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

Find the largest value of the function y=(7x^2-56x+56)e^x on the segment [-3; 2].

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Solution

Let's find the derivative of the original function using the product derivative formula y"=(7x^2-56x+56)"e^x\,+ (7x^2-56x+56)\left(e^x\right)"= (14x-56)e^x+(7x^2-56x+56)e^x= (7x^2-42x)e^x= 7x(x-6)e^x. Let's calculate the zeros of the derivative: y"=0;

7x(x-6)e^x=0,

x_1=0, x_2=6.

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function on a given segment.

From the figure it is clear that on the segment [-3; 0] the original function increases, and on the segment it decreases. Thus, the largest value on the segment [-3; 2] is achieved at x=0 and is equal to y(0)= 7\cdot 0^2-56\cdot 0+56=56.

Answer

Condition

Find the greatest value of the function y=12x-12tg x-18 on the segment \left.

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Solution

y"= (12x)"-12(tg x)"-(18)"= 12-\frac(12)(\cos ^2x)= \frac(12\cos ^2x-12)(\cos ^2x)\leqslant0. This means that the original function is non-increasing on the interval under consideration and takes the greatest value at the left end of the interval, that is, at x=0. The largest value is y(0)= 12\cdot 0-12 tg (0)-18= -18.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the minimum point of the function y=(x+8)^2e^(x+52).

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Solution

We will find the minimum point of the function using the derivative. Let's find the derivative of a given function using the formulas for the derivative of the product, the derivative of x^\alpha and e^x:

y"(x)= \left((x+8)^2\right)"e^(x+52)+(x+8)^2\left(e^(x+52)\right)"= 2(x+8)e^(x+52)+(x+8)^2e^(x+52)= (x+8)e^(x+52)(2+x+8)= (x+8)(x+10)e^(x+52).

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function. e^(x+52)>0 for any x. y"=0 at x=-8, x=-10.

The figure shows that the function y=(x+8)^2e^(x+52) has a single minimum point x=-8.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the maximum point of the function y=8x-\frac23x^\tfrac32-106.

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Solution

ODZ: x \geqslant 0. Let's find the derivative of the original function:

y"=8-\frac23\cdot\frac32x^\tfrac12=8-\sqrt x.

Let's calculate the zeros of the derivative:

8-\sqrt x=0;

\sqrt x=8;

x=64.

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function.

The figure shows that point x=64 is the only maximum point of the given function.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the smallest value of the function y=5x^2-12x+2\ln x+37 on the segment \left[\frac35; \frac75\right].

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Solution

ODZ: x>0.

Let's find the derivative of the original function:

y"(x)= 10x-12+\frac(2)(x)= \frac(10x^2-12x+2)(x).

Let's define the zeros of the derivative: y"(x)=0;

\frac(10x^2-12x+2)(x)=0,

5x^2-6x+1=0,

x_(1,2)= \frac(3\pm\sqrt(3^2-5\cdot1))(5)= \frac(3\pm2)(5),

x_1=\frac15\notin\left[\frac35; \frac75\right],

x_2=1\in\left[\frac35; \frac75\right].

Let us arrange the signs of the derivative and determine the intervals of monotonicity of the original function on the interval under consideration.

From the figure it is clear that on the segment \left[\frac35; 1\right] the original function decreases, and on the segment \left increases. Thus, the smallest value on the segment \left[\frac35; \frac75\right] is achieved at x=1 and is equal to y(1)= 5\cdot 1^2-12\cdot 1+2 \ln 1+37= 30.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Find the greatest value of the function y=(x+4)^2(x+1)+19 on the segment [-5; -3].

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Solution

Let's find the derivative of the original function using the product derivative formula.

A simple algorithm for finding extrema..

Finding the derivative of the function
We equate this derivative to zero
We find the values of the variable of the resulting expression (the values of the variable at which the derivative is converted to zero)
Using these values, we divide the coordinate line into intervals (do 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 which of these intervals the derivative will be positive and which will be negative. To do this, you need to substitute the value from the interval into the derivative.

Of the points suspicious for an extremum, it is necessary to find . To do this, we look at our intervals 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 values of a function, you need to calculate the value of the function at the ends of the segment and at the extremum points. Then select the largest and smallest value.

Let's look at an example
Find the derivative and equate it to zero:

We plot the obtained values of the variables on the coordinate line and calculate the sign of the derivative on each of the intervals. Well, for example, for the first one let's take-2 , then the derivative will be equal-0,24 , for the second we’ll 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, this will be the minimum point, and when passing through 1, it will change sign from plus to minus, respectively, this will be the maximum point.

The 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 understand this difficult topic. So what is the best way to find the maximum and minimum points of a function?

Function: definition

Any variable that in some way depends 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 many different types: logical, vector, logarithmic, trigonometric, numeric and others. They were studied by such outstanding minds as Lacroix, Lagrange, Leibniz and Bernoulli. Their works serve as a mainstay 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 depend 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 falls or rises along the ordinate axis (this is the whole set of “y” numbers along the vertical graph). So, determining the maximum and minimum points of a function is precisely related to these “oscillations”. Let us explain what this relationship is.

