The maximum and minimum points of a function. What are extrema of a function: critical points of maximum and minimum

Many problems require calculating the maximum or minimum value of a quadratic function. The maximum or minimum can be found if the original function is written in standard form: or through the coordinates of the vertex of the parabola: f (x) = a (x − h) 2 + k (\displaystyle f(x)=a(x-h)^(2)+k). Moreover, the maximum or minimum of any quadratic function can be calculated using mathematical operations.

Steps

The quadratic function is written in standard form

Write the function in standard form. A quadratic function is a function whose equation involves a variable x 2 (\displaystyle x^(2)). The equation may or may not include a variable x (\displaystyle x). If an equation includes a variable with an exponent greater than 2, it does not describe a quadratic function. If necessary, provide similar terms and rearrange them to write the function in standard form.

• For example, given the function f (x) = 3 x + 2 x − x 2 + 3 x 2 + 4 (\displaystyle f(x)=3x+2x-x^(2)+3x^(2)+4). Add terms with variable x 2 (\displaystyle x^(2)) and members with variable x (\displaystyle x) to write the equation in standard form:
  • f (x) = 2 x 2 + 5 x + 4 (\displaystyle f(x)=2x^(2)+5x+4)

1. The graph of a quadratic function is a parabola. The branches of the parabola are directed up or down. If the coefficient a (\displaystyle a) with variable x 2 (\displaystyle x^(2)) a (\displaystyle a)

   • f (x) = 2 x 2 + 4 x − 6 (\displaystyle f(x)=2x^(2)+4x-6). Here a = 2 (\displaystyle a=2)
   • f (x) = − 3 x 2 + 2 x + 8 (\displaystyle f(x)=-3x^(2)+2x+8). Here, therefore, the parabola is directed downward.
   • f (x) = x 2 + 6 (\displaystyle f(x)=x^(2)+6). Here a = 1 (\displaystyle a=1), so the parabola is directed upward.
   • If the parabola is directed upward, you need to look for its minimum. If the parabola is pointing down, look for its maximum.

2. Compute -b/2a. Meaning − b 2 a (\displaystyle -(\frac (b)(2a))) is the coordinate x (\displaystyle x) vertices of the parabola. If a quadratic function is written in standard form a x 2 + b x + c (\displaystyle ax^(2)+bx+c), use the coefficients for x (\displaystyle x) And x 2 (\displaystyle x^(2)) as follows:

   • In the function coefficients a = 1 (\displaystyle a=1) And b = 10 (\displaystyle b=10)
     • x = − 10 (2) (1) (\displaystyle x=-(\frac (10)((2)(1))))
     • x = − 10 2 (\displaystyle x=-(\frac (10)(2)))
   • As a second example, consider the function. Here a = − 3 (\displaystyle a=-3) And b = 6 (\displaystyle b=6). Therefore, calculate the “x” coordinate of the vertex of the parabola as follows:
     • x = − b 2 a (\displaystyle x=-(\frac (b)(2a)))
     • x = − 6 (2) (− 3) (\displaystyle x=-(\frac (6)((2)(-3))))
     • x = − 6 − 6 (\displaystyle x=-(\frac (6)(-6)))
     • x = − (− 1) (\displaystyle x=-(-1))
     • x = 1 (\displaystyle x=1)

3. Find the corresponding value of f(x). Plug the found value of “x” into the original function to find the corresponding value of f(x). This way you will find the minimum or maximum of the function.

