The function can be specified as mathematical expressions or graphical image. If the polynomial written in canonical form, and the schedule is recognizable curve, it is possible to determine the value of the argument in different parts of the coordinate plane. For example, if you specify a function Y=√x, then the argument can take only positive values. And for the function F=1/x is unacceptable the value of the argument x=0.
If the function is specified graphically some arbitrary curve, conclusions about the values of the argument can be done only on the visible part of graphics in the coordinate space. It is possible that different intervals have different functional dependencies. To find the value of the argument corresponding to a particular value of the function, find the specified number to the axis OY. From this point, draw a perpendicular to the intersection with the given curve. From the resulting point drop a perpendicular on the axis OX. The number on the x-axis is the required value of the argument. It is possible that the perpendicular y-axis crosses at several points. In this case from each point of intersection drop perpendiculars to the x-axis and record was found numeric values of the argument. They all match the specified numeric value to the function.
If the function is specified by a mathematical expression, first simplify the entry. Then to find the argument to solve the equation, equating a mathematical expression for a given value of the function. For example, for the function Y=x2 the value of the function Y=4 correspond to the value of the argument h=2 and h=-2. These values are obtained from the solution of the equation x2 =4.
Advice 2: How to find intermediate value
To determine the unknown intermediate values of a function or a table of data in mathematics we use the technique of interpolation. A discrete set of known parameters can be specified with the arguments x0, x1 . . . xn and function values yj=f(xj) (where j=0, 1, . . . , n). In a simple particular case of the task of finding intermediate values of a given row can be solved using the linear interpolation.
The essence of linear interpolation can be described by the following assumption: in the interval between the known neighbouring tabular values of the argument xi and xj consider the function y=f(x) can be approximately regarded as linear. In other words, in this range the value function is proportional to the change in the argument.
More clearly this assumption can be shown in graphic form in the Cartesian coordinate system. Consider the cut functions u and u appears to be a continuous straight line with known coordinates. When searching for intermediate values of the function Y, unknown argument X is between adjacent values XI and xj. Thus, we can write the following inequalities XI < X < xj, yi < Y < yj.
Express written terms of proportions of the following: (yj – yi)/(xj – XI) = (Y – yi)/(X – XI). Here yj and xj is the target value, yi, XI – initial value of the segment, X and Y – the desired intermediate values.
As can be seen from the proportions for a given increment of the argument X - XI it is easy to find the corresponding change of the function Y – yi. Express the increment: Y – yi = ((yj – yi)/(xj – XI))*(X – XI).
Thus, the intermediate function values can be determined, knowing only the increment on which there is a change of argument. Calculate the difference yj – yi xj – XI at a given step of the argument X – XI. Substituting these values in the formula increment, find the rate of change of a function.
Find the intermediate value Y. For this to the increment value add to the initial figure o functions on the considered segment. Likewise, is any intermediate value with a specified step increment.
If there is a problem in defining X for given values of the function y=f(x) is the inverse linear interpolation. Its essence is finding values of X using the same proportion, but now as a known parameter is the increment of the function Y – u. Using a similar transformation is unknown intermediate value of the argument X = ((yj – yi)/(xj – XI))/(Y – u) + XI.
Advice 3: How to find the minimum value of the function
The need to find the minimum value of a mathematical function is a practical interest in the solution of applied tasks, for example, in the economy. Great value for business is to minimize losses.
To find the minimum value of the function, you need to determine what value of the argument x0 to run the inequality y(x0) ≤ y(x), where x ≠ x0. Usually this problem is solved at a certain interval or in the whole region of values of the function, if one is not specified. One aspect of the solution is the location of the stationary points.
A stationary point is called a value argument, in which the derivative of the function vanishes. According to the Fermat's theorem if a differentiable function takes the extreme value at some point (in this case a local minimum), then this point is stationary.
The minimum value of a function often accepts it at this point, however, it is possible to determine not always. Moreover, it is not always possible to say with certainty what is the minimum of the function or it takes an infinitely small value. Then, as a rule, find the limit to which it tends when descending.
In order to determine the minimum value of the function, you need to perform a sequence of actions, consisting of four stages: finding domain of a function, obtaining stationary points, analysis values of the function at these points and at the ends of the interval, identifying a minimum.
So, imagine you are given some function y(x) on the interval with boundary points A and B. Find the area of its definition and find out whether the interval is its subset.
Calculate the derivative of the function. Paranaita the resulting expression to zero and find the roots of the equation. Check whether these stationary points in the interval. If not, then at the next stage, they are ignored.
Consider the interval subject to boundary types: open, closed, combined or infinite. It depends on how you look for the minimum value. For example, the interval [A, b] is a closed interval. Substitute them into the function and calculate values. Do the same with a stationary point. Select the minimum result.
With open and infinite intervals the situation is more complicated. There will have to find one-sided limits that do not always give an unambiguous result. For example, for the interval with one closed and one-punctured boundary of [A, b) should find the function at x = A and one-sided limit lim y as x → -0.