The derivative of any function is graphed in order to study its basic characteristics and calculate how quickly the function changes (i.e. 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 can begin to decrease, and then the graph of the derivative will decrease. Those points at which the derivative changes from a minus sign to a plus sign are called minimum points. In order to know how to find minimum points, you should better understand

How to calculate derivative?

The definition and functions imply several concepts from In general, the very definition of a derivative can be expressed as follows: this is the quantity that shows the rate of change of the function.

The mathematical way of determining it seems complicated for many students, but in reality everything is much simpler. You just need to follow the standard plan for finding the derivative of any function. Below we describe how you can find the minimum point of a function without applying the rules of differentiation and without memorizing the table of derivatives.

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 x-axis and the tangent form a certain angle a. To calculate the value of how quickly a function increases, you need to calculate the tangent of this angle a.
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 method for determining the derivative.

Methods for studying function

In the school mathematics curriculum, it is possible to find the minimum point of a function in two ways. We have already discussed the first method using a graph, but how can we 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 types of functions (both geometric and logarithmic).

It is necessary to equate the function to the derivative function, and then simplify the expression using the rules of differentiation.
In some cases, when given a function in which the variable “x” is in the divisor, it is necessary to determine the range of acceptable values, excluding the point “0” from it (for the simple reason that in mathematics one should never divide by zero).
After this, you should transform the original form of the function into a simple equation, equating the entire expression to zero. For example, if the function looked like this: f(x) = 2x 3 +38x, then according to the rules of differentiation its derivative is equal to f"(x) = 3x 2 +1. Then we transform this expression into an equation of the following form: 3x 2 +1 = 0 .
After solving the equation and finding the “x” points, you should plot them on the x-axis and determine whether the derivative in these sections 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, changes sign from minus to the opposite. It is in this way that you can find both the minimum and maximum points.

Rules of differentiation

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

The derivative of any constant is equal to zero (f(x) = 0). That is, the derivative f(x) = x 5 + x - 160 will take the following form: f" (x) = 5x 4 +1.
Derivative of the sum of two terms: (f+w)" = f"w + fw".
Derivative of a logarithmic function: (log a d)" = d/ln a*d. This formula applies to all types of logarithms.
Derivative of the power: (x n)"= n*x n-1. For example, (9x 2)" = 9*2x = 18x.
The 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 minimum points, but there is also the concept of maximum points of a function. If the minimum denotes those points at which the function changes from a minus sign to a 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 using the method described above, but you should take into account that they indicate those areas in 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 a task asks you to identify these points, it means that you need to calculate the derivative of a given function and find the minimum and maximum points.

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 a function is the following: 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 go to zero, infinity, or not exist without the function having an extremum at this point.

What is the sufficient condition for the extremum of a 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 ) has maximum provided that the function f(x ) is continuous here.

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

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

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

About the case f?? (a) = 0 can be read 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., 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 discontinuities.

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

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

Derivative of a function: y? (x) = 6 x + 2

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

6x + 2 = 0.6x = -2,x = -2/6 = -1/3

In this case, the critical point is x 0 = -1/3. It is with this argument value that the function has extremum. To him find, 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 when passing through the critical point x 0 changes from “plus” to “minus”, 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 point x 0 there is neither a maximum nor a minimum.

For the example considered:

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

At x = -1 the value of the derivative will be y? (-1) = 6*(-1) + 2 = -6 + 2 = -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

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

As you can see, the derivative changed sign from minus to plus when passing through the critical point. This means that at the critical value x 0 we have a minimum point.

Largest and smallest value of a function on the interval(on a segment) are found using 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) = 3 sin (x) - 0.5x

at intervals:

a) [-9; 9]

b) [ -6; -3 ]

So, the derivative of the function is

y? (x) = 3 cos (x) - 0.5

Solving Equation 3 cos(x) - 0.5 = 0

3cos(x) = 0.5

cos(x) = 0.5/3 = 0.16667

x = ±arccos(0.16667) + 2πk.

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

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

x = - arccos (0.16667) - 2 π *1 = -7.687

x = arccos (0.16667) - 2 π *1 = -4.88

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

x = arccos (0.16667) + 2 π *0 = 1.403

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

x = arccos (0.16667) + 2 π *1 = 7.687

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

We find the function values 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 when 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 equal to y = 5.398.

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 greatest value of the function

y = 5.398 at x = -4.88

smallest value -

y = 1.077 at x = -3

How to find the inflection points of a function graph and determine the convex and concave sides?

To find all inflection points of a line y = f(x ), we 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 doesn’t change, then there is no bend.

The roots of the equation f? (x ) = 0, as well as possible discontinuity points of the function and the second derivative, divide the domain of definition of the function into a number of intervals. The convexity on 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 ) here the concavity is turned upward, and if it is negative, then downward.

How to find the extrema of a function of two variables?

To find the extrema of a function f(x, y ), differentiable in the domain of its specification, you need:

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

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

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

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

for all points (x;y) sufficiently close to P 0 . If the difference remains positive, then at point P 0 we have a minimum, if negative, then we have a maximum. If the difference does not retain its sign, then there is no extremum at point P 0.

The extrema of a function are determined similarly for a larger number of arguments.

Sources:

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