In practice, it is quite common to use the derivative in order to calculate the largest and smallest value of a function. We perform this action when we figure out how to minimize costs, increase profits, calculate the optimal load on production, etc., that is, in those cases when it is necessary to determine the optimal value of a parameter. To solve such problems correctly, one must have a good understanding of what the largest and smallest value of a function are.
Usually we define these values within some interval x , which in turn can correspond to the entire scope of the function or part of it. It can be either a segment [ a ; b ] , and open interval (a ; b) , (a ; b ] , [ a ; b) , infinite interval (a ; b) , (a ; b ] , [ a ; b) or infinite interval - ∞ ; a , (- ∞ ; a ] , [ a ; + ∞) , (- ∞ ; + ∞) .
In this article, we will describe how the largest and smallest value of an explicitly given function with one variable y=f(x) y = f (x) is calculated.
Basic definitions
We begin, as always, with the formulation of the main definitions.
Definition 1
The largest value of the function y = f (x) on some interval x is the value m a x y = f (x 0) x ∈ X , which, for any value x x ∈ X , x ≠ x 0, makes the inequality f (x) ≤ f (x 0) .
Definition 2
The smallest value of the function y = f (x) on some interval x is the value m i n x ∈ X y = f (x 0) , which, for any value x ∈ X , x ≠ x 0, makes the inequality f(X f (x) ≥ f(x0) .
These definitions are fairly obvious. It can be even simpler to say this: the largest value of a function is its largest value on a known interval at the abscissa x 0, and the smallest is the smallest accepted value on the same interval at x 0.
Definition 3
Stationary points are such values of the function argument at which its derivative becomes 0.
Why do we need to know what stationary points are? To answer this question, we need to remember Fermat's theorem. It follows from it that a stationary point is a point at which the extremum of a differentiable function is located (i.e., its local minimum or maximum). Consequently, the function will take the smallest or largest value on a certain interval exactly at one of the stationary points.
Another function can take on the largest or smallest value at those points at which the function itself is definite, and its first derivative does not exist.
The first question that arises when studying this topic is: in all cases, can we determine the maximum or minimum value of a function on a given interval? No, we cannot do this when the boundaries of the given interval will coincide with the boundaries of the domain of definition, or if we are dealing with an infinite interval. It also happens that a function in a given interval or at infinity will take on infinitely small or infinitely large values. In these cases, it is not possible to determine the largest and/or smallest value.
These moments will become more understandable after the image on the graphs:
The first figure shows us a function that takes on the largest and smallest values (m a x y and m i n y) at stationary points located on the interval [ - 6 ; 6].
Let us examine in detail the case indicated in the second graph. Let's change the value of the segment to [ 1 ; 6] and we get that the largest value of the function will be achieved at the point with the abscissa in the right boundary of the interval, and the smallest - at the stationary point.
In the third figure, the abscissas of the points represent the boundary points of the segment [ - 3 ; 2]. They correspond to the largest and smallest value of the given function.
Now let's look at the fourth picture. In it, the function takes m a x y (the largest value) and m i n y (the smallest value) at stationary points in the open interval (- 6 ; 6) .
If we take the interval [ 1 ; 6) , then we can say that the smallest value of the function on it will be reached at a stationary point. We will not know the maximum value. The function could take the largest value at x equal to 6 if x = 6 belonged to the interval. It is this case that is shown in Figure 5.
On graph 6, this function acquires the smallest value in the right border of the interval (- 3 ; 2 ] , and we cannot draw definite conclusions about the largest value.
In figure 7, we see that the function will have m a x y at the stationary point, having an abscissa equal to 1 . The function reaches its minimum value at the interval boundary on the right side. At minus infinity, the values of the function will asymptotically approach y = 3 .
If we take an interval x ∈ 2 ; + ∞ , then we will see that the given function will not take on it either the smallest or the largest value. If x tends to 2, then the values of the function will tend to minus infinity, since the straight line x = 2 is a vertical asymptote. If the abscissa tends to plus infinity, then the values of the function will asymptotically approach y = 3. This is the case shown in Figure 8.
In this paragraph, we will give a sequence of actions that must be performed to find the largest or smallest value of a function on a certain interval.
- First, let's find the domain of the function. Let's check whether the segment specified in the condition is included in it.
- Now let's calculate the points contained in this segment at which the first derivative does not exist. Most often, they can be found in functions whose argument is written under the modulus sign, or in power functions, the exponent of which is a fractionally rational number.
- Next, we find out which stationary points fall into a given segment. To do this, you need to calculate the derivative of the function, then equate it to 0 and solve the resulting equation, and then choose the appropriate roots. If we do not get a single stationary point or they do not fall into a given segment, then we proceed to the next step.
