The smallest value of the function f x. The largest and smallest value of the function. Scheme for finding the largest and smallest values of the function $f(x)$ on the segment $$

07.08.2020

The figures below show where the function can reach its smallest and largest value. In the left figure, the smallest and largest values are fixed at the points of the local minimum and maximum of the function. In the right figure - at the ends of the segment.

If the function y = f(x) continuous on the segment [ a, b] , then it reaches on this segment least and highest values . This, as already mentioned, 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 of them.

Let, for example, it is required to determine the maximum value of the function f(x) on the segment [ a, b] . To do this, find all its critical points lying on [ a, b] .

critical point is called the point at which function defined, and her derivative is either zero or does not exist. Then you should calculate the values of the function at 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] .

The problem of finding the smallest values of the function .

We are looking 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] .

Decision. We find the derivative of this function. 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 the following: , , . It follows that smallest function value(marked in red on the graph below), equal to -7, is reached at the right end of the segment - at the point , and greatest(also red on the graph), is equal to 9, - at the critical point .

If the 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 be the smallest and largest. So, for example, the function depicted in the figure below is continuous on ]-∞, +∞[ and does not have the largest value.

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

For self-checking during calculations, you can use online derivatives calculator .

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

Decision. 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 interval [-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 the point and the greatest value equal to 1 at the point .

We continue to search 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 more complicated than those just considered, 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 lovers of making students think in full (table of derivatives). Therefore, the logarithm and the trigonometric function will be used.

Example 8. Find the smallest and largest values of a function on the segment .

Decision. 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:

The result of all actions: the function reaches its minimum value, equal to 0, at a point and at a point and the greatest value equal to e² , at the point .

For self-checking during calculations, you can use online derivatives calculator .

Example 9. Find the smallest and largest values of a function on the segment .

Decision. We find the derivative of this function:

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 the greatest value, equal to , at the point .

In applied extremal problems, finding the smallest (largest) function values, as a rule, is reduced to finding the minimum (maximum). But it is not the minima or maxima themselves that are of greater practical interest, but the values of the argument at which they are achieved. When solving applied problems, an additional difficulty arises - the compilation of functions that describe the phenomenon or process under consideration.

Example 10 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 should be the dimensions of the tank in order to cover it with the least amount of material?

Decision. Let be 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 , whence . Substituting the found expression h into the formula for S:

Let us examine this function for an extremum. It is defined and differentiable everywhere in ]0, +∞[ , and

We equate the derivative to zero () and find the critical point. In addition, at , the derivative does not exist, but this value is not included in the domain of definition and therefore cannot be an extremum point. So, - 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 . Because this minimum - the only extremum of this function, it is its smallest value. So, the side of the base of the tank should be equal to 2 m, and its height.

For self-checking during calculations, you can use

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 smaller 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, 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 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.

Consider the real tasks from the trial exam in mathematics - this is where this technique is most common. At the same time, we will make sure that in this way many problems of B15 become almost verbal.

Under the root is a 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 monotone, so the largest 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 квадратный корень из отрицательного числа не существует.

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 \u003d 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. 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

In task B14 from the USE in mathematics, you need to find the smallest or largest value of a function of one variable. This is a fairly trivial problem from mathematical analysis, and it is for this reason that every high school graduate can and should learn how to solve it normally. Let's analyze a few examples that schoolchildren solved at the diagnostic work in mathematics, which took place in Moscow on December 7, 2011.

Depending on the interval on which you want to find the maximum or minimum value of the function, one of the following standard algorithms is used to solve this problem.

I. Algorithm for finding the largest or smallest value of a function on a segment:

Find the derivative of a function.
Select from the points suspected of an extremum those that belong to a given segment and the domain of the function.
Calculate values functions(not a derivative!) at these points.
Among the obtained values, choose the largest or smallest, it will be the desired one.

Example 1 Find the smallest value of a function
y = x 3 – 18x 2 + 81x+ 23 on the segment .

Decision: we act according to the algorithm for finding the smallest value of a function on a segment:

The scope of the function is not limited: D(y) = R.
The derivative of the function is: y' = 3x 2 – 36x+ 81. The scope of the derivative of a function is also not limited: D(y') = R.
Zeros of the derivative: y' = 3x 2 – 36x+ 81 = 0, so x 2 – 12x+ 27 = 0, whence x= 3 and x= 9, our interval includes only x= 9 (one point suspicious for an extremum).
We find the value of the function at a point suspicious of an extremum and at the edges of the interval. For the convenience of calculations, we represent the function in the form: y = x 3 – 18x 2 + 81x + 23 = x(x-9) 2 +23:
- y(8) \u003d 8 (8-9) 2 +23 \u003d 31;
- y(9) = 9 (9-9) 2 +23 = 23;
- y(13) = 13 (13-9) 2 +23 = 231.

So, from the obtained values, the smallest is 23. Answer: 23.

II. The algorithm for finding the largest or smallest value of a function:

Find the scope of the function.
Find the derivative of a function.
Determine the points that are suspicious of an extremum (those points at which the derivative of the function vanishes, and the points at which there is no two-sided finite derivative).
Mark these points and the domain of the function on the number line and determine the signs derivative(not functions!) on the resulting intervals.
Define values functions(not a derivative!) at the minimum points (those points at which the sign of the derivative changes from minus to plus), the smallest of these values will be the smallest value of the function. If there are no minimum points, then the function does not have a minimum value.
Define values functions(not a derivative!) at the maximum points (those points at which the sign of the derivative changes from plus to minus), the largest of these values will be the largest value of the function. If there are no maximum points, then the function does not have a maximum value.

Example 2 Find the largest value of the function.

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 .
Decision.

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) .
Decision.
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.
Decision. 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.
Decision. Let x be the first term. Then (49-x) is the second term.
The product will be maximal: x (49-x) → max

In July 2020, NASA launches an expedition to Mars. The spacecraft will deliver to Mars an electronic carrier with the names of all registered members of the expedition.

If this post solved your problem or you just liked it, share the link to it with your friends on social networks.

One of these code options needs to be copied and pasted into the code of your web page, preferably between the tags and or right after the tag . According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you paste the second code, then the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the load code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not at all necessary , since the MathJax script is loaded asynchronously). That's all. Now learn the MathML, LaTeX, and ASCIIMathML markup syntax and you're ready to embed math formulas into your web pages.

Another New Year's Eve... frosty weather and snowflakes on the window pane... All this prompted me to write again about... fractals, and what Wolfram Alpha knows about it. On this occasion, there is an interesting article in which there are examples of two-dimensional fractal structures. Here we will consider more complex examples of three-dimensional fractals.

A fractal can be visually represented (described) as a geometric figure or body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, it is a self-similar structure, considering the details of which, when magnified, we will see the same shape as without magnification. Whereas in the case of the usual geometric figure(not a fractal), when zoomed in, we will see details that have a simpler shape than the original figure itself. For example, at a sufficiently high magnification, part of an ellipse looks like a straight line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which with each increase will be repeated again and again.

Benoit Mandelbrot, the founder of the science of fractals, in his article Fractals and Art for Science wrote: "Fractals are geometric shapes that are as complex in their details as they are in their overall form. That is, if part of the fractal will be enlarged to the size of the whole, it will look like the whole, or exactly, or perhaps with a slight deformation.

We are looking for the smallest and largest values ​​​​of the function together