Square trinomial is called a trinomial of the form a*x 2 +b*x+c, where a,b,c are some arbitrary real numbers, and x is a variable. Moreover, the number a should not be equal to zero.
The numbers a,b,c are called coefficients. The number a is called the leading coefficient, the number b is the coefficient of x, and the number c is called the free term.
Root quadratic trinomial a*x 2 +b*x+c is any value of the variable x such that the square trinomial a*x 2 +b*x+c vanishes.
In order to find the roots of a quadratic trinomial it is necessary to solve quadratic equation of the form a*x 2 +b*x+c=0.
How to find the roots of a quadratic trinomial
To solve this, you can use one of the known methods.
- 1 way.
Finding the roots of a square trinomial using the formula.
1. Find the value of the discriminant using the formula D =b 2 -4*a*c.
2. Depending on the value of the discriminant, calculate the roots using the formulas:
If D > 0, then the square trinomial has two roots.
x = -b±√D / 2*a
If D< 0, then the square trinomial has one root.
If the discriminant is negative, then the quadratic trinomial has no roots.
- Method 2.
Finding the roots of a quadratic trinomial by isolating full square. Let's look at the example of the given quadratic trinomial. A reduced quadratic equation whose leading coefficient is equal to one.
Let's find the roots of the quadratic trinomial x 2 +2*x-3. To do this, we solve the following quadratic equation: x 2 +2*x-3=0;
Let's transform this equation:
On the left side of the equation there is a polynomial x 2 +2*x, in order to represent it as a square of the sum we need there to be another coefficient equal to 1. Add and subtract 1 from this expression, we get:
(x 2 +2*x+1) -1=3
What can be represented in parentheses as the square of a binomial
This equation breaks down into two cases: either x+1=2 or x+1=-2.
In the first case, we get the answer x=1, and in the second, x=-3.
Answer: x=1, x=-3.
As a result of the transformations, we need to get the square of the binomial on the left side, and a certain number on the right side. The right side should not contain a variable.
Study of such an object mathematical analysis as a function has great meaning and in other fields of science. For example, in economic analysis behavior is constantly required to be assessed functions profit, namely to determine its greatest meaning and develop a strategy to achieve it.
Instructions
The study of any behavior should always begin with a search for the domain of definition. Usually by condition specific task it is necessary to determine the greatest meaning functions either over this entire area, or over a specific interval of it with open or closed borders.
Based on , the largest is meaning functions y(x0), in which for any point in the domain of definition the inequality y(x0) ≥ y(x) (x ≠ x0) holds. Graphically, this point will be the highest if the argument values are placed along the abscissa axis, and the function itself along the ordinate axis.
To determine the greatest meaning functions, follow the three-step algorithm. Please note that you must be able to work with one-sided and , as well as calculate the derivative. So, let some function y(x) be given and you need to find its greatest meaning on a certain interval with boundary values A and B.
Find out whether this interval is within the scope of the definition functions. To do this, you need to find it by considering all possible restrictions: the presence of a fraction in the expression, square root etc. The domain of definition is the set of argument values for which the function makes sense. Determine whether given interval its subset. If yes, then go to next stage.
Find the derivative functions and solve the resulting equation by equating the derivative to zero. This way you will get the values of the so-called stationary points. Evaluate whether at least one of them belongs to the interval A, B.
At the third stage, consider these points and substitute their values into the function. Depending on the interval type, perform the following additional steps. If there is a segment of the form [A, B], the boundary points are included in the interval; this is indicated by parentheses. Calculate Values functions for x = A and x = B. If the interval is open (A, B), the boundary values are punctured, i.e. are not included in it. Solve one-sided limits for x→A and x→B. A combined interval of the form [A, B) or (A, B), one of whose boundaries belongs to it, the other does not. Find the one-sided limit as x tends to the punctured value, and substitute the other into the function. Infinite two-sided interval (-∞, +∞) or one-sided infinite intervals of the form: , (-∞, B).For real limits A and B, proceed according to the principles already described, and for infinite ones, look for limits for x→-∞ and x→+∞, respectively.
The task at this stage
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 explicitly given function one variable y=f(x) .
Page navigation.
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 stationary point, and the greatest - 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 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 highest 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 straight line x=2 is 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 a function is the entire set real numbers, except for 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 one 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): 
Page 1
Theoretical facts:
The square trinomial = ax2+ bx + c has an extreme value that it takes when
This value is the smallest if a > 0, and the largest if a< 0. Если существует y(макс), то y(мин) не существует, и наоборот.
No. 1. Expand this positive number And into two terms so that their product is greatest.
Solution. Let us denote one of the required terms by x. Then the second term will be equal to A - x, and their product 
or. 
Thus, the question led to finding the value of x at which this quadratic trinomial will receive the greatest value. According to Theorem 4, such a value certainly exists (since here the leading coefficient is equal to - 1, i.e. negative) and is equal to In this case, and, therefore, both terms must be equal to each other.
For example, the number 30 allows the following expansions:
All products received are less than
No. 2. There is a wire of length L. You need to bend it so that you get a rectangle that limits the largest possible area.
Solution. Let us denote (Fig. 1) one of the sides of the rectangle by x. Then, obviously, its other side will be an area 
or 
. This function takes its maximum value at, which will be the desired value of one of the sides of the rectangle. Then its other side will be , i.e. our rectangle turns out to be a square. The resulting solution to the problem can be summarized in the form of the following theorem.
Of all rectangles that have the same perimeter, the square has the largest area.
Comment.
Our problem can also be easily solved using the result obtained when solving Problem 1.
In fact, we see that the area of the rectangle we are interested in is In other words, there is a product of two factors x and But the sum of these factors is 
,T. i.e. a number that does not depend on the choice of x. This means that the matter comes down to decomposing the number into two terms so that their product is greatest. As we know, this product will be greatest when both terms are equal, i.e.
No. 3. From the existing boards you can build a fence 200 m long. You need to enclose a rectangular yard with this fence largest area, using a factory wall for one side of the yard.
trinomial theorem derivative function
Solution. Let us denote (Fig. 2) one of the sides of the yard by x. Then its other side will be equal and its area will be
According to the theorem, the greatest value of this function is achieved by it when
So, the side of the yard perpendicular to the factory wall should be equal to 50 m, from where the value for the side parallel to the wall is 100 m, i.e. the yard should have the shape of half a square.