In task B14 from the Unified State Examination in mathematics, you need to find the smallest or highest value functions of one variable. This is a fairly trivial problem from mathematical analysis, and it is for this reason that every graduate can and should learn to solve it normally high school. Let's look at a few examples that schoolchildren solved during diagnostic work in mathematics, held in Moscow on December 7, 2011.
Depending on the interval over which you want to find the maximum or minimum value of a function, one of the following standard algorithms is used to solve this problem.
I. Algorithm for finding the largest or smallest value of a function on a segment:
- Find the derivative of the function.
- Select from the points suspected of being an extremum those that belong to the given segment and domain of definition of the function.
- Calculate values functions(not derivative!) at these points.
- Among the obtained values, select the largest or smallest, it will be the desired one.
Example 1. Find the smallest value of the function
y = x 3 – 18x 2 + 81x+ 23 on the segment.
Solution: We follow the algorithm for finding the smallest value of a function on a segment:
- The scope of a function is not limited: D(y) = R.
- The derivative of the function is equal to: y' = 3x 2 – 36x+ 81. The domain of definition of the derivative of a function is also not limited: D(y’) = R.
- Zeros of the derivative: y' = 3x 2 – 36x+ 81 = 0, which means x 2 – 12x+ 27 = 0, whence x= 3 and x= 9, our interval includes only x= 9 (one point suspicious for an extremum).
- We find the value of the function at a point suspicious of an extremum and at the edges of the gap. For ease of calculation, we present the function in the form: y = x 3 – 18x 2 + 81x + 23 = x(x-9) 2 +23:
- y(8) = 8 · (8-9) 2 +23 = 31;
- y(9) = 9 · (9-9) 2 +23 = 23;
- y(13) = 13 · (13-9) 2 +23 = 231.
So, of the obtained values, the smallest is 23. Answer: 23.
II. Algorithm for finding the largest or smallest value of a function:
- Find the domain of definition of the function.
- Find the derivative of the function.
- Identify points suspicious for extremum (those points at which the derivative of the function vanishes, and points at which there is no two-sided finite derivative).
- Mark these points and the domain of definition of the function on the number line and determine the signs derivative(not functions!) on the resulting intervals.
- Define values functions(not the derivative!) at the minimum points (those points at which the sign of the derivative changes from minus to plus), the smallest of these values will be the smallest value of the function. If there are no minimum points, then the function does not have a minimum value.
- Define values functions(not the derivative!) at the maximum points (those points at which the sign of the derivative changes from plus to minus), the largest of these values will be the largest value of the function. If there are no maximum points, then the function does not have the greatest value.
Example 2. Find the largest value of the function.
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.
Rules for entering functions:
Necessary condition for the extremum of a function of one variable
The equation f" 0 (x *) = 0 is necessary condition 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 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.
Example No. 4. Divide the number 49 into two terms whose product will be greatest.
Solution. Let us denote x as the first term. Then (49-x) is the second term.
The product will be maximum: x·(49-x) → max
From a practical point of view, the greatest interest is in using the derivative to find the largest and smallest values of a function. What is this connected with? Maximizing profits, minimizing costs, determining the optimal load of equipment... In other words, in many areas of life we have to solve problems of optimizing some parameters. And these are the tasks of finding the largest and smallest values of a function.
It should be noted that the largest and smallest values of a function are usually sought on a certain interval X, which is either the entire domain of the function or part of the domain of definition. The interval X itself can be a segment, an open interval 
, an infinite interval.
In this article we will talk about finding the largest and smallest values of an explicitly specified function of one variable y=f(x) .
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The largest and smallest value of a function - definitions, illustrations.
Let's briefly look at the main definitions.
The largest value of the function 
that for anyone 
inequality is true.
The smallest value of the function y=f(x) on the interval X is called such a value 
that for anyone inequality is true.
These definitions are intuitive: the largest (smallest) value of a function is the largest (smallest) accepted value on the interval under consideration at the abscissa.
