Examine a function for monotonicity online. Study of functions for monotonicity and extrema

Date of: 19.11.2023

Extrema and convexity.

Asymptotes of the graph of a function

Definition.Critical point functions at = f(X) is the point at which the derivative is zero or does not exist.

Theorem. If in the interval (a; b) the derivative positive/negative, then the function increases/decreases in this interval.

Theorem. If, upon passing through the critical point, the derivative changes sign from “+” to “−” (from “−” to “+”), then − is the maximum (minimum) point of the function

Definition. Function called convex up(down) in the interval (a; b), if in this interval the points of the graph lie under (above) the tangents constructed at these points. Inflection point is a point in the graph of a function that divides it into parts with different directions of convexity.

Example 2.3.

Explore function for monotony and extrema, convexity.

1. We examine the function for monotonicity and extrema.

Let's make a drawing ( rice. 2.1).

y′′

−

issue down

issue up

issue down

Rice. 2.2. Study of a function for convexity

Let's calculate the ordinates of the inflection points of the graph:

Coordinates of inflection points: (0; 0), (1; −1).

2.32. Examine the function for monotonicity and extrema:

1) 2) 3)

4) 5) 6)

2.33. Find the smallest and largest values of the function:

1) on the interval;

2) on the interval [−1; 1];

3) on the interval [−4; 4];

4) on the interval [−2; 1].

2.34. Production costs C (cu) depend on the volume of output X(units): Find the highest production costs if X changes over the interval . Find value X, at which the profit will be maximum if the revenue from the sale of a unit of production is equal to 15 c.u. e.

2.35. It is required to allocate a rectangular plot of land of 512 m2, fence it and divide it with a fence into three equal parts parallel to one of the sides of the site. What should be the size of the site so that the least amount of material is used for the fencing?

2.36. Given the perimeter of a rectangular window, find its dimensions such that it lets in the greatest amount of light.

2.37. Find the maximum profit if income R and costs C are determined by the formulas: where X− quantity of goods sold.

2.38. Dependence of production volume W from capital costs TO determined by the function
Find the change interval TO, where increasing capital costs is ineffective.

2.39. The cost function has the form Income from the sale of a unit of production is equal to 200. Find the optimal value of output for the manufacturer.

2.40. The dependence of the volume of output (in monetary units) on capital costs is determined by the function Find the interval of values at which increasing capital costs is ineffective.

2.41. It is believed that the increase in sales from advertising costs (million rubles) is determined by the ratio Income from the sale of a unit of production is equal to 20 thousand rubles. Find the level of advertising costs at which the company will receive maximum profit.

2.42. The income from the production of products using resource units is equal to The cost of a resource unit is 10 den. units How much of a resource should be purchased in order for the profit to be greatest?

2.43. The cost function has the form The income from the sale of a unit of production is 50. Find the maximum profit value that the manufacturer can receive.

2.44. The dependence of the monopoly's income on the quantity of output is defined as: The cost function in this interval has the form Find the optimal output value for the monopoly.

2.45. The price for the products of a monopoly producer is set in accordance with the ratio identified as . At what value of product output will the income from its sales be greatest?

2.46. The cost function has the following form at at . Currently the level of production Under what condition on the parameter p Is it profitable for a company to reduce output if the income from the sale of a unit of output is 50?

The derivative also helps in studying a function for increasing and decreasing functions. Let us first recall the corresponding definition.

Definition . Let the function be defined on the interval . They say that it increases (decreases) on the interval if such that .

Theorem. If a function is differentiable on the interval and , then it increases (decreases) on the interval .

Let the derivative of the function be continuous on the interval. To study its increase and decrease, the following plan is usually followed:

1) Find points from , where . These points are called stationary.

2) In all intervals into which stationary points are divided, determine the sign. To do this, it is enough to determine the sign at one point of each interval (the sign inside each interval does not change, since otherwise, according to the Bolzano-Cauchy theorem, inside this interval there must be a zero derivative, which is impossible). If inside the interval, then according to the theorem it increases. If , then it decreases.