- Let us determine what values the function will take at the given stationary points (if any), or at those points where the first derivative does not exist (if any), or we calculate the values for x = a and x = b .
- 5. We have a series of function values, from which we now need to choose the largest and smallest. This will be the largest and smallest values of the function that we need to find.
Let's see how to apply this algorithm correctly when solving problems.
Example 1
Condition: the function y = x 3 + 4 x 2 is given. Determine its largest and smallest value on the segments [ 1 ; 4 ] and [ - 4 ; - one ] .
Decision:
Let's start by finding the domain of this function. In this case, it will be the set of all real numbers except 0 . In other words, D (y) : x ∈ (- ∞ ; 0) ∪ 0 ; +∞ . Both segments specified in the condition will be inside the definition area.
Now we calculate the derivative of the function according to the rule of differentiation of a fraction:
y "= x 3 + 4 x 2" = x 3 + 4 " x 2 - x 3 + 4 x 2" x 4 = = 3 x 2 x 2 - (x 3 - 4) 2 x x 4 = x 3 - 8 x 3
We learned that the derivative of the function will exist at all points of the segments [ 1 ; 4 ] and [ - 4 ; - one ] .
Now we need to determine the stationary points of the function. Let's do this with the equation x 3 - 8 x 3 = 0. It has only one real root, which is 2. It will be a stationary point of the function and will fall into the first segment [ 1 ; 4 ] .
Let us calculate the values of the function at the ends of the first segment and at the given point, i.e. for x = 1 , x = 2 and x = 4:
y(1) = 1 3 + 4 1 2 = 5 y(2) = 2 3 + 4 2 2 = 3 y(4) = 4 3 + 4 4 2 = 4 1 4
We have obtained that the largest value of the function m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 will be achieved at x = 1 , and the smallest m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 – at x = 2 .
The second segment does not include any stationary points, so we need to calculate the function values only at the ends of the given segment:
y (- 1) = (- 1) 3 + 4 (- 1) 2 = 3
Hence, m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
Answer: For the segment [ 1 ; 4 ] - m a x y x ∈ [ 1 ; 4 ] = y (2) = 3 , m i n y x ∈ [ 1 ; 4 ] = y (2) = 3 , for the segment [ - 4 ; - 1 ] - m a x y x ∈ [ - 4 ; - 1 ] = y (- 1) = 3 , m i n y x ∈ [ - 4 ; - 1 ] = y (- 4) = - 3 3 4 .
See picture:
Before learning this method, we advise you to review how to correctly calculate the one-sided limit and the limit at infinity, as well as learn the basic methods for finding them. To find the largest and / or smallest value of a function on an open or infinite interval, we perform the following steps in sequence.
- First you need to check whether the given interval will be a subset of the domain of the given function.
- Let us determine all the points that are contained in the required interval and at which the first derivative does not exist. Usually they occur in functions where the argument is enclosed in the sign of the module, and in power functions with a fractionally rational exponent. If these points are missing, then you can proceed to the next step.
- Now we determine which stationary points fall into a given interval. First, we equate the derivative to 0, solve the equation and find suitable roots. If we do not have a single stationary point or they do not fall within the specified interval, then we immediately proceed to further actions. They are determined by the type of interval.
- If the interval looks like [ a ; b) , then we need to calculate the value of the function at the point x = a and the one-sided limit lim x → b - 0 f (x) .
- If the interval has the form (a ; b ] , then we need to calculate the value of the function at the point x = b and the one-sided limit lim x → a + 0 f (x) .
- If the interval has the form (a ; b) , then we need to calculate the one-sided limits lim x → b - 0 f (x) , lim x → a + 0 f (x) .
- If the interval looks like [ a ; + ∞) , then it is necessary to calculate the value at the point x = a and the limit to plus infinity lim x → + ∞ f (x) .
- If the interval looks like (- ∞ ; b ] , we calculate the value at the point x = b and the limit at minus infinity lim x → - ∞ f (x) .
- If - ∞ ; b , then we consider the one-sided limit lim x → b - 0 f (x) and the limit at minus infinity lim x → - ∞ f (x)
- If - ∞ ; + ∞ , then we consider the limits to minus and plus infinity lim x → + ∞ f (x) , lim x → - ∞ f (x) .
- At the end, you need to draw a conclusion based on the obtained values of the function and limits. There are many options here. So, if the one-sided limit is equal to minus infinity or plus infinity, then it is immediately clear that nothing can be said about the smallest and largest value of the function. Below we will consider one typical example. Detailed descriptions will help you understand what's what. If necessary, you can return to figures 4 - 8 in the first part of the material.