Stationary points– these are the values of the argument at which the derivative of the function becomes zero.
Why do we need stationary points when finding the largest and smallest values? The answer to this question is given by Fermat's theorem. From this theorem it follows 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 its largest and smallest values at points at which 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 definition of the function, or the interval X is infinite. And some functions at infinity and at the boundaries of the domain of definition can take on 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 will give a graphic illustration. Look at the pictures and a lot will become clearer.
On the segment
In the first figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside the segment [-6;6].
Consider the case depicted in the second figure. Let's change the segment to . In this example, the smallest value of the function is achieved at a stationary point, and the largest at the point with the abscissa corresponding to the right boundary of the interval.
In Figure 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.
On an open interval
In the fourth figure, the function takes the largest (max y) and smallest (min y) values at stationary points located inside the open interval (-6;6).
On the interval , no conclusions can be drawn about the largest value.
At infinity
In the example presented in the seventh figure, the function takes the largest value (max y) at a stationary point with abscissa x=1, and the smallest value (min y) is achieved on the right boundary of the interval. At minus infinity, the function values asymptotically approach y=3.
Over the interval, the function reaches neither the smallest nor the largest value. As x=2 approaches from the right, the function values tend to minus infinity (the 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 a segment.
Let us write an algorithm that allows us to find the largest and smallest values of a function on a segment.
- We find the domain of definition of the function and check whether it contains the entire segment.
- We find all the points at which the first derivative does not exist and which are contained in the segment (usually such points are found in functions with an argument under the modulus sign and in power functions with a fractional-rational exponent). If there are no such points, then move on to the next point.
- We determine all stationary points falling within the segment. To do this, we equate it to zero, solve the resulting equation and select suitable roots. If there are no stationary points or none of them fall into the segment, then move on to the next point.
- We calculate the values of the function at selected stationary points (if any), at points at which the first derivative does not exist (if any), as well as at x=a and x=b.
- From the obtained values of the function, we select the largest and smallest - they will be the required largest and smallest values of the function, respectively.
Let's analyze the algorithm for solving an example to find 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 segment [-4;-1] .
Solution.
The domain of definition of a function is the entire set of real numbers, with the exception of zero, that is. Both segments fall within the definition domain.
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].
We determine stationary points 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 the stationary point, that is, for x=1, x=2 and x=4: 
Therefore, the greatest value of the function 
is achieved at x=1, and the smallest value 
– at x=2.
For the second case, we calculate the function values only at the ends of the segment [-4;-1] (since it does not contain a single stationary point): 
Solution.
Let's start with the domain of the function. The square trinomial in the denominator of the fraction must not vanish: 
It is easy to check that all intervals from the problem statement belong to the domain of definition of the function.
Let's differentiate the function: 
Obviously, the derivative exists throughout the entire domain of definition of the function.
Let's find stationary points. The derivative goes to zero at . This stationary point falls within the intervals (-3;1] and (-3;2).
Now you can compare the results obtained at each point with the graph of the function. Blue dotted lines indicate asymptotes.
At this point we can finish with finding the largest and smallest values 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 values of the function on any interval. This gives a clearer picture and rigorous justification for the results.
Sometimes in problems B15 there are “bad” functions for which it is difficult to find a derivative. Previously, this only happened during sample tests, but now these tasks are so common that they can no longer be ignored when preparing for the real Unified State Exam.
In this case, other techniques work, one of which is monotone.
A function f (x) is said to be monotonically increasing on the segment if for any points x 1 and x 2 of this segment the following holds:
x 1< x 2 ⇒ f (x 1) < f (x 2).
A function f (x) is said to be monotonically decreasing on the segment if for any points x 1 and x 2 of this segment the following holds:
x 1< x 2 ⇒ f (x 1) > f ( x 2).
In other words, for an increasing function, the larger x, the larger f(x). For a decreasing function the opposite is true: the larger x, the less f(x).
For example, the logarithm increases monotonically if the base a > 1, and monotonically decreases 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 the monotony is broken.
