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Critical points of a function. How to find the maximum and minimum points of a function How to find the maximum point of the derivative of a function

A function and the study of its features occupies one of the key chapters in modern mathematics. The main component of any function is graphs depicting not only its properties, but also the parameters of the derivative of this function. Let's take a look at this tricky topic. So what is the best way to find the maximum and minimum points of a function?

Function: Definition

Any variable that depends in some way on the values ​​of another quantity can be called a function. For example, the function f(x 2) is quadratic and determines the values ​​for the entire set x. Let's say that x = 9, then the value of our function will be equal to 9 2 = 81.

Functions come in a variety of types: logical, vector, logarithmic, trigonometric, numeric, and others. Such outstanding minds as Lacroix, Lagrange, Leibniz and Bernoulli were engaged in their study. Their writings serve as a bulwark in modern ways of studying functions. Before finding the minimum points, it is very important to understand the very meaning of the function and its derivative.

Derivative and its role

All functions are dependent on their variables, which means that they can change their value at any time. On the graph, this will be depicted as a curve that either descends or rises along the y-axis (this is the whole set of numbers "y" along the vertical of the graph). And so definition of a point of a maximum and a minimum of function just is connected with these "oscillations". Let us explain what this relationship is.

The derivative of any function is drawn on a graph in order to study its main characteristics and calculate how quickly the function changes (ie changes its value depending on the variable "x"). At the moment when the function increases, the graph of its derivative will also increase, but at any second the function may begin to decrease, and then the graph of the derivative will decrease. Those points at which the derivative goes from minus to plus are called minimum points. In order to know how to find minimum points, you should better understand

How to calculate the derivative?

Definition and functions implies several concepts from In general, the very definition of the derivative can be expressed as follows: this is the value that shows the rate of change of the function.

The mathematical way of defining it for many students seems complicated, but in fact everything is much simpler. It is only necessary to follow the standard plan for finding the derivative of any function. The following describes how you can find the minimum point of a function without applying the rules of differentiation and without memorizing the table of derivatives.

  1. You can calculate the derivative of a function using a graph. To do this, you need to depict the function itself, then take one point on it (point A in the figure), draw a line vertically down to the abscissa axis (point x 0), and at point A draw a tangent to the graph of the function. The abscissa axis and the tangent form an angle a. To calculate the value of how fast the function increases, you need to calculate the tangent of this angle a.
  2. It turns out that the tangent of the angle between the tangent and the direction of the x-axis is the derivative of the function in a small area with point A. This method is considered a geometric way to determine the derivative.

Methods for examining a function

In the school curriculum of mathematics, it is possible to find the minimum point of a function in two ways. We have already analyzed the first method using the graph, but how to determine the numerical value of the derivative? To do this, you will need to learn several formulas that describe the properties of the derivative and help convert variables like "x" into numbers. The following method is universal, so it can be applied to almost all kinds of functions (both geometric and logarithmic).

  1. It is necessary to equate the function to the derivative function, and then simplify the expression using the rules of differentiation.
  2. In some cases, when a function is given in which the variable "x" is a divisor, it is necessary to determine the range of acceptable values ​​by excluding the point "0" from it (for the simple reason that in mathematics one cannot divide by zero in any case).
  3. After that, the original form of the function should be converted into a simple equation, equating the entire expression to zero. For example, if the function looked like this: f (x) \u003d 2x 3 + 38x, then according to the rules of differentiation, its derivative is equal to f "(x) \u003d 3x 2 + 1. Then we transform this expression into an equation of the following form: 3x 2 +1 \u003d 0 .
  4. After solving the equation and finding the points "x", you should depict them on the x-axis and determine whether the derivative in these areas between the marked points is positive or negative. After the designation, it will become clear at what point the function begins to decrease, that is, it changes sign from minus to the opposite. It is in this way that you can find both the minimum and maximum points.

Differentiation rules

The most basic component in the study of a function and its derivative is the knowledge of the rules of differentiation. Only with their help it is possible to transform cumbersome expressions and large complex functions. Let's get acquainted with them, there are a lot of them, but they are all very simple due to the regular properties of both power and logarithmic functions.