Definition . Points at which the derivative of a function is equal to zero are called stationary. The points at which the derivative of a function is zero or does not exist are called critical.

Example. Examine increasing and decreasing function

This function is differentiable on the entire number line.

1) . Let's find stationary points: . The roots of the equation are the numbers , .

2) Points , divide the number line into three intervals: , , .

On the first interval we take .

Therefore, it increases over the interval. In the interval we take , . Therefore it decreases. On the interval we take , . Therefore, it increases over the interval.

Definition. Let the function be defined in . A point is called a local maximum (minimum) point if there is such that

If inequalities (1) are strict for , then the point is called a point of strict local maximum (minimum). The points of local maximum and minimum are called extremum points.

Theorem(a necessary condition for an extremum). If a function is differentiable at a point and is an extremum point, then

The proof of the theorem is not difficult to obtain from the definition of the derivative.

Comment. It follows from the theorem that the extremum points of the function must be sought among stationary points and points where the derivative does not exist. One of the sufficient conditions for an extremum follows directly from the following theorem.

Comment. A necessary condition is not sufficient. For example, for a function we have , but the point is not an extremum, since the function increases along the entire number line.

Theorem(sufficient condition for extremum). Let the function be continuous at a point and differentiable at . Then:

a) if the derivative, when passing through a point, changes sign from plus to minus, then the point is a point of local maximum;

b) if the derivative, when passing through a point, changes sign from minus to plus, then the point is a local minimum point of the function.

Note that from the theorem it follows that in the previous example the point is a local maximum point, and the point is a local minimum point of the function.

Often, when solving various problems, it is necessary to find the largest and smallest values of a function on a certain set.

Let's consider how this problem is solved first for the case when this is a segment. Let the function be continuous on the segment and differentiable on the interval except, perhaps, for a finite number of points. Then, according to Weierstrass’s theorem, the function reaches its largest and smallest values on the segment.

From the above theorems, the following plan for finding the largest and smallest values of the function follows.

1) Find the derivative and zeros of the derivative of .

2) Find values

a) at the zeros of the derivative of ;

b) at the ends of the segment;

c) at points where the derivative does not exist.

3) From the resulting numbers, choose the largest and smallest.

Note 1. Note that it is not at all necessary to find intervals of increasing and decreasing here.

Note 2. If it is an interval, half-interval or an infinite interval, then the above plan cannot be used. In this case, to solve the problem of the largest and smallest values, you need to find the intervals of increase and decrease of the function, the limits at the boundary points and, using simple analysis, obtain the answer.

Example 3. Find the largest and smallest values of the function on the interval.

Let's find the intervals of increasing and decreasing. To do this, we find the derivative:

The dot splits the interval into two intervals: and . Let us find the sign of the derivative in these intervals. To do this, let's calculate

Thus, the function decreases on a half-interval, and increases on the interval. That's why There is no greatest value because . In this case they write: .

Lesson and presentation in algebra in 10th grade on the topic: "Investigation of a function for monotonicity. Research algorithm"

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What we will study:
1. Decreasing and increasing functions.
2. Relationship between derivative and monotonicity of a function.
3. Two important theorems on monotonicity.
4. Examples.

Guys, earlier we looked at many different functions and plotted them. Now let's introduce new rules that work for all the functions that we have considered and will continue to consider.

Decreasing and increasing functions

Let's look at the concept of increasing and decreasing functions. Guys, what is a function?

A function is a correspondence y= f(x), in which each value of x is associated with a single value of y.

Let's look at the graph of some function:

Our graph shows: the larger x, the smaller y. So let's define a decreasing function. A function is called decreasing if a larger value of the argument corresponds to a smaller value of the function.

If x2 > x1, then f(x2) Now let's look at the graph of this function:
This graph shows that the larger x, the larger y. So let's define an increasing function. A function is called increasing if a larger value of the argument corresponds to a larger value of the function.
If x2 > x1, then f(x2 > f(x1) or: the greater x, the greater y.

If a function increases or decreases over a certain interval, then it is said that it is monotonic on this interval.