All these functions are never found in their pure form. They add polynomials, fractions and other nonsense, which makes it difficult to calculate the derivative. Let's look at what happens in this case.
Parabola vertex coordinates
Most often the function argument is replaced with quadratic trinomial of the form y = ax 2 + bx + c. Its graph is a standard parabola in which we are interested in:
- The branches of a parabola can go up (for a > 0) or down (a< 0). Задают направление, в котором функция может принимать бесконечные значения;
- The vertex of a parabola is the extremum point of a quadratic function at which this function takes its minimum (for a > 0) or maximum (a< 0) значение.
Of greatest interest is vertex of 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, let us formulate the key rule:
The extremum points of a quadratic trinomial and the complex function in which it is included coincide. Therefore, you can look for x 0 for a quadratic trinomial, and forget about the function.
From the above reasoning, it remains unclear which point we get: maximum or minimum. However, the tasks are specifically designed so that this does not matter. Judge for yourself:
- There is no segment in the problem statement. Therefore, there is no need 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 vertex of the parabola x 0, the coordinates of which are calculated literally orally and without any derivatives.
Thus, solving the problem is greatly simplified and comes down to just two steps:
- Write out the equation of the parabola 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 rationale may seem complex. I deliberately do not post a “bare” solution diagram, since thoughtless application of such rules is fraught with errors.
Let's look at real problems from trial Unified State Exam in mathematics - this is where this technique is found most often. At the same time, we will make sure that in this way many B15 problems become almost oral.
Under the root stands quadratic function y = x 2 + 6x + 13. The graph of this function is a parabola with branches upward, since the coefficient a = 1 > 0.
Vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 1) = −6/2 = −3
Since the branches of the parabola are directed upward, at the point x 0 = −3 the function y = x 2 + 6x + 13 takes on its minimum value.
The root increases monotonically, which means 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 there is again a quadratic function: y = x 2 + 2x + 9. The graph is a parabola with branches up, because a = 1 > 0.
Vertex of the parabola:
x 0 = −b /(2a ) = −2/(2 1) = −2/2 = −1
So, at the point x 0 = −1 the quadratic function takes on its minimum value. But the function y = log 2 x is monotonic, so:
y min = y (−1) = log 2 ((−1) 2 + 2 · (−1) + 9) = ... = log 2 8 = 3
The exponent contains the 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 monotonic, so the greatest value will be at the found point x 0 = −2:
An attentive reader will probably notice that we did not write out the range of permissible values of the root and logarithm. But this was not required: inside there are functions whose values are always positive.
Corollaries from the domain of a function
Sometimes simply finding the vertex of the parabola is not enough to solve Problem B15. The value you are looking for may lie at the end of the segment, and not at all at the extremum point. If the problem does not indicate a segment at all, look at range of acceptable values original function. Namely:
Please note again: zero may well be under the root, but never in the logarithm or denominator of a fraction. Let's see how this works with specific examples:
Task. Find the largest value of the function:
Under the root is again a quadratic function: y = 3 − 2x − x 2 . Its graph is a parabola, but branches down because a = −1< 0. Значит, парабола уходит на минус бесконечность, что недопустимо, поскольку арифметический square root of a negative number does not exist.
We write out the range of permissible values (APV):
3 − 2x − x 2 ≥ 0 ⇒ x 2 + 2x − 3 ≤ 0 ⇒ (x + 3)(x − 1) ≤ 0 ⇒ x ∈ [−3; 1]
Now let's 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 calculate 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 = 6x − x 2 − 5. This is a parabola with branches down, but in a logarithm there cannot be negative numbers, 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. This differs the logarithm from the root, where the ends of the segment suit us quite well.
We are looking for the vertex of the parabola:
x 0 = −b /(2a ) = −6/(2 · (−1)) = −6/(−2) = 3
The vertex of the parabola fits according to the ODZ: x 0 = 3 ∈ (1; 5). But since we are not interested in the ends of the segment, we calculate 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