + 2π * 0 \u003d 1.403

x \u003d -arccos (0.16667) + 2π * 1 \u003d 4.88

x \u003d arccos (0.16667) + 2π * 1 \u003d 7.687

x \u003d -arccos (0.16667) + 2π * 2 \u003d 11.163 (not included in the interval)

We find the values of the function 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 y = 5.398.

We 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 largest value of the function

y = 5.398 at x = -4.88

the smallest value is

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the sides of convexity and concavity?

To find all the inflection points of the line y \u003d 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 does not change, then there is no inflection.

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 the function into a number of intervals. The convexity at 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 upwards here, and if it is negative, then downwards.

How to find extrema of a function of two variables?

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

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

fx? (x,y) = 0, fy? (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 retains a positive sign, then at the point P0 we have a minimum, if negative, then a maximum. If the difference does not retain its sign, then there is no extremum at the point Р0.

Similarly, the extrema of the function are determined for a larger number of arguments.

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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.

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

What is the sufficient condition for the extremum of the 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 itself, the function f(x) has maximum

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

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

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

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

Function derivative: y?(x) = 6x + 2

We 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 for this value of the argument that the function has extremum. To get it find, we 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?

For the considered example:

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

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

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

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

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

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

y (x) \u003d 3 sin (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 \u003d ± arccos (0.16667) + 2πk.

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

x \u003d arccos (0.16667) - 2π * 2 \u003d -11.163 (not included in the interval)

x \u003d -arccos (0.16667) - 2π * 1 \u003d -7.687

x \u003d arccos (0.16667) - 2π * 1 \u003d -4.88

x \u003d -arccos (0.16667) + 2π * 0 \u003d -1.403

x \u003d arccos (0.16667) + 2π * 0 \u003d 1.403

x \u003d -arccos (0.16667) + 2π * 1 \u003d 4.88

x \u003d arccos (0.16667) + 2π * 1 \u003d 7.687

x \u003d -arccos (0.16667) + 2π * 2 \u003d 11.163 (not included in the interval)

We find the values of the function 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 y = 5.398.

We 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 largest value of the function

y = 5.398 at x = -4.88

the smallest value is

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the sides of convexity and concavity?

How to find extrema of a function of two variables?

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

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

fx? (x,y) = 0, fy? (x,y) = 0

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

Similarly, the extrema of the function are determined for a larger number of arguments.

How to find the largest and smallest values of a function on a segment?

For this we follow the well-known algorithm:

1 . We find ODZ functions.

2 . Finding the derivative of a function

3 . Equate the derivative to zero

4 . We find the intervals at which the derivative retains its sign, and from them we determine the intervals of increase and decrease of the function:

If on the interval I the derivative of the function 0" title="f^(prime)(x)>0">, то функция !} increases over this interval.

If on the interval I the derivative of the function , then the function decreases over this interval.

5 . We find maximum and minimum points of the function.

IN the function maximum point, the derivative changes sign from "+" to "-".

IN minimum point of the functionderivative changes sign from "-" to "+".

6 . We find the value of the function at the ends of the segment,

then we compare the value of the function at the ends of the segment and at the maximum points, and choose the largest of them if you need to find the largest value of the function
or we compare the value of the function at the ends of the segment and at the minimum points, and choose the smallest of them if you need to find the smallest value of the function

However, depending on how the function behaves on the interval, this algorithm can be significantly reduced.

Consider the function . The graph of this function looks like this:

Let's look at some examples of solving problems from open bank assignments for

1 . Task B15 (#26695)

On the cut.

1. The function is defined for all real values of x

Obviously, this equation has no solutions, and the derivative is positive for all values of x. Therefore, the function increases and takes on the largest value at the right end of the interval, that is, at x=0.

Answer: 5.

2 . Task B15 (No. 26702)

Find the largest value of a function on the segment.

1.ODZ function title="x(pi)/2+(pi)k, k(in)(bbZ)">!}

The derivative is zero at , however, at these points it does not change sign:

Therefore, title="3/(cos^2(x))>=3">, значит, title="3/(cos^2(x))-3>=0">, то есть производная при всех допустимых значених х неотрицательна, следовательно, функция !} increases and takes the greatest value at the right end of the interval, at .

To make it clear why the derivative does not change sign, we transform the expression for the derivative as follows:

Title="y^(prime)=3/(cos^2(x))-3=(3-3cos^2(x))/(cos^2(x))=(3sin^2 (x))/(cos^2(x))=3tg^2(x)>=0">!}

Answer: 5.

3 . Task B15 (#26708)

Find the smallest value of the function on the interval .

1. ODZ functions: title="x(pi)/2+(pi)k, k(in)(bbZ)">!}

Let's place the roots of this equation on a trigonometric circle.

The interval contains two numbers: and

Let's put up the signs. To do this, we determine the sign of the derivative at the point x=0: . When passing through the points and the derivative changes sign.

Let's depict the change of signs of the derivative of the function on the coordinate line:

Obviously, the point is a minimum point (where the derivative changes sign from "-" to "+"), and in order to find the smallest value of the function on the interval, you need to compare the values of the function at the minimum point and at the left end of the segment, .

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