   • In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) you have calculated that the x coordinate of the vertex of the parabola is x = − 5 (\displaystyle x=-5). In the original function, instead of x (\displaystyle x) substitute − 5 (\displaystyle -5)
     • f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1)
     • f (x) = (− 5) 2 + 10 (− 5) − 1 (\displaystyle f(x)=(-5)^(2)+10(-5)-1)
     • f (x) = 25 − 50 − 1 (\displaystyle f(x)=25-50-1)
     • f (x) = − 26 (\displaystyle f(x)=-26)
   • In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) you found that the x coordinate of the vertex of the parabola is x = 1 (\displaystyle x=1). In the original function, instead of x (\displaystyle x) substitute 1 (\displaystyle 1) to find its maximum value:
     • f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4)
     • f (x) = − 3 (1) 2 + 6 (1) − 4 (\displaystyle f(x)=-3(1)^(2)+6(1)-4)
     • f (x) = − 3 + 6 − 4 (\displaystyle f(x)=-3+6-4)
     • f (x) = − 1 (\displaystyle f(x)=-1)

4. Write down your answer. Re-read the problem statement. If you need to find the coordinates of the vertex of a parabola, write down both values ​​in your answer x (\displaystyle x) And y (\displaystyle y)(or f (x) (\displaystyle f(x))). If you need to calculate the maximum or minimum of a function, write down only the value in your answer y (\displaystyle y)(or f (x) (\displaystyle f(x))). Look again at the sign of the coefficient a (\displaystyle a) to check whether you calculated the maximum or minimum.

   • In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) meaning a (\displaystyle a) positive, so you have calculated the minimum. The vertex of the parabola lies at the point with coordinates (− 5 , − 26) (\displaystyle (-5,-26)), and the minimum value of the function is − 26 (\displaystyle -26).
   • In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) meaning a (\displaystyle a) negative, so you have found the maximum. The vertex of the parabola lies at the point with coordinates (1 , − 1) (\displaystyle (1,-1)), and the maximum value of the function is − 1 (\displaystyle -1).

5. Determine the direction of the parabola. To do this, look at the sign of the coefficient a (\displaystyle a). If the coefficient a (\displaystyle a) positive, the parabola is directed upward. If the coefficient a (\displaystyle a) negative, the parabola is directed downward. For example:

   • . Here a = 2 (\displaystyle a=2), that is, the coefficient is positive, so the parabola is directed upward.
   • . Here a = − 3 (\displaystyle a=-3), that is, the coefficient is negative, so the parabola is directed downward.
   • If the parabola is directed upward, you need to calculate the minimum value of the function. If the parabola is directed downward, you need to find the maximum value of the function.

6. Find the minimum or maximum value of the function. If the function is written through the coordinates of the vertex of the parabola, the minimum or maximum is equal to the value of the coefficient k (\displaystyle k). In the above examples:

   • f (x) = 2 (x + 1) 2 − 4 (\displaystyle f(x)=2(x+1)^(2)-4). Here k = − 4 (\displaystyle k=-4). This is the minimum value of the function because the parabola is directed upward.
   • f (x) = − 3 (x − 2) 2 + 2 (\displaystyle f(x)=-3(x-2)^(2)+2). Here k = 2 (\displaystyle k=2). This is the maximum value of the function because the parabola is directed downward.

7. Find the coordinates of the vertex of the parabola. If the problem requires finding the vertex of a parabola, its coordinates are (h , k) (\displaystyle (h,k)). Please note that when a quadratic function is written through the coordinates of the vertex of a parabola, the subtraction operation must be enclosed in parentheses (x − h) (\displaystyle (x-h)), so the value h (\displaystyle h) is taken with the opposite sign.

   • f (x) = 2 (x + 1) 2 − 4 (\displaystyle f(x)=2(x+1)^(2)-4). Here the addition operation (x+1) is enclosed in parentheses, which can be rewritten as follows: (x-(-1)). Thus, h = − 1 (\displaystyle h=-1). Therefore, the coordinates of the vertex of the parabola of this function are equal to (− 1 , − 4) (\displaystyle (-1,-4)).
   • f (x) = − 3 (x − 2) 2 + 2 (\displaystyle f(x)=-3(x-2)^(2)+2). Here in brackets is the expression (x-2). Hence, h = 2 (\displaystyle h=2). The coordinates of the vertex are (2,2).