Condition: given a function y = 3 e 1 x 2 + x - 6 - 4 . Calculate its largest and smallest value in the intervals - ∞ ; - 4 , - ∞ ; - 3 , (- 3 ; 1 ] , (- 3 ; 2) , [ 1 ; 2) , 2 ; + ∞ , [ 4 ; +∞) .
Decision
First of all, we find the domain of the function. The denominator of the fraction is a square trinomial, which should not go to 0:
x 2 + x - 6 = 0 D = 1 2 - 4 1 (- 6) = 25 x 1 = - 1 - 5 2 = - 3 x 2 = - 1 + 5 2 = 2 ⇒ D (y) : x ∈ (- ∞ ; - 3) ∪ (- 3 ; 2) ∪ (2 ; + ∞)
We have obtained the scope of the function, to which all the intervals specified in the condition belong.
Now let's differentiate the function and get:
y "= 3 e 1 x 2 + x - 6 - 4" = 3 e 1 x 2 + x - 6 " = 3 e 1 x 2 + x - 6 1 x 2 + x - 6 " == 3 e 1 x 2 + x - 6 1 "x 2 + x - 6 - 1 x 2 + x - 6" (x 2 + x - 6) 2 = - 3 (2 x + 1) e 1 x 2 + x - 6 x 2 + x - 6 2
Consequently, derivatives of a function exist on the entire domain of its definition.
Let's move on to finding stationary points. The derivative of the function becomes 0 at x = - 1 2 . This is a stationary point that is in the intervals (- 3 ; 1 ] and (- 3 ; 2) .
Let's calculate the value of the function at x = - 4 for the interval (- ∞ ; - 4 ] , as well as the limit at minus infinity:
y (- 4) \u003d 3 e 1 (- 4) 2 + (- 4) - 6 - 4 \u003d 3 e 1 6 - 4 ≈ - 0. 456 lim x → - ∞ 3 e 1 x 2 + x - 6 = 3 e 0 - 4 = - 1
Since 3 e 1 6 - 4 > - 1 , then m a x y x ∈ (- ∞ ; - 4 ] = y (- 4) = 3 e 1 6 - 4 . This does not allow us to uniquely determine the smallest value of the function. We can only to conclude that there is a limit below - 1 , since it is to this value that the function approaches asymptotically at minus infinity.
A feature of the second interval is that it does not have a single stationary point and not a single strict boundary. Therefore, we cannot calculate either the largest or the smallest value of the function. By defining the limit at minus infinity and as the argument tends to - 3 on the left side, we get only the range of values:
lim x → - 3 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 - 0 3 e 1 (x + 3) (x - 3) - 4 = 3 e 1 (- 3 - 0 + 3) (- 3 - 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → - ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
This means that the function values will be located in the interval - 1 ; +∞
To find the maximum value of the function in the third interval, we determine its value at the stationary point x = - 1 2 if x = 1 . We also need to know the one-sided limit for the case when the argument tends to - 3 on the right side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e 4 25 - 4 ≈ - 1 . 444 y (1) = 3 e 1 1 2 + 1 - 6 - 4 ≈ - 1 . 644 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 - 3 + 0 + 3 (- 3 + 0 - 2) - 4 = = 3 e 1 (- 0) - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
It turned out that the function will take the largest value at a stationary point m a x y x ∈ (3 ; 1 ] = y - 1 2 = 3 e - 4 25 - 4. As for the smallest value, we cannot determine it. All that we know , is the presence of a lower limit to - 4 .
For the interval (- 3 ; 2), let's take the results of the previous calculation and once again calculate what the one-sided limit is equal to when tending to 2 from the left side:
y - 1 2 = 3 e 1 - 1 2 2 + - 1 2 - 6 - 4 = 3 e - 4 25 - 4 ≈ - 1 . 444 lim x → - 3 + 0 3 e 1 x 2 + x - 6 - 4 = - 4 lim x → 2 - 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 - 0 + 3) (2 - 0 - 2) - 4 = = 3 e 1 - 0 - 4 = 3 e - ∞ - 4 = 3 0 - 4 = - 4
Hence, m a x y x ∈ (- 3 ; 2) = y - 1 2 = 3 e - 4 25 - 4 , and the smallest value cannot be determined, and the values of the function are bounded from below by the number - 4 .
Based on what we did in the two previous calculations, we can assert that on the interval [ 1 ; 2) the function will take the largest value at x = 1, and it is impossible to find the smallest.
