Find the smallest value of the function f x x. The smallest and largest values ​​of a function on a segment

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 rather 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.

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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 an ordinary 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.

Let the function y=f(X) 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 an interior 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, for x=a and x = b;

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

Investigation of a function for convexity and an inflection point.

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

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

Algorithm for studying for convexity and inflection point:

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

2. Put critical points on the number line, breaking it into intervals. Find the sign of the second derivative on each interval; if , then the function is convex upwards, if, then the function is convex downwards.

3. If, when passing through a critical point of the second kind, it changes sign 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. Investigation of a function into asymptotes.

Definition. The asymptote of the graph of a function is called straight, which has the property that the distance from any point of the graph to this line tends to zero with an unlimited removal of the graph point from the origin.

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

Definition. Direct called vertical asymptote function graph 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 - breaking point.

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

Example.

x

y

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

General scheme for the study of functions and plotting.

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

3. Investigate for even and odd functions ( y (‒ x) = y (x) ‒ parity; y(‒ x) = ‒ y (x) ‒ odd).

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

5. Find 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 graph of the function.

8. On the basis of the conducted research, construct a graph of the function.

Example. Investigate the function and plot its graph.

1) D (y) =

x= 4 - breaking point.

2) When x = 0,

(0; – 5) – point of intersection with oy.

At y = 0,

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

4) We investigate for asymptotes.

a) vertical

b) horizontal

c) find oblique asymptotes where

‒oblique asymptote equation

5) In this equation, it is not required to find intervals of monotonicity of the function.

6)

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

no extra.

It can be seen from the table that the point X= ‒2‒maximum point, at the point X= 4‒ no extremum, X= 10 – minimum point.

Substitute the value (‒ 3) into the equation:

9 + 24 ‒ 20 > 0

25 ‒ 40 ‒ 20 < 0

121 ‒ 88 ‒ 20 > 0

The maximum of this function is

(– 2; – 4) – maximum extremum.

The minimum of this function is

(10; 20) is the minimum extremum.

7) examine the convexity and inflection point of the graph of the function


The concept of the largest and smallest values ​​of a function.

The concept of the largest and smallest values ​​is closely related to the concept of the critical point of a function.

Definition 1

$x_0$ is called a critical point of the function $f(x)$ if:

1) $x_0$ - internal point of the domain of definition;

2) $f"\left(x_0\right)=0$ or does not exist.

Let us now introduce the definitions of the largest and smallest values ​​of a function.

Definition 2

A function $y=f(x)$ defined on the interval $X$ reaches its maximum value if there exists a point $x_0\in X$ such that for all $x\in X$ the inequality

Definition 3

A function $y=f(x)$ defined on the interval $X$ reaches its minimum value if there exists a point $x_0\in X$ such that for all $x\in X$ the inequality

Weierstrass' theorem on a function continuous on an interval

Let us first introduce the concept of a function continuous on an interval:

Definition 4

A function $f\left(x\right)$ is called continuous on a segment $$ if it is continuous at every point of the interval $(a,b)$, and also continuous on the right at the point $x=a$ and on the left at the point $x =b$.

Let us formulate a theorem on a function continuous on an interval.

Theorem 1

Weierstrass theorem

The function $f\left(x\right)$, which is continuous on the interval $$, reaches its maximum and minimum values ​​on this interval, that is, there are points $\alpha ,\beta \in $ such that for all $x\in $ inequality $f(\alpha)\le f(x)\le f(\beta)$.

The geometric interpretation of the theorem is shown in Figure 1.

Here the function $f(x)$ reaches its minimum value at the point $x=\alpha $ reaches its maximum value at the point $x=\beta $.

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

1) Find the derivative $f"(x)$;

2) Find the points where the derivative $f"\left(x\right)=0$;

3) Find points where the derivative $f"(x)$ does not exist;

4) Choose from the points obtained in paragraphs 2 and 3 those that belong to the segment $$;

5) Calculate the value of the function at the points obtained in step 4, as well as at the ends of the segment $$;

6) Choose from the obtained values ​​the largest and smallest value.

Problems for finding the largest and smallest values ​​of a function on a segment

Example 1

Find the largest and smallest value of a function on the segment : $f(x)=(2x)^3-15x^2+36x+1$

Decision.

1) $f"\left(x\right)=6x^2-30x+36$;

2) $f"\left(x\right)=0$;

\ \ \

4) $2\in \left,\ 3\in $;

5) Values:

\ \ \ \

6) The largest of the found values ​​is $33$, the smallest of the found values ​​is $1$. Thus, we get:

Answer:$max=33,\ min=1$.

Example 2

Find the largest and smallest value of a function on the segment : $f\left(x\right)=x^3-3x^2-45x+225$

Decision.

The solution will be carried out according to the above scheme.

1) $f"\left(x\right)=3x^2-6x-45$;

2) $f"\left(x\right)=0$;

\ \ \

3) $f"(x)$ exists at all points of the domain of definition;

4) $-3\notin\left,\5\in $;

5) Values:

\ \ \

6) The largest of the found values ​​is $225$, the smallest of the found values ​​is $50$. Thus, we get:

Answer:$max=225,\ min=50$.

Example 3

Find the largest and smallest value of a function on the interval [-2,2]: $f\left(x\right)=\frac(x^2-6x+9)(x-1)$

Decision.

The solution will be carried out according to the above scheme.

1) $f"\left(x\right)=\frac(\left(2x-6\right)\left(x-1\right)-(x^2-6x+9))(((x- 1))^2)=\frac(x^2-2x-3)(((x-1))^2)$;

2) $f"\left(x\right)=0$;

\[\frac(x^2-2x-3)(((x-1))^2)=0\] \ \

3) $f"(x)$ does not exist at the point $x=1$

4) $3\notin \left[-2,2\right],\ -1\in \left[-2,2\right],\ 1\in \left[-2,2\right]$, however 1 does not belong to the scope;

5) Values:

\ \ \

6) The largest of the found values ​​is $1$, the smallest of the found values ​​is $-8\frac(1)(3)$. Thus, we get: \end(enumerate)

Answer:$max=1,\ min==-8\frac(1)(3)$.

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 interval [ 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 segment [ 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

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