How to Calculate Minimum or Maximum Using Math Operations

1. First, let's look at the standard form of the equation. Write the quadratic function in standard form: f (x) = a x 2 + b x + c (\displaystyle f(x)=ax^(2)+bx+c). If necessary, add similar terms and rearrange them to obtain the standard equation.

   • For example: .

2. Find the first derivative. The first derivative of a quadratic function, which is written in standard form, is equal to f ′ (x) = 2 a x + b (\displaystyle f^(\prime )(x)=2ax+b).

   • f (x) = 2 x 2 − 4 x + 1 (\displaystyle f(x)=2x^(2)-4x+1). The first derivative of this function is calculated as follows:
     • f ′ (x) = 4 x − 4 (\displaystyle f^(\prime )(x)=4x-4)

3. Equate the derivative to zero. Recall that the derivative of a function is equal to the slope of the function at a certain point. At minimum or maximum, the slope is zero. Therefore, to find the minimum or maximum value of a function, the derivative must be set to zero. In our example.

77419.Find the maximum point of the function y=x 3 –48x+17

Let's find the zeros of the derivative:

Let's get the roots:

Let's determine the signs of the derivative of the function by substituting values ​​from the intervals into the resulting derivative, and depict the behavior of the function in the figure:

We found that at point –4 the derivative changes its sign from positive to negative. Thus, point x=–4 is the desired maximum point.

Answer: –4

77423. Find the maximum point of the function y=x 3 –3x 2 +2

Let's find the derivative of the given function:

Let's equate the derivative to zero and solve the equation:

At the point x=0, the derivative changes sign from positive to negative, which means this is the maximum point.

77427. Find the maximum point of the function y=x 3 +2x 2 +x+3

Let's find the derivative of the given function:

When we equalize the derivative to zero and solve the equation:

Let's determine the signs of the derivative of the function and depict in the figure the intervals of increase and decrease of the function by substituting the values ​​from each interval into the expression of the derivative:

At the point x=–1, the derivative changes sign from positive to negative, which means this is the desired maximum point.

Answer: –1

77431. Find the maximum point of the function y=x 3 –5x 2 +7x–5

Let's find the derivative of the function:

Let's find the zeros of the derivative:

3x 2 – 10x + 7 = 0

3∙0 2 – 10∙0 + 7 = 7 > 0

3∙2 2 – 10∙2 + 7 = – 1< 0

3∙3 2 – 10∙3 + 7 = 4 > 0

At the point x = 1, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

77435. Find the maximum point of the function y=7+12x–x 3

Let's find the derivative of the function:

Let's find the zeros of the derivative:

12 – 3x 2 = 0

Solving the quadratic equation we get:

*These are points of possible maximum (minimum) of the function.

Let's construct a number line and mark the zeros of the derivative. Let's determine the signs of the derivative by substituting an arbitrary value from each interval into the expression of the derivative of the function and schematically depict the increase and decrease on the intervals:

12 – 3∙(–3) 2 = –15 < 0

12 – 3∙0 2 = 12 > 0

12 – 3∙3 2 = –15 < 0

At the point x = 2, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

*For the same function, the minimum point is the point x = – 2.

77439. Find the maximum point of the function y=9x 2 – x 3

Let's find the derivative of the function:

Let's find the zeros of the derivative:

18x –3x 2 = 0

3x(6 – x) = 0

Solving the equation we get:

*These are points of possible maximum (minimum) of the function.

Let's construct a number line and mark the zeros of the derivative. Let's determine the signs of the derivative by substituting an arbitrary value from each interval into the expression of the derivative of the function and schematically depict the increase and decrease on the intervals:

18 (–1) –3 (–1) 2 = –21< 0

18∙1 –3∙1 2 = 15 > 0

18∙7 –3∙7 2 = –1 < 0

At the point x=6, the derivative changes its sign from positive to negative, which means this is the desired maximum point.

*For the same function, the minimum point is the point x = 0.

With this service you can find the largest and smallest value of a function one variable f(x) with the solution formatted 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 increasing and decreasing functions.