Relationship between derivative and monotonicity of a function

Guys, now let's think about how you can apply the concept of derivative when studying function graphs. Let's draw a graph of an increasing differentiable function and draw a couple of tangents to our graph.

If you look at our tangents or visually draw any other tangent, you will notice that the angle between the tangent and the positive direction of the x-axis will be acute. This means that the tangent has a positive slope. The angle coefficient of the tangent is equal to the value of the derivative in the abscissa of the point of tangency. Thus, the value of the derivative is positive at all points in our graph. For an increasing function, the following inequality holds: f"(x) ≥ 0, for any point x.

Guys, now let's look at the graph of some decreasing function and construct tangents to the graph of the function.

Let's look at the tangents and visually draw any other tangent. We will notice that the angle between the tangent and the positive direction of the x-axis is obtuse, which means the tangent has a negative slope. Thus, the value of the derivative is negative at all points in our graph. For a decreasing function, the following inequality holds: f"(x) ≤ 0, for any point x.

So, the monotonicity of a function depends on the sign of the derivative:

If a function increases on an interval and has a derivative on this interval, then this derivative will not be negative.

If a function decreases on an interval and has a derivative on this interval, then this derivative will not be positive.

Important, so that the intervals on which we consider the function are open!

Two important theorems on monotonicity

Theorem 1. If the inequality f'(x) ≥ 0 holds at all points of an open interval X (and the equality of the derivative to zero either does not hold or holds, but only at a finite set of points), then the function y= f(x) increases on the interval X.

Theorem 2. If the inequality f'(x) ≤ 0 holds at all points of an open interval X (and the equality of the derivative to zero either does not hold or holds, but only at a finite set of points), then the function y= f(x) decreases on the interval X.

Theorem 3. If at all points of the open interval X the equality
f’(x)= 0, then the function y= f(x) is constant on this interval.

Examples of studying a function for monotonicity

1) Prove that the function y= x 7 + 3x 5 + 2x - 1 is increasing on the entire number line.

Solution: Let's find the derivative of our function: y"= 7 6 + 15x 4 + 2. Since the degree at x is even, the power function takes only positive values. Then y" > 0 for any x, which means by Theorem 1, our function increases along the entire number line.

2) Prove that the function is decreasing: y= sin(2x) - 3x.

Let's find the derivative of our function: y"= 2cos(2x) - 3.
Let's solve the inequality:
2cos(2x) - 3 ≤ 0,
2cos(2x) ≤ 3,
cos(2x) ≤ 3/2.
Because -1 ≤ cos(x) ≤ 1, which means our inequality is satisfied for any x, then by Theorem 2 the function y= sin(2x) - 3x decreases.

3) Examine the monotonicity of the function: y= x 2 + 3x - 1.

Solution: Let's find the derivative of our function: y"= 2x + 3.
Let's solve the inequality:
2x + 3 ≥ 0,
x ≥ -3/2.
Then our function increases for x ≥ -3/2, and decreases for x ≤ -3/2.
Answer: For x ≥ -3/2, the function increases, for x ≤ -3/2, the function decreases.

4) Examine the monotonicity of the function: y= $\sqrt(3x - 1)$.

Solution: Let's find the derivative of our function: y"= $\frac(3)(2\sqrt(3x - 1))$.
Let's solve the inequality: $\frac(3)(2\sqrt(3x - 1))$ ≥ 0.

Our inequality is greater than or equal to zero:
$\sqrt(3x - 1)$ ≥ 0,
3x - 1 ≥ 0,
x ≥ 1/3.
Let's solve the inequality:
$\frac(3)(2\sqrt(3x-1))$ ≤ 0,

$\sqrt(3x-1)$ ≤ 0,
3x - 1 ≤ 0.
But this is impossible, because The square root is defined only for positive expressions, which means our function has no decreasing intervals.
Answer: for x ≥ 1/3 the function increases.