On the interval (2 ; + ∞), the function will not reach either the largest or the smallest value, i.e. it will take values from the interval - 1 ; +∞ .
lim x → 2 + 0 3 e 1 x 2 + x - 6 - 4 = lim x → - 3 + 0 3 e 1 (x + 3) (x - 2) - 4 = 3 e 1 (2 + 0 + 3 ) (2 + 0 - 2) - 4 = = 3 e 1 (+ 0) - 4 = 3 e + ∞ - 4 = + ∞ lim x → + ∞ 3 e 1 x 2 + x - 6 - 4 = 3 e 0 - 4 = - 1
Having calculated what the value of the function will be equal to at x = 4 , we find out that m a x y x ∈ [ 4 ; + ∞) = y (4) = 3 e 1 14 - 4 , and the given function at plus infinity will asymptotically approach the line y = - 1 .
Let's compare what we got in each calculation with the graph of the given function. In the figure, the asymptotes are shown by dotted lines.
That's all we wanted to talk about finding the largest and smallest value of a function. Those sequences of actions that we have given will help you make the necessary calculations as quickly and simply as possible. But remember that it is often useful to first find out on which intervals the function will decrease and on which it will increase, after which further conclusions can be drawn. So you can more accurately determine the largest and smallest value of the function and justify the results.
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Let's see how to explore a function using a graph. It turns out that looking at the graph, you can find out everything that interests us, namely:
- function scope
- function range
- function zeros
- periods of increase and decrease
- high and low points
- the largest and smallest value of the function on the segment.
Let's clarify the terminology:
Abscissa is the horizontal coordinate of the point.
Ordinate- vertical coordinate.
abscissa- the horizontal axis, most often called the axis.
Y-axis- vertical axis, or axis.
Argument is an independent variable on which the values of the function depend. Most often indicated.
In other words, we ourselves choose , substitute in the function formula and get .
Domain functions - the set of those (and only those) values of the argument for which the function exists.
Denoted: or .
In our figure, the domain of the function is a segment. It is on this segment that the graph of the function is drawn. Only here this function exists.
Function range is the set of values that the variable takes. In our figure, this is a segment - from the lowest to the highest value.
Function zeros- points where the value of the function is equal to zero, i.e. . In our figure, these are the points and .
Function values are positive where . In our figure, these are the intervals and .
Function values are negative where . We have this interval (or interval) from to.
The most important concepts - increasing and decreasing function on some set. As a set, you can take a segment, an interval, a union of intervals, or the entire number line.
Function increases
In other words, the more , the more , that is, the graph goes to the right and up.
Function decreasing on the set if for any and belonging to the set the inequality implies the inequality .
For a decreasing function, a larger value corresponds to a smaller value. The graph goes right and down.
In our figure, the function increases on the interval and decreases on the intervals and .
Let's define what is maximum and minimum points of the function.
Maximum point- this is an internal point of the domain of definition, such that the value of the function in it is greater than in all points sufficiently close to it.
In other words, the maximum point is such a point, the value of the function at which more than in neighboring ones. This is a local "hill" on the chart.
In our figure - the maximum point.
Low point- an internal point of the domain of definition, such that the value of the function in it is less than in all points sufficiently close to it.
That is, the minimum point is such that the value of the function in it is less than in neighboring ones. On the graph, this is a local “hole”.
In our figure - the minimum point.
The point is the boundary. It is not an interior point of the domain of definition and therefore does not fit the definition of a maximum point. After all, she has no neighbors on the left. In the same way, there can be no minimum point on our chart.
The maximum and minimum points are collectively called extremum points of the function. In our case, this is and .
But what if you need to find, for example, function minimum on the cut? In this case, the answer is: because function minimum is its value at the minimum point.
Similarly, the maximum of our function is . It is reached at the point .
We can say that the extrema of the function are equal to and .
Sometimes in tasks you need to find the largest and smallest values of the function on a given segment. They do not necessarily coincide with extremes.
In our case smallest function value on the interval is equal to and coincides with the minimum of the function. But its largest value on this segment is equal to . It is reached at the left end of the segment.
In any case, the largest and smallest values of a continuous function on a segment are achieved either at the extremum points or at the ends of the segment.
Sometimes in problems B15 there are "bad" functions for which it is difficult to find the derivative. Previously, this was only on probes, but now these tasks are so common that they can no longer be ignored when preparing for this exam.
In this case, other tricks work, one of which is - monotone.
The function f (x) is called monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following is true:
x 1< x 2 ⇒ f (x 1) < f (x2).
The function f (x) is called monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following is true:
x 1< x 2 ⇒ f (x 1) > f( x2).
In other words, for an increasing function, the larger x is, the larger f(x) is. For a decreasing function, the opposite is true: the more x , the smaller f(x).