Find the largest and smallest value of a function

y=

on the segment [ ;]

Include theory

Rules for entering functions:

Necessary condition for the extremum of a function of one variable

The equation f" 0 (x *) = 0 is a necessary condition for the extremum of a function of one variable, i.e. at point x * the first derivative of the function must vanish. It identifies stationary points x c at which the function does not increase or decrease .

Sufficient condition for the 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 point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *) > 0

Then point x * is the local (global) minimum point of the function.

If at point x * the condition is met:

F" 0 (x *) = 0
f"" 0 (x *)< 0

Then point x * is a local (global) maximum.

Example No. 1. Find the largest and smallest values ​​of the function: on the segment.
Solution.

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 at x=2; f max =9 at x=1

Example No. 2. Using higher order derivatives, find the extremum of the function y=x-2sin(x) .
Solution.
Find the derivative of the function: y’=1-2cos(x) . Let's find the critical points: 1-cos(x)=2, cos(x)=½, x=± π / 3 +2πk, k∈Z. We find y’’=2sin(x), calculate , which means x= π / 3 +2πk, k∈Z are the minimum points of the function; , which means x=- π / 3 +2πk, k∈Z are the maximum points of the function.

Example No. 3. Investigate the extremum function in the vicinity of the point x=0.
Solution. 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 can 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 it is necessary to use other methods to study functions at an extremum.

Increasing, decreasing and extrema of a function

Finding the intervals of increase, decrease and extrema of a function is both an independent task and an essential part of other tasks, in particular, full function study. Initial information about the increase, decrease and extrema of the function is given in theoretical chapter on derivative, which I highly recommend for preliminary study (or repetition)– also for the reason that the following material is based on the very essentially derivative, being a harmonious continuation of this article. Although, if time is short, then a purely formal practice of examples from today’s lesson is also possible.

And today there is a spirit of rare unanimity in the air, and I can directly feel that everyone present is burning with desire learn to explore a function using its derivative. Therefore, reasonable, good, eternal terminology immediately appears on your monitor screens.

For what? One of the reasons is the most practical: so that it is clear what is generally required of you in a particular task!

Monotonicity of the function. Extremum points and extrema of a function

Let's consider some function. To put it simply, we assume that she continuous on the entire number line:

Just in case, let’s immediately get rid of possible illusions, especially for those readers who have recently become acquainted with intervals of constant sign of the function. Now we NOT INTERESTED, how the graph of the function is located relative to the axis (above, below, where the axis intersects). To be convincing, mentally erase the axes and leave one graph. Because that’s where the interest lies.

Function increases on an interval if for any two points of this interval connected by the relation , the inequality is true. That is, a larger value of the argument corresponds to a larger value of the function, and its graph goes “from bottom to top”. The demonstration function grows over the interval.

Likewise, the function decreases on an interval if for any two points of a given interval such that , the inequality is true. That is, a larger value of the argument corresponds to a smaller value of the function, and its graph goes “from top to bottom”. Our function decreases on intervals .

If a function increases or decreases over an interval, then it is called strictly monotonous at this interval. What is monotony? Take it literally – monotony.

You can also define non-decreasing function (relaxed condition in the first definition) and non-increasing function (softened condition in the 2nd definition). A non-decreasing or non-increasing function on an interval is called a monotonic function on a given interval (strict monotonicity is a special case of “simply” monotonicity).

The theory also considers other approaches to determining the increase/decrease of a function, including on half-intervals, segments, but in order not to pour oil-oil-oil on your head, we will agree to operate with open intervals with categorical definitions - this is clearer, and for solving many practical problems quite enough.

Thus, in my articles the wording “monotonicity of a function” will almost always be hidden intervals strict monotony(strictly increasing or strictly decreasing function).