Problems to solve independently

a) Prove that the function y= x 9 + 4x 3 + 1x - 10 is increasing along the entire number line.
b) Prove that the function is decreasing: y= cos(5x) - 7x.
c) Examine the monotonicity of the function: y= 2x 3 + 3x 2 - x + 5.
d) Examine the monotonicity of the function: y = $\frac(3x-1)(3x+1)$.

We first met in a 7th grade algebra course. Looking at the graph of the function, we took down the corresponding information: if, moving along the graph from left to right, we at the same time move from bottom to top (as if climbing a hill), then we declared the function to be increasing (Fig. 124); if we move from top to bottom (go down a hill), then we declared the function to be decreasing (Fig. 125).

However, mathematicians are not very fond of this method of studying the properties of a function. They believe that definitions of concepts should not be based on a drawing - the drawing should only illustrate one or another property of a function on its graphics. Let us give strict definitions of the concepts of increasing and decreasing functions.

Definition 1. The function y = f(x) is said to be increasing on the interval X if, from the inequality x 1< х 2 - где хг и х2 - любые две точки промежутка X, следует неравенство f(x 1) < f(x 2).

Definition 2. The function y = f(x) is said to be decreasing on the interval X if the inequality x 1< х 2 , где х 1 и х 2 - любые две точки промежутка X, следует inequality f(x 1) > f(x 2).

In practice, it is more convenient to use the following formulations:

a function increases if a larger value of the argument corresponds to a larger value of the function;
a function decreases if a larger value of the argument corresponds to a smaller value of the function.

Using these definitions and the properties of numerical inequalities established in § 33, we will be able to substantiate conclusions about the increase or decrease of previously studied functions.

1. Linear function y = kx +m

If k > 0, then the function increases throughout (Fig. 126); if k< 0, то функция убывает на всей числовой прямой (рис. 127).

Proof. Let f(x) = kx +m. If x 1< х 2 и k >Oh, then, according to the property of 3 numerical inequalities (see § 33), kx 1< kx 2 . Далее, согласно свойству 2, из kx 1 < kx 2 следует, что kx 1 + m < kx 2 + m, т. е. f(х 1) < f(х 2).

So, from the inequality x 1< х 2 следует, что f(х 1) < f(x 2). Это и означает возрастание функции у = f(х), т.е. linear functions y = kx+ m.

If x 1< х 2 и k < 0, то, согласно свойству 3 числовых неравенств, kx 1 >kx 2 , and according to property 2, from kx 1 > kx 2 it follows that kx 1 + m> kx 2 + i.e.

So, from the inequality x 1< х 2 следует, что f(х 1) >f(x 2). This means a decrease in the function y = f(x), i.e., the linear function y = kx + m.

If a function increases (decreases) throughout its entire domain of definition, then it can be called increasing (decreasing) without indicating the interval. For example, about the function y = 2x - 3 we can say that it is increasing along the entire number line, but we can also say it more briefly: y = 2x - 3 - increasing
function.

2. Function y = x2

1. Consider the function y = x 2 on the ray. Let's take two non-positive numbers x 1 and x 2 such that x 1< х 2 . Тогда, согласно свойству 3 числовых неравенств, выполняется неравенство - х 1 >- x 2. Since the numbers - x 1 and - x 2 are non-negative, then by squaring both sides of the last inequality, we obtain an inequality of the same meaning (-x 1) 2 > (-x 2) 2, i.e. This means that f(x 1) > f(x 2).

So, from the inequality x 1< х 2 следует, что f(х 1) >f(x 2).

Therefore, the function y = x 2 decreases on the ray (- 00, 0] (Fig. 128).

1. Consider a function on the interval (0, + 00).
Let x1< х 2 . Так как х 1 и х 2 - , то из х 1 < x 2 следует (см. пример 1 из § 33), т. е. f(x 1) >f(x 2).

So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2). This means that the function decreases on the open ray (0, + 00) (Fig. 129).

2. Consider a function on the interval (-oo, 0). Let x 1< х 2 , х 1 и х 2 - отрицательные числа. Тогда - х 1 >- x 2, and both sides of the last inequality are positive numbers, and therefore (we again used the inequality proven in example 1 from § 33). Next we have, where we get from.