  1. The derivative of any constant is zero (f(x) = 0). That is, the derivative f (x) \u003d x 5 + x - 160 will take the following form: f "(x) \u003d 5x 4 +1.
  2. The derivative of the sum of two terms: (f+w)" = f"w + fw".
  3. Derivative of a logarithmic function: (log a d)" = d/ln a*d. This formula applies to all kinds of logarithms.
  4. Power derivative: (x n)"= n*x n-1. For example, (9x 2)" = 9*2x = 18x.
  5. Derivative of the sinusoidal function: (sin a)" = cos a. If the sin of angle a is 0.5, then its derivative is √3/2.

extremum points

We have already discussed how to find the minimum points, however, there is the concept of maximum points of a function. If the minimum denotes those points at which the function goes from minus to plus, then the maximum points are those points on the x-axis at which the derivative of the function changes from plus to the opposite - minus.

You can find it by the method described above, only it should be taken into account that they denote those sections on which the function begins to decrease, that is, the derivative will be less than zero.

In mathematics, it is customary to generalize both concepts, replacing them with the phrase "points of extrema". When the task asks to determine these points, this means that it is necessary to calculate the derivative of this function and find the minimum and maximum points.

Consider the following figure.

It shows the graph of the function y = x^3 - 3*x^2. Consider some interval containing the point x = 0, for example, from -1 to 1. Such an interval is also called the neighborhood of the point x = 0. As can be seen on the graph, in this neighborhood the function y = x^3 - 3*x^2 takes the largest value exactly at the point x = 0.

Maximum and minimum of a function

In this case, the point x = 0 is called the maximum point of the function. By analogy with this, the point x = 2 is called the minimum point of the function y = x^3 - 3*x^2. Because there is such a neighborhood of this point in which the value at this point will be minimal among all other values ​​from this neighborhood.

dot maximum function f(x) is called a point x0, provided that there is a neighborhood of the point x0 such that for all x not equal to x0 from this neighborhood, the inequality f(x)< f(x0).

dot minimum function f(x) is called a point x0, provided that there is a neighborhood of the point x0 such that for all x not equal to x0 from this neighborhood, the inequality f(x) > f(x0) is satisfied.

At the maximum and minimum points of the functions, the value of the derivative of the function is equal to zero. But this is not a sufficient condition for the existence of a function at a maximum or minimum point.

For example, the function y = x^3 at the point x = 0 has a derivative equal to zero. But the point x = 0 is not the minimum or maximum point of the function. As you know, the function y = x^3 increases on the entire real axis.

Thus, the minimum and maximum points will always be among the root of the equation f’(x) = 0. But not all roots of this equation will be maximum or minimum points.

Stationary and critical points

The points at which the value of the derivative of a function is equal to zero are called stationary points. There can also be points of maximum or minimum at points where the derivative of the function does not exist at all. For example, y = |x| at the point x = 0 has a minimum, but the derivative does not exist at this point. This point will be the critical point of the function.

The critical points of a function are the points at which the derivative is equal to zero, or the derivative does not exist at this point, that is, the function at this point is non-differentiable. In order to find the maximum or minimum of a function, a sufficient condition must be satisfied.

Let f(x) be some function differentiable on the interval (a;b). The point x0 belongs to this interval and f'(x0) = 0. Then:

1. if, when passing through the stationary point x0, the function f (x) and its derivative changes sign, from “plus” to “minus”, then the point x0 is the maximum point of the function.

2. if, when passing through the stationary point x0, the function f (x) and its derivative changes sign, from “minus” to “plus”, then the point x0 is the minimum point of the function.

A simple algorithm for finding extrema..

  • Finding the derivative of a function
  • Equate this derivative to zero
  • We find the values ​​of the variable of the resulting expression (values ​​of the variable at which the derivative is converted to zero)
  • We divide the coordinate line into intervals with these values ​​(at the same time, we should not forget about the break points, which also need to be plotted on the line), all these points are called “suspicious” points for the extremum
  • We calculate on which of these intervals the derivative will be positive, and on which it will be negative. To do this, you need to substitute the value from the interval into the derivative.