For example, the logarithm increases monotonically if the base a > 1 and decreases monotonically if 0< a < 1. Не забывайте про область допустимых значений логарифма: x > 0.
f (x) = log a x (a > 0; a ≠ 1; x > 0)
The arithmetic square (and not only square) root increases monotonically over the entire domain of definition:
The exponential function behaves similarly to the logarithm: it increases for a > 1 and decreases for 0< a < 1. Но в отличие от логарифма, показательная функция определена для всех чисел, а не только для x > 0:
f (x) = a x (a > 0)
Finally, degrees with a negative exponent. You can write them as a fraction. They have a break point where monotony is broken.
All these functions are never found in their pure form. Polynomials, fractions and other nonsense are added to them, because of which it becomes difficult to calculate the derivative. What happens in this case - now we will analyze.
Parabola vertex coordinates
Most often, the function argument is replaced with square trinomial of the form y = ax 2 + bx + c . Its graph is a standard parabola, in which we are interested in:
- Parabola branches - can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
- The vertex of the parabola is the extremum point of a quadratic function, at which this function takes its smallest (for a > 0) or largest (a< 0) значение.
Of greatest interest is top of a parabola, the abscissa of which is calculated by the formula:
So, we have found the extremum point of the quadratic function. But if the original function is monotonic, for it the point x 0 will also be an extremum point. Thus, we formulate the key rule:
The extremum points of the square trinomial and the complex function it enters into coincide. Therefore, you can look for x 0 for a square trinomial, and forget about the function.
From the above reasoning, it remains unclear what kind of point we get: a maximum or a minimum. However, the tasks are specifically designed so that it does not matter. Judge for yourself:
- There is no segment in the condition of the problem. Therefore, it is not required to calculate f(a) and f(b). It remains to consider only the extremum points;
- But there is only one such point - this is the top of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.
Thus, the solution of the problem is greatly simplified and reduced to just two steps:
- Write out the parabola equation y = ax 2 + bx + c and find its vertex using the formula: x 0 = −b /2a;
- Find the value of the original function at this point: f (x 0). If there are no additional conditions, this will be the answer.
At first glance, this algorithm and its justification may seem complicated. I deliberately do not post a "bare" solution scheme, since the thoughtless application of such rules is fraught with errors.
Consider the real tasks from the trial exam in mathematics - this is where this technique is most common. At the same time, we will make sure that in this way many problems of B15 become almost verbal.
Under the root is a quadratic function y \u003d x 2 + 6x + 13. The graph of this function is a parabola with branches up, since the coefficient a \u003d 1\u003e 0.
Top of the parabola:
x 0 \u003d -b / (2a) \u003d -6 / (2 1) \u003d -6 / 2 \u003d -3
Since the branches of the parabola are directed upwards, at the point x 0 \u003d −3, the function y \u003d x 2 + 6x + 13 takes on the smallest value.
The root is monotonically increasing, so x 0 is the minimum point of the entire function. We have:
Task. Find the smallest value of the function:
y = log 2 (x 2 + 2x + 9)
Under the logarithm is again a quadratic function: y \u003d x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.
Top of the parabola:
x 0 \u003d -b / (2a) \u003d -2 / (2 1) \u003d -2/2 \u003d -1
So, at the point x 0 = −1, the quadratic function takes on the smallest value. But the function y = log 2 x is monotone, so:
y min = y (−1) = log 2 ((−1) 2 + 2 (−1) + 9) = ... = log 2 8 = 3
The exponent is a quadratic function y = 1 − 4x − x 2 . Let's rewrite it in normal form: y = −x 2 − 4x + 1.
Obviously, the graph of this function is a parabola, branches down (a = −1< 0). Поэтому вершина будет точкой максимума:
x 0 = −b /(2a ) = −(−4)/(2 (−1)) = 4/(−2) = −2
The original function is exponential, it is monotone, so the largest value will be at the found point x 0 = −2:
An attentive reader will surely notice that we did not write out the area of \u200b\u200bpermissible values of the root and logarithm. But this was not required: inside there are functions whose values are always positive.
Consequences from the scope of a function
Sometimes, to solve problem B15, it is not enough just to find the vertex of the parabola. The desired value may lie at the end of the segment, but not at the extremum point. If the task does not specify a segment at all, look at tolerance range original function. Namely:
Pay attention again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how it works with specific examples:
Task. Find the largest value of the function:
Under the root is again a quadratic function: y \u003d 3 - 2x - x 2. Its graph is a parabola, but branches down since a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический квадратный корень из отрицательного числа не существует.