So, from the inequality x 1< х 2 следует, что f(x 1) >f(x 2) i.e. function decreases on the open ray (- 00 , 0)

Usually the terms “increasing function” and “decreasing function” are combined under the general name monotonic function, and the study of a function for increasing and decreasing is called the study of a function for monotonicity.

Solution.

1) Let’s plot the function y = 2x2 and take the branch of this parabola at x< 0 (рис. 130).

2) Construct and select its part on the segment (Fig. 131).

3) Let's construct a hyperbola and select its part on the open ray (4, + 00) (Fig. 132).
4) Let us depict all three “pieces” in one coordinate system - this is the graph of the function y = f(x) (Fig. 133).

Let's read the graph of the function y = f(x).

1. The domain of definition of the function is the entire number line.

2. y = 0 at x = 0; y > 0 for x > 0.

3. The function decreases on the ray (-oo, 0], increases on the segment , decreases on the ray, is convex upward on the segment, convex downward on the ray, Lagrange’s theorem applies: there is a point x 0 of ( x 1 ;x 2) such that f(x 2) -f(x 1) = (x 2 -x 1)×f¢( x 0). But, according to the condition, f"(x 0) = 0, therefore, f(x 2) =f(x 1), i.e. function f(x) is constant on ( a; b). This means that sufficiency has been proven. The theorem has been proven.

Theorem 4 (necessary condition for the monotonicity of a function). Let in the interval (a; b) function f(x) differentiable. Then:

A)if f(x) increases, then its derivative in(a; b) not negative, i.e. f ¢( x) ³ 0;

b) if f(x) decreases, then its derivative in (a; b) not positive, i.e. f ¢( x) £ 0.

Proof. A). Let the function f(x) increases in ( a; b), i.e. for any x 1 ,x 2 of ( a; b) the following relation holds: x 1 < x 2® f(x 1) < f(x 2). Then, for the indicated points x 1 ,x 2 the following relation is positive:

It follows that the derivative f ¢( x 1) ³ 0. Statement A b).

Theorem 5 (sufficient condition for the monotonicity of a function). Let in the interval (a; b) function f(x) differentiable. Then:

A)if f ¢( x) > 0 on (a; b), then f(x)increases by (a; b);

b) if f ¢( x) < 0on(a; b),then f(x) decreases by (a ; b).

Proof. A). Let f ¢( x) > 0 on ( a; b) and points x 1 , x 2 of ( a; b) such that x 1 < x 2. According to Lagrange's theorem, there is a point x 0 of ( x 1 ;x 2) such that f(x 2) -f(x 1) = (x 2 -x 1)×f¢( x 0). Here the right side of the equality is positive, so f(x 2) -f(x 1) > 0, i.e. f(x 2) > f(x 1) . It means that f(x) increases by ( a; b). Statement A) has been proven. The statement is proved in a similar way b).

Example 9. Function at= X 3 increases everywhere, since with increasing values X The cubes of these values increase. Derivative of this function at¢= 3 X 2 is non-negative everywhere, i.e. the necessary monotonicity condition is satisfied.

Example 10. Find the intervals of increasing and decreasing function y= 0,25X 4 - 0,5X 2 .

Solution. The derivative of this function is found at¢ = X 3 - X, and intervals are constructed in which X 3 - X positive or negative. To do this, we first find critical points at which at¢ = 0: X 3 - X = 0 ® X(X + 1)(X-1) = 0 ® X 1 = 0, X 2 = -1 X 3 = 1. These points divide the number line into 4 spaces:

- + - + X

-¥ -2 -1 0 1 2 3 +¥

Damn.36.

In general, to determine the signs of the derivative, take one point in each interval and calculate the values of the derivative at these points. But sometimes it is enough to take only one point in the rightmost interval, determine the sign of the derivative at this point, and alternate the signs in the remaining intervals. In this example, let X= 2, then at¢(2) = 2 3 – 2 = 6 > 0. A + sign is placed in the right interval, and then the signs alternate. Received at¢ > 0 on the intervals (-1; 0) and (1; +¥), therefore, the function under study increases on these intervals. Further, at¢< 0 на (- ¥; -1) и (0; 1), следовательно, исследуемая функция на этих промежутках убывает. Ниже на чертеже 37 построен график этой функции.

