# Advice 1: How to solve a function

To compute the value functions are used different techniques: using the formula, where it is set, graph or table. All these methods have a certain algorithm execution.
Instruction
1
If you want to find the value of the function, using the formula, substitute in this formula instead of the argument (x), its valid values, i.e. values that are included in its scope. For this you need to find a domain of valid values for this function.
2
To find the domain of the function, determine what type she has. If the function of the form y = a/b, its domain would be all values except zero. The number a is any number. To find the area of function definition, radical expressions, provided the even indicator, this expression must be greater than or equal to zero. Finding the domain of the function the same expression, but with an odd index, keep in mind that x can be any number in that case, if the radical expression is not a fraction. Finding the domain of logarithmic functions, use the rule that expression which stands under the sign of logarithm must be a positive value.
3
Finding the domain of the function, go to it. For example, to solve the function: y = 2.5 x – 10 when x = 100, substitute in this formula x is the number 100. This operation will be as follows: y = 2.5 x 100 – 10; y = 240. This number is the required value of the function.
4
To find the value of the function using the graph, put in a rectangular coordinate system on the axis OX of the value of the argument (the point corresponding to the argument). Then, from the point of the guide perpendicular to the intersection of it with the plot function. From the resulting points of intersection of the perpendicular with the plot function, we drop a perpendicular on the axis OU. The base is constructed perpendicular will meet the desired value of the function.
5
If the function is set to table, then each argument value there is a corresponding value of the function.

# Advice 2: How to build a logarithmic function

Is called the logarithmic function, which is inverse exponential. Such a function has the form: y = logax, where a is a positive number (not zero). The appearance of the graph of the logarithmic function depends on the value of a.
You will need
• mathematical Handbook;
• - the range;
• - pencil;
• - notebook;
• - handle.
Instruction
1
Before you can begin to build the graph of the logarithmic function notice that the domain of definition of this function is the set of positive numbers: this value is denoted by R+. However, the logarithmic function is the domain of values, which is represented by real numbers.
2
Carefully read the conditions of the job. If a>1, the flow chart depicted an increasing logarithmic function. To prove this feature of the logarithm function easy. For example, take two arbitrary positive values of x1 and x2, and x2>x1. Prove that loga x2>x1 loga (this can be done by contradiction).
3
Assume that loga x2≤loga x1. Given that the exponential function of the form y=Ah if set and>1 increases, the inequality will be as follows: aloga x2≤x1 aloga. On the known definition of the logarithm aloga x2=x2, while aloga x1=x1. Because of this, the inequality becomes: x2≤x1, and this is directly contrary to initial assumptions, in accord with which x2>x1. So, you have come to the conclusion that we wanted to prove: if a>1 the logarithmic function is increasing.
4
Draw the graph of the logarithmic function. The graph of the function y = logax will pass through the point (1;0). If a>1, the function will be increasing. Therefore, if 0
Note
If a job log will be denoted lg x, don't think that the authors of the mathematical benefits made a mistake by omitting the letter "o" in front of you is the decimal logarithm.
For accuracy of graphing logarithmic functions calculate, what will be equal to y at different values of x (0,5; 2; 4, 8). Based on these data put of the plot.

# Advice 3: How to solve a function f x

The term of the decision function itself is not used in mathematics. Under this wording should be understood to perform some actions on the given function to find any specific characteristics, as well as clarification of the data required for plotting function.
Instruction
1
It is possible to consider an approximate scheme in which it is appropriate to explore the behavior of functions and to build its graph.
Find the domain of the function. Determine whether the function is even and odd. In the case of finding the right answer, continue testing, only the desired axis. Determine whether the function is periodic. In the case of a positive answer will continue to study just for one period. Find the break points of the function and determine its behavior in the vicinity of these points.
2
Find the point of intersection of the function with the coordinate axes. Find the asymptotes, if any. Investigate using a first derivative function for extrema and intervals of monotonicity. Also do your research by using the second derivative for convexity, concavity and points of inflection. Select points to specify the behaviour of the function and calculate the values of the function. Plot the function, considering the results obtained for all conducted researches.
3
On the axis 0X should highlight the salient point: the break point x=0 , the zeros of the function, extreme points, inflection points. In these pointx and calculate the function values (if they exist) and on the plane 0xy, select the corresponding points of the graph and the points selected for updating. A line drawn through all points is constructed, taking account of the intervals of monotony, convexity and asymptotes, and give a sketch of the graph of a function.
4
So, a specific example of the function y=((x^2)+1)/(x-1) do your research using the first derivative. Rewrite the function in the form y=x+1+2/(x-1). The first derivative is equal to y’=1-2/((x-1)^2).
Find the critical points of the rst kind: y’=0, (x-1)^2=2, the result is two points: x1=1-sqrt2, x2=1+sqrt2. Note the measured value on the scope of the function (Fig. 1).
Determine the sign of the derivative on each of the intervals. Based on the rule of alternating signs from "+" to "-" and "-" to "+", we get that the maximum point of the function x1=1-sqrt2, and the minimum point x2=1+sqrt2. The same conclusion can be drawn and the sign of the second derivative.
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