Of the points suspected of an extremum, it is necessary to find exactly . To do this, we look at our gaps on the coordinate line. If, when passing through some point, the sign of the derivative changes from plus to minus, then this point will be maximum, and if from minus to plus, then minimum.

To find the largest and smallest value of a function, you need to calculate the value of the function at the ends of the segment and at the extremum points. Then choose the largest and smallest value.

Consider an example
We find the derivative and equate it to zero:

We apply the obtained values ​​of the variables to the coordinate line and calculate the sign of the derivative on each of the intervals. Well, for example, for the first take-2 , then the derivative will be-0,24 , for the second take0 , then the derivative will be2 , and for the third we take2 , then the derivative will be-0.24. We put down the appropriate signs.

We see that when passing through point -1, the derivative changes sign from minus to plus, that is, it will be a minimum point, and when passing through 1, from plus to minus, respectively, this is a maximum point.

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:

f? (x ) is positive to the left of a and negative to the right of a, then at the very point x = a the function f(x ) It has maximum provided that the function f(x ) is continuous here.

If in sufficient proximity to the point x \u003d a, the derivative f? (x ) is negative to the left of a and positive to the right of a, then at the very point x = a the function f(x ) It has minimum provided that the function f(x ) is continuous here.

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

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

About case f?? (a) = 0 can be found in the Handbook of Higher Mathematics by M.Ya. Vygodsky.

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 functions 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 can be an extremum. They can be easily identified by looking at derivative graph: we are interested in those values ​​of the argument at which the graph of the function intersects the abscissa axis (Ox axis) and those at which the graph suffers breaks.

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

Function y (x) \u003d 3 x 2 + 2 x - 50.

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

We solve the equation: y? (x) = 0

6x + 2 \u003d 0.6x \u003d -2, x \u003d -2/6 \u003d -1/3

In this case, the critical point is x 0 = -1/3. It is for this value of the argument that the function has extremum. To get it to find, we substitute the found number in the expression for the function instead of "x":

y 0 = 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 x 0, then x 0 is maximum point; if the sign of the derivative changes from minus to plus, then x 0 is minimum point; if the sign does not change, then at the point x 0 there is neither a maximum nor a minimum.

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) \u003d 6 * (-1) + 2 \u003d -6 + 2 \u003d -4 (i.e. the sign is “minus”).

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

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 x 0 we have a minimum point.

The largest and smallest value of the function on the interval(on the segment) are found according to 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 have either a maximum or a minimum. In this case, to determine the largest and smallest values ​​of the function, we also take into account the values ​​of the function at the ends of the interval.

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

y (x) \u003d 3 sin (x) - 0.5x

at intervals:

a) [-9; nine]

b) [-6; -3]

So the derivative of the function is

y? (x) \u003d 3 cos (x) - 0.5

Solving Equation 3 cos (x) - 0.5 \u003d 0

3cos(x) = 0.5

cos(x) = 0.5/3 = 0.16667

x \u003d ± arccos (0.16667) + 2πk.

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

x \u003d arccos (0.16667) - 2 π *2 = -11.163 (out of range)

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

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

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

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

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

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

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

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

y (-3) = 3 cos (-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 breakpoints of a line y=f(x ), you need to find the second derivative, equate it to zero (solve the equation) and test all those values ​​of x for which the second derivative is zero, infinite or does not exist. If, when passing through one of these values, the second derivative changes sign, then the graph of the function has an inflection at this point. If it 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 turned here by concavity upwards, and if it is negative, then downwards.

How to find extrema of a function of two variables?

To find the extrema of a function f (x, y ), differentiable in the domain of its assignment, it is necessary:

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

f x ? (x, y) \u003d 0, f y? (x, y) = 0

2) for each critical point Р 0 ( a; b ) to investigate whether the sign of the difference remains unchanged

f (x, y) - f (a, b)

for all points (x; y) sufficiently close to Р 0 . If the difference retains a positive sign, then at the point P 0 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 Р 0.

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

Sources:

  • Vygodsky M.Ya. Handbook of Higher Mathematics
  • Chernenko V.D. Higher mathematics in examples and tasks. In 3 volumes. Volume 1