Definition 3. 1). Dot X o is called maximum point functions f(x), if there is an interval ( a; b), containing X oh in which meaning f(x o) the greatest, i.e. f(x o) > f(x) for all X from ( a; b).

2). Dot X o is called minimum point functions f(x), if there is an interval ( a; b), containing X oh in which meaning f(x o) the smallest, i.e. f(x O)< f(x) for all X from ( a; b). The maximum and minimum points are called extremum points.

Theorem 6(necessary condition for the extremum of the function). If x O is the extremum point of the function f(x)and there is a derivative

f ¢( x 0),then f "(x 0) = 0.

The proof is similar to the proof of Rolle's theorem.

Dot x 0 , in which f ¢( x 0) = 0 or f ¢( x 0) does not exist, called critical point functions f (x). They say that critical points suspicious of extremes, i.e. they may or may not be maximum or minimum points.

Theorem 7 (sufficient condition for the extremum of a function). Let f(x)differentiable in some interval containing the critical point x O ( except, perhaps, the point x itself O) . Then:

A) if when passing through x O left to right derivative f ¢( x) changes sign from + to -,then x O is the maximum point of the function f (x);

b) if when passing through x O left to right derivative f ¢( x) changes sign from - to+,then x O is the minimum point of the function f (x).

Proof. Let all the conditions of the paragraph be met A). Let's take a point X(from the specified interval) such that X <X oh, and apply Lagrange’s theorem to the interval ( X; X O). We get: f(x 0) -f(x) = (x 0 -x)×f¢( x 1), where x 1 – some point from ( X; X O). By condition, f¢( x 1) > 0 and ( x 0 -x) > 0, therefore f(x 0) >f(x) . Similarly, it is proved that for any point X >X oh too f(x 0) >f(x). From these statements it follows that is the maximum point, the statement A) has been proven. The statement is proved in a similar way b).

Example 11. Example 9 shows that the function at= X 3 increases everywhere, therefore, it has no extrema. Indeed, its derivative y"= 3X 2 is equal to zero only when X o = 0, i.e. at this point the necessary condition for the extremum of the function is satisfied. But when passing through 0 its derivative y"= 3X 2 does not change sign, so X o = 0 is not the extremum point of this function.

Example 12. Example 10 shows that the function at = 0,25X 4 - 0,5X 2 has critical points X 1 = 0, X 2 = -1, X 3 = 1. In drawing 34 it is indicated that when passing through these points its derivative changes sign, therefore, X 1 , X 2 , X 3 - extremum points, while X 1 = 0 is the maximum point, and X 2 = -1, X 3 = 1 - minimum points.

Next, a drawing is made for this example. Function f(x) = 0,25X 4 - 0,5X 2 is being studied on parity: f(-x) = 0,25(-X) 4 - 0,5(-X) 2 = f(x), therefore, this function is even, and its graph is symmetrical about the axis OY. The graph points found above and some auxiliary points lying on the graph are plotted and connected by a smooth line.

y= 0,25x 4 - 0,5x 2 0,5 -0,11

1 0 max 1 x Ö`1/3 –0,14 A B

Damn.37.

Theorem 8 (second sufficient condition for extremum). Let x 0 – critical point of function f(x), and there is a second-order derivative f¢¢( X 0). Then:

a) if f ¢¢( X 0) < 0, then x 0 – maximum point of function f(x);

b) if f ¢¢( X 0) > 0, then x 0 - minimum point of function f(x).

The proof of this theorem is not considered (see).

Example 13. Examine the extremum of the function y= 2x 2 - x 4 .