We write out the area of permissible values (ODZ):
3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; one]
Now find the vertex of the parabola:
x 0 = −b /(2a ) = −(−2)/(2 (−1)) = 2/(−2) = −1
The point x 0 = −1 belongs to the ODZ segment - and this is good. Now we consider the value of the function at the point x 0, as well as at the ends of the ODZ:
y(−3) = y(1) = 0
So, we got the numbers 2 and 0. We are asked to find the largest - this is the number 2.
Task. Find the smallest value of the function:
y = log 0.5 (6x - x 2 - 5)
Inside the logarithm there is a quadratic function y \u003d 6x - x 2 - 5. This is a parabola with branches down, but there cannot be negative numbers in the logarithm, so we write out the ODZ:
6x − x 2 − 5 > 0 ⇒ x 2 − 6x + 5< 0 ⇒ (x − 1)(x − 5) < 0 ⇒ x ∈ (1; 5)
Please note: the inequality is strict, so the ends do not belong to the ODZ. In this way, the logarithm differs from the root, where the ends of the segment suit us quite well.
Looking for the vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 (−1)) = −6/(−2) = 3
The top of the parabola fits along the ODZ: x 0 = 3 ∈ (1; 5). But since the ends of the segment do not interest us, we consider the value of the function only at the point x 0:
y min = y (3) = log 0.5 (6 3 − 3 2 − 5) = log 0.5 (18 − 9 − 5) = log 0.5 4 = −2
\(\blacktriangleright\) In order to find the largest/smallest value of a function on the segment \(\) , it is necessary to schematically depict the graph of the function on this segment.
In the problems from this subtopic, this can be done using the derivative: find the intervals of increase (\(f">0\) ) and decrease (\(f"<0\)
) функции, критические точки (где \(f"=0\)
или \(f"\)
не существует).
\(\blacktriangleright\) Do not forget that the function can take the maximum/smallest value not only at the internal points of the segment \(\) , but also at its ends.
\(\blacktriangleright\) The largest/smallest value of the function is the value of the coordinate \(y=f(x)\) .
\(\blacktriangleright\) The derivative of a complex function \(f(t(x))\) is searched according to the rule: \[(\Large(f"(x)=f"(t)\cdot t"(x)))\]
\[\begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x)\\ \hline \textbf(1) & c & 0\\&&\\ \textbf(2) & x^a & a\cdot x^(a-1)\\&&\\ \textbf(3) & \ln x & \dfrac1x\\&&\\ \ textbf(4) & \log_ax & \dfrac1(x\cdot \ln a)\\&&\\ \textbf(5) & e^x & e^x\\&&\\ \textbf(6) & a^x & a^x\cdot \ln a\\&&\\ \textbf(7) & \sin x & \cos x\\&&\\ \textbf(8) & \cos x & -\sin x\\ \hline \end(array) \quad \quad \quad \quad \begin(array)(|r|c|c|) \hline & \text(Function ) f(x) & \text(Derivative ) f"(x) \\ \hline \textbf(9) & \mathrm(tg)\, x & \dfrac1(\cos^2 x)\\&&\\ \textbf(10) & \mathrm(ctg)\, x & -\ ,\dfrac1(\sin^2 x)\\&&\\ \textbf(11) & \arcsin x & \dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(12) & \ arccos x & -\,\dfrac1(\sqrt(1-x^2))\\&&\\ \textbf(13) & \mathrm(arctg)\, x & \dfrac1(1+x^2)\\ &&\\ \textbf(14) & \mathrm(arcctg)\, x & -\,\dfrac1(1+x^2)\\ \hline \end(array)\]
Task 1 #2357
Task level: Equal to the Unified State Examination
Find the smallest value of the function \(y = e^(x^2 - 4)\) on the interval \([-10; -2]\) .
ODZ: \(x\) - arbitrary.
1) \
\ So \(y" = 0\) when \(x = 0\) .
3) Let's find intervals of constant sign \(y"\) on the considered segment \([-10; -2]\) :
4) Sketch of the graph on the segment \([-10; -2]\) :
Thus, the function reaches its smallest value on \([-10; -2]\) at \(x = -2\) .
\ Total: \(1\) is the smallest value of the function \(y\) on \([-10; -2]\) .
Answer: 1
Task 2 #2355
Task level: Equal to the Unified State Examination
\(y = \sqrt(2)\cdot\sqrt(x^2 + 1)\) on the segment \([-1; 1]\) .
ODZ: \(x\) - arbitrary.
1) \
Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[\sqrt(2)\cdot\dfrac(x)(\sqrt(x^2 + 1)) = 0\qquad\Leftrightarrow\qquad x = 0\,.\] The derivative exists for any \(x\) .