Neighborhood of a point. Words after which students run away wherever they can and hide in horror in the corners. ...Although after the post Cauchy limits They’re probably no longer hiding, but just shuddering slightly =) Don’t worry, now there will be no proofs of the theorems of mathematical analysis - I needed the surroundings to formulate the definitions more strictly extremum points. Let's remember:

Neighborhood of a point an interval that contains a given point is called, and for convenience the interval is often assumed to be symmetrical. For example, a point and its standard neighborhood:

Actually, the definitions:

The point is called strict maximum point, If exists her neighborhood, for everyone values ​​of which, except for the point itself, the inequality . In our specific example, this is a dot.

The point is called strict minimum point, If exists her neighborhood, for everyone values ​​of which, except for the point itself, the inequality . In the drawing there is point “a”.

Note : the requirement of neighborhood symmetry is not at all necessary. In addition, it is important the very fact of existence neighborhood (whether tiny or microscopic) that satisfies the specified conditions

The points are called strictly extremum points or just extremum points functions. That is, it is a generalized term for maximum points and minimum points.

How do we understand the word “extreme”? Yes, just as directly as monotony. Extreme points of roller coasters.

As in the case of monotonicity, loose postulates exist and are even more common in theory (which, of course, the strict cases considered fall under!):

The point is called maximum point, If exists its surroundings are such that for everyone
The point is called minimum point, If exists its surroundings are such that for everyone values ​​of this neighborhood, the inequality holds.

Note that according to the last two definitions, any point of a constant function (or a “flat section” of a function) is considered both a maximum and a minimum point! The function, by the way, is both non-increasing and non-decreasing, that is, monotonic. However, we will leave these considerations to theorists, since in practice we almost always contemplate traditional “hills” and “hollows” (see drawing) with a unique “king of the hill” or “princess of the swamp”. As a variety, it occurs tip, directed up or down, for example, the minimum of the function at the point.

Oh, and speaking of royalty:
– the meaning is called maximum functions;
– the meaning is called minimum functions.

Common name – extremes functions.

Please be careful with your words!

Extremum points– these are “X” values.
Extremes– “game” meanings.

! Note : sometimes the listed terms refer to the “X-Y” points that lie directly on the GRAPH OF the function ITSELF.

How many extrema can a function have?

None, 1, 2, 3, ... etc. ad infinitum. For example, sine has infinitely many minima and maxima.

IMPORTANT! The term "maximum of function" not identical the term “maximum value of a function”. It is easy to notice that the value is maximum only in a local neighborhood, and there are “cooler comrades” at the top left. Likewise, “minimum of a function” is not the same as “minimum value of a function,” and in the drawing we see that the value is minimum only in a certain area. In this regard, extremum points are also called local extremum points, and the extrema – local extremes. They walk and wander nearby and global brethren. So, any parabola has at its vertex global minimum or global maximum. Further, I will not distinguish between types of extremes, and the explanation is voiced more for general educational purposes - the additional adjectives “local”/“global” should not take you by surprise.

Let’s summarize our short excursion into the theory with a test shot: what does the task “find the monotonicity intervals and extremum points of the function” mean?

The wording encourages you to find:

– intervals of increasing/decreasing function (non-decreasing, non-increasing appears much less often);

– maximum and/or minimum points (if any exist). Well, to avoid failure, it’s better to find the minimums/maximums themselves ;-)

How to determine all this? Using the derivative function!

How to find intervals of increasing, decreasing,
extremum points and extrema of the function?

Many rules, in fact, are already known and understood from lesson about the meaning of a derivative.

Tangent derivative brings the cheerful news that function is increasing throughout domain of definition.

With cotangent and its derivative the situation is exactly the opposite.

The arcsine increases over the interval - the derivative here is positive: .
When the function is defined, but not differentiable. However, at the critical point there is a right-handed derivative and a right-handed tangent, and at the other edge there are their left-handed counterparts.

I think it won’t be too difficult for you to carry out similar reasoning for the arc cosine and its derivative.

All of the above cases, many of which are tabular derivatives, I remind you, follow directly from derivative definitions.

Why explore a function using its derivative?