Solution. The derivative is found y¢ and critical points at which

y¢= 9: y¢= 4 x - 4x 3 ; 4x - 4x 3 = 0 ® x 1 = 0, x 2 = 1, x 3 = -1 - critical points. The second order derivative is found y¢¢ and its values at critical points are calculated: y¢¢= 4 –12 X 2 ; y¢¢(0) = 4, y¢¢(1) = –8, y¢¢(-1) = –8. Because y¢¢(0) > 0, then x 1 = 0 - minimum point; and since y¢¢(1)< 0, y¢¢(-1)< 0, то x 2 = 1, x 3 = -1 - maximum points of this function.

Absolute extrema of a function on a segment [a; b] are called the largest and smallest values f(x) on [ a; b]. These extrema are reached either at critical points of the function f(x), or at the ends of the segment [ a; b].

Example 14. Determine the largest and smallest values of the function y = X 2× lnx on the interval .

Solution. The derivative of this function and its critical points are found: at¢ = 2 x× lnx + x 2 ×(1/ x) = x×(2 lnx+1); x×(2× lnx+1) = 0 ® a) X 1 = 0; b) 2× lnx+ 1 = 0 ® ln x= -0.5 ® X 2 = e - 0,5 = 1/Ö `e» 0.607. Critical point X 1 = 0 is not included in the interval under consideration, therefore the function values are found at the point X 2 = e- 0.5 and at the ends A= 0,5, b = e. at(e -0,5) = (e- 0.5) 2 × ln(e - 0,5) =e - 1 (-0,5) = -0,5/e» -0.184; at(0.5) = 0.25× ln 0.5 » 0.25(-0.693) = -0.17325; at(e) = e 2× lne = e 2 × 1" 7.389. The largest and smallest values among the found values are selected: the largest value "7.389 in X = e, smallest value "-0.184 V at X = e - 0,5 .

Extremum problems.

In such problems, two variables are considered X And at, and you need to find such a value X, at which the value at is the largest or smallest. Solving this problem contains the following steps:

1) an extreme value is selected y, the maximum or minimum of which must be found;

2) a variable is selected X, And y expressed through X;

3) the derivative is calculated at"and there are critical points at which at" is 0 or does not exist;

4) critical points at the extremum are investigated;

5) values are considered y at the ends, and the value required in the problem is calculated.

Example 15. It has been experimentally established that gasoline consumption

at(l) on 100 km by car GAZ-69 depending on speed x(km/h) described by the function y = 18 - 0,3X + 0,003X 2 . Determine the most economical speed.

Solution. Here the first two steps 1) and 2) are completed in the problem statement. Therefore, the derivative is immediately calculated: y"= -0,3 +0,006X, and the critical point is found: -0.3 + 0.006 X = 0 ® X o = 50. Now, the second sufficient condition for the extremum applies: y""= 0.006 > 0 at any point, therefore, X o = 50 - minimum point. Conclusion: the most economical speed is 50 km/h, while gasoline consumption is 18 - 0.3 × 50 + 0.003 × 50 2 = 10.5 liters. per 100 km.

Example 16. From a square sheet of cardboard with a side of 60 cm, identical squares are cut out at the corners and a rectangular box is glued from the remaining part. What should be the side of the cut square so that the volume of the box is greatest?.

Solution. The above steps to solve the problem are carried out.

1). By condition, the volume of the box should be the largest, so let y- volume of the box.

2). Behind X(cm) take the side of the square being cut out. Then the height of the box will be equal to X and the base of the box will be a square with a side

(60 – 2X), its area is (60 – 2 X) 2 . Therefore, the volume of the box is y= X(60 – 2X) 2 = 3600X - 240X 2 + 4X 3 .

3). The derivative is calculated and the critical points are found: y"= 3600 - 480X + 12X 2 ; X 2 - 40X+300 = 0 ® X 1 =10, X 2 =30 - critical points.

4). The 2nd order derivative is equal to y""= - 480 + 24X And y""(10) = -240, y""(30) = 240. By Theorem 8, X 1 =10 - maximum point and y max = 400 (cm 3).

5). Besides, X can take an extreme value X 3 = 0. But at(0) = 0 - this is less than y max.

Answer: The side of the cut square is 10 cm.