2) Find the intervals of constant sign \(y"\) :
3) Let's find intervals of constant sign \(y"\) on the considered segment \([-1; 1]\) :
4) Sketch of the graph on the segment \([-1; 1]\) :
Thus, the function reaches its maximum value on \([-1; 1]\) in \(x = -1\) or in \(x = 1\) . Let's compare the values of the function at these points.
\ Total: \(2\) is the largest value of the function \(y\) on \([-1; 1]\) .
Answer: 2
Task 3 #2356
Task level: Equal to the Unified State Examination
Find the smallest value of the function \(y = \cos 2x\) on the interval \(\) .
ODZ: \(x\) - arbitrary.
1) \
Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[-2\cdot \sin 2x = 0\qquad\Leftrightarrow\qquad 2x = \pi n, n\in\mathbb(Z)\qquad\Leftrightarrow\qquad x = \dfrac(\pi n)(2), n\in\mathbb(Z)\,.\] The derivative exists for any \(x\) .
2) Find the intervals of constant sign \(y"\) :
(here there is an infinite number of intervals in which the signs of the derivative alternate).
3) Let's find intervals of constancy \(y"\) on the considered segment \(\) :
4) Sketch of the graph on the segment \(\) :
Thus, the function reaches its smallest value on \(\) at \(x = \dfrac(\pi)(2)\) .
\ Total: \(-1\) is the smallest value of the function \(y\) on \(\) .
Answer: -1
Task 4 #915
Task level: Equal to the Unified State Examination
Find the largest value of a function
\(y = -\log_(17)(2x^2 - 2\sqrt(2)x + 2)\).
ODZ: \(2x^2 - 2\sqrt(2)x + 2 > 0\) . Let's decide on ODZ:
1) Denote \(2x^2-2\sqrt(2)x+2=t(x)\) , then \(y(t)=-\log_(17)t\) .
Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[-\dfrac(1)(\ln 17)\cdot\dfrac(4x-2\sqrt(2))(2x^2-2\sqrt(2)x+2) = 0\qquad\Leftrightarrow\qquad 4x-2\sqrt(2) = 0\]– on the ODZ, from where we find the root \(x = \dfrac(\sqrt(2))(2)\) . The derivative of the function \(y\) does not exist for \(2x^2-2\sqrt(2)x+2 = 0\) , but this equation has a negative discriminant, therefore, it has no solutions. In order to find the largest / smallest value of a function, you need to understand how its graph looks schematically.
2) Find the intervals of constant sign \(y"\) :
3) Graphic sketch:
Thus, the function reaches its maximum value at \(x = \dfrac(\sqrt(2))(2)\) :
\(y\left(\dfrac(\sqrt(2))(2)\right) = -\log_(17)1 = 0\),
Total: \(0\) is the largest value of the function \(y\) .
Answer: 0
Task 5 #2344
Task level: Equal to the Unified State Examination
Find the smallest value of a function
\(y = \log_(3)(x^2 + 8x + 19)\).
ODZ: \(x^2 + 8x + 19 > 0\) . Let's decide on ODZ:
1) Denote \(x^2 + 8x + 19=t(x)\) , then \(y(t)=\log_(3)t\) .
Let's find the critical points (that is, the internal points of the function's domain, in which its derivative is equal to \(0\) or does not exist): \[\dfrac(1)(\ln 3)\cdot\dfrac(2x+8)(x^2 + 8x + 19) = 0\qquad\Leftrightarrow\qquad 2x+8 = 0\]- on the ODZ, from where we find the root \ (x \u003d -4 \) . The derivative of the function \(y\) does not exist for \(x^2 + 8x + 19 = 0\) , but this equation has a negative discriminant, therefore, it has no solutions. In order to find the largest / smallest value of a function, you need to understand how its graph looks schematically.
2) Find the intervals of constant sign \(y"\) :
3) Graphic sketch:
Thus, \(x = -4\) is the minimum point of the function \(y\) and the smallest value is reached in it:
\(y(-4) = \log_(3)3 = 1\) .
Total: \(1\) is the smallest value of the function \(y\) .
Answer: 1
Task 6 #917
Task level: More difficult than the exam
Find the largest value of a function
\(y = -e^((x^2 - 12x + 36 + 2\ln 2))\).
From a practical point of view, the most interesting is the use of the derivative to find the largest and smallest value of a function. What is it connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life, one has to solve the problem of optimizing some parameters. And this is the problem of finding the largest and smallest values of the function.
It should be noted that the largest and smallest value of a function is usually sought on some interval X , which is either the entire domain of the function or part of the domain. The interval X itself can be a line segment, an open interval 
, an infinite interval .
In this article, we will talk about finding the largest and smallest values of an explicitly given function of one variable y=f(x) .