To better understand what the graph of this function looks like: where it goes “bottom up”, where “top down”, where it reaches minimums and maximums (if it reaches at all). Not all functions are so simple - in most cases we have no idea at all about the graph of a particular function.

It's time to move on to more meaningful examples and consider algorithm for finding intervals of monotonicity and extrema of a function:

Example 1

Find intervals of increase/decrease and extrema of the function

Solution:

1) The first step is to find domain of a function, and also take note of breakpoints (if they exist). In this case, the function is continuous on the entire number line, and this action is to a certain extent formal. But in a number of cases, serious passions flare up here, so let’s treat the paragraph without disdain.

2) The second point of the algorithm is due to

a necessary condition for an extremum:

If there is an extremum at a point, then either the value does not exist.

Confused by the ending? Extremum of the “modulus x” function .

The condition is necessary, but not enough, and the converse is not always true. So, it does not yet follow from the equality that the function reaches a maximum or minimum at point . A classic example has already been highlighted above - this is a cubic parabola and its critical point.

But be that as it may, the necessary condition for an extremum dictates the need to find suspicious points. To do this, find the derivative and solve the equation:

At the beginning of the first article about function graphs I told you how to quickly build a parabola using an example : “...we take the first derivative and equate it to zero: ...So, the solution to our equation: - it is at this point that the vertex of the parabola is located...”. Now, I think, everyone understands why the vertex of the parabola is located exactly at this point =) In general, we should start with a similar example here, but it is too simple (even for a dummies). In addition, there is an analogue at the very end of the lesson about derivative of a function. Therefore, let's increase the degree:

Example 2

Find intervals of monotonicity and extrema of the function

This is an example for you to solve on your own. A complete solution and an approximate final sample of the problem at the end of the lesson.

The long-awaited moment of meeting with fractional-rational functions has arrived:

Example 3

Explore a function using the first derivative

Pay attention to how variably one and the same task can be reformulated.

Solution:

1) The function suffers infinite discontinuities at points.

2) We detect critical points. Let's find the first derivative and equate it to zero:

Let's solve the equation. A fraction is zero when its numerator is zero:

Thus, we get three critical points:

3) We plot ALL detected points on the number line and interval method we define the signs of the DERIVATIVE:

I remind you that you need to take some point in the interval and calculate the value of the derivative at it and determine its sign. It’s more profitable not to even count, but to “estimate” verbally. Let's take, for example, a point belonging to the interval and perform the substitution: .

Two “pluses” and one “minus” give a “minus”, therefore, which means that the derivative is negative over the entire interval.

The action, as you understand, needs to be carried out for each of the six intervals. By the way, note that the numerator factor and denominator are strictly positive for any point in any interval, which greatly simplifies the task.

So, the derivative told us that the FUNCTION ITSELF increases by and decreases by . It is convenient to connect intervals of the same type with the join icon.

At the point the function reaches its maximum:
At the point the function reaches a minimum:

Think about why you don't have to recalculate the second value ;-)

When passing through a point, the derivative does not change sign, so the function has NO EXTREMUM there - it both decreased and remained decreasing.

! Let's repeat an important point: points are not considered critical - they contain a function not defined. Accordingly, here In principle there can be no extremes(even if the derivative changes sign).

Answer: function increases by and decreases by At the point the maximum of the function is reached: , and at the point – the minimum: .

Knowledge of monotonicity intervals and extrema, coupled with established asymptotes already gives a very good idea of ​​the appearance of the function graph. A person of average training is able to verbally determine that the graph of a function has two vertical asymptotes and an oblique asymptote. Here is our hero:

Try once again to correlate the results of the study with the graph of this function.
There is no extremum at the critical point, but there is graph inflection(which, as a rule, happens in similar cases).

Example 4

Find the extrema of the function

Example 5

Find monotonicity intervals, maxima and minima of the function

…it’s almost like some kind of “X in a cube” holiday today....
Soooo, who in the gallery offered to drink for this? =)

Each task has its own substantive nuances and technical subtleties, which are commented on at the end of the lesson.