Page navigation.
The largest and smallest value of a function - definitions, illustrations.
Let us briefly dwell on the main definitions.
The largest value of the function 
, which for any 
the inequality is true.
The smallest value of the function y=f(x) on the interval X is called such a value 
, which for any the inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) value accepted in the interval under consideration with the abscissa.
Stationary points are the values of the argument at which the derivative of the function vanishes.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. It follows from this theorem that if a differentiable function has an extremum (local minimum or local maximum) at some point, then this point is stationary. Thus, the function often takes its largest (smallest) value on the interval X at one of the stationary points from this interval.
Also, a function can often take on the largest and smallest values at points where the first derivative of this function does not exist, and the function itself is defined.
Let's immediately answer one of the most common questions on this topic: "Is it always possible to determine the largest (smallest) value of a function"? No not always. Sometimes the boundaries of the interval X coincide with the boundaries of the domain of the function, or the interval X is infinite. And some functions at infinity and on the boundaries of the domain of definition can take both infinitely large and infinitely small values. In these cases, nothing can be said about the largest and smallest value of the function.
For clarity, we give a graphic illustration. Look at the pictures - and much will become clear.
On the segment
In the first figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points inside the segment [-6;6] .
Consider the case shown in the second figure. Change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest - at a point with an abscissa corresponding to the right boundary of the interval.
In figure No. 3, the boundary points of the segment [-3; 2] are the abscissas of the points corresponding to the largest and smallest value of the function.
In the open range
In the fourth figure, the function takes the largest (max y ) and smallest (min y ) values at stationary points within the open interval (-6;6) .
On the interval , no conclusions can be drawn about the largest value.
At infinity
In the example shown in the seventh figure, the function takes the largest value (max y ) at a stationary point with the abscissa x=1 , and the smallest value (min y ) is reached at the right boundary of the interval. At minus infinity, the values of the function asymptotically approach y=3 .
On the interval, the function does not reach either the smallest or the largest value. As x=2 tends to the right, the function values tend to minus infinity (the straight line x=2 is a vertical asymptote), and as the abscissa tends to plus infinity, the function values asymptotically approach y=3 . A graphic illustration of this example is shown in Figure 8.
Algorithm for finding the largest and smallest values of a continuous function on the segment .
We write an algorithm that allows us to find the largest and smallest value of a function on a segment.
- We find the domain of the function and check if it contains the entire segment .
- We find all points at which the first derivative does not exist and which are contained in the segment (usually such points occur in functions with an argument under the module sign and in power functions with a fractional-rational exponent). If there are no such points, then go to the next point.
- We determine all stationary points that fall into the segment. To do this, we equate it to zero, solve the resulting equation and choose the appropriate roots. If there are no stationary points or none of them fall into the segment, then go to the next step.
- We calculate the values of the function at the selected stationary points (if any), at points where the first derivative does not exist (if any), and also at x=a and x=b .
- From the obtained values of the function, we select the largest and smallest - they will be the desired maximum and smallest values of the function, respectively.
Let's analyze the algorithm when solving an example for finding the largest and smallest values of a function on a segment.
Example.
Find the largest and smallest value of a function
- on the segment;
- on the interval [-4;-1] .
Decision.
The domain of the function is the entire set of real numbers, except for zero, that is, . Both segments fall within the domain of definition.
We find the derivative of the function with respect to: 
Obviously, the derivative of the function exists at all points of the segments and [-4;-1] .
Stationary points are determined from the equation . The only real root is x=2 . This stationary point falls into the first segment.
For the first case, we calculate the values of the function at the ends of the segment and at a stationary point, that is, for x=1 , x=2 and x=4 : 
Therefore, the largest value of the function 
is reached at x=1 , and the smallest value 
– at x=2 .
For the second case, we calculate the values of the function only at the ends of the segment [-4;-1] (since it does not contain any stationary points): 
Decision.
Let's start with the scope of the function. The square trinomial in the denominator of a fraction must not vanish: 
It is easy to check that all intervals from the condition of the problem belong to the domain of the function.
Let's differentiate the function: 
Obviously, the derivative exists on the entire domain of the function.
Let's find stationary points. The derivative vanishes at . This stationary point falls within the intervals (-3;1] and (-3;2) .
And now you can compare the results obtained at each point with the graph of the function. The blue dotted lines indicate the asymptotes.
This can end with finding the largest and smallest value of the function. The algorithms discussed in this article allow you to get results with a minimum of actions. However, it can be useful to first determine the intervals of increase and decrease of the function and only after that draw conclusions about the largest and smallest value of the function on any interval. This gives a clearer picture and a rigorous justification of the results.
