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## Theory of Functions and function evaluation

Basic Definitions ofFunctions:It is a rule which associates a value in domain through a value in the range.Relation:The function is a relation (or rule) which assigns each and every element in the domain to precisely one element in the range.Function:: It is the set of all values which might be input to a function. That is, the set of all values the independent variable might suppose. Graphically, domain is the set of all x-coordinates.Domain: It is the set of all values that are output whenever the function is computed at all the input values from domain. That is, the set of all values the dependent variable might suppose. Graphically, the range is the set of all y-coordinates.Range: Usually, the independent variable is x. Though, the independent variable is the variable that is free to suppose distinct values independently of the other variable. Most of what we are going to discuss will only include one independent variable, however realize that it is possible to encompass more than one independent variable.Independent Variable: Usually, the dependent variable is y. Though, the dependent variable is the variable that is found out based on the value of independent variable(s). When a function is written as y = 3x + 2, then y depends on x, however x does not depend on y (in the form it is written in). Therefore, x is independent of y; however y is dependent of x.Dependent Variable: It is the set of all real numbers for which the expression is stated.Implied DomainHow is a function different from a relation?Some of the guidelines are given below for finding out whether a relation is a function or not.

a) All elements in the domain (x) should be matched by an element in the range (y).

b) Each and every element in the domain (x) can merely be matched with one element in the range (y).

c) Various elements in the domain can go to the similar element in the range. The y-coordinate might be repeated; however an x-coordinate might not.

d) Some of the values in the range do not encompass to be employed at all.

Function Notation:Function notation is employed to name functions for simple reference. Suppose if each and every function in the world had to begin off with y=. Quite soon, you would become perplexed regarding which y= you were talking about. This is the manner antecedents work in English. When you just state ‘it’ is red, you really do not include any clue which ‘it’ you are talking about. You require some other way of naming the things. Enter function notation.

Function Definition:f(x) = 3x + 2

g(x,y) = x

^{2}+ 3yIn this illustration, the f is a function of x. That is, x is independent variable, and the value of f based on what x is. As well, g is the function of both x and y.

The notation f(x) doesn’t mean f times x. This means the ‘value of f computed at x’ or ‘value of f at x’ or just ‘f of x’.

Function Evaluation:f(3) = 3(3) + 2 = 9 + 2 = 11

f(3) doesn’t mean f times 3. This means the ‘value of f computed whenever x is 3’.

f(t) = 3(t) + 2 = 3t + 2

Whatever is in the parentheses on the left side of function (t in this case) is replaced for the value of independent variable on the right side.

f(x+h) = 3(x+h) + 2 = 3x + 3h + 2

Each and every occurrence of the independent variable is substituted by the quantity in parenthesis. The common mistake is to take a quantity and apply linear or affine the conversions to it.

:f(x+h)a) f(x+h) doesn’t equal to f(x) + h = 3x+2 + h

b) f(x+h) doesn’t equal to f(x) + f(h) = 3x+2 + 3h+2 = 3x + 3h + 4

c) f(x+h) does equal to 3(x+h) + 2 = 3x + 3h + 2

:f(3x)a) f(3x) doesn’t equal to 3 * f(x) = 3( 3x+2) = 9x + 6

b) This does equal to 3(3x) + 2 = 9x + 2

You as well specify which function you wish to use whenever you use the function notation. Consider:

g(2,1) = 2

^{2}+ 3(1) = 4 + 3 = 7As the order of independent variables in the original definition was x and then y, then the function g is computed when x = 2 and y = 1.

g(1,2) is something totally different. In this condition, x = 1 and y = 2.

g(1,2) = 1

^{2}+ 3(2) = 1 + 6 = 7We can say that it is not entirely different, however it is found in a different way.

Piecewise Definitions:At times, functions are a little more complicated than the simple functions we have explained so far. When different rules are used for distinct values of the independent variable, then we can employ a piecewise definition.

Let consider the function shown and the following computations:

f(2.1) = 2(2.1) = 4.2

As 2.1 is in the interval [1,3), we utilize the second piece of definition, f(x) = 2x.

f(-2) = 3 - (-2)2 =3 - 4 = -1

As -2 is in the interval (-∞,-1), we utilize the first piece of definition, f(x) = 3 - x

^{2}f(3) = 5 - (3) = 5 - 3 = 2

As 3 is not involved in the second piece, however it is comprised in the third piece on the interval [3,+∞) we employ the third piece of definition, f(x) = 5 - x.

f(0) is undefined

As 0 does not fall to any of domains, the function is undefined there.

If all the domains are joined, then domain of the function f, is the set of all real numbers apart from [-1,1). You could as well write it as (-∞,-1)U[1,+∞). The symbol U signifies the union of two sets.

Piecewise Functions and the Calculator:We can place piecewise functions to the graphing calculator. TI82 and TI83 calculators encompass a [Test] key obtained by pressing [2nd] [Math]. Beneath this key, we will find the various test operators (equivalent to, not equal to, greater than, less than, greater than or equal to, and less than or equal to).

If the calculator computes a test expression (x<1), this will return the value of 1 when the statement is true and value of 0 if the statement is false. This works actually well with multiplication, as multiplying by 1 won't modify the expression, and multiplying by 0 will form the expression 0.

y = (1 - x

^{2})*(x<-1)Let consider just the first portion of the piecewise definition from the above. When x<-1, then the calculator will return (1-x^2). When x is not less than negative one, then the calculator will return 0. That is not the proper value to return (it must be undefined), however it will graph on the x-axis and won't illustrate up as the axis is already there. Therefore, it consists of the semblance on graphing properly.

The complete piecewise function can be stated for the calculator as:

y = (1-x

^{2})*(x<-1) + 5x*(1≤x and x<3) + (5-3x)*(x≥3)The ‘and’ keyword can be found beneath [Test] Logic menu. Likewise, the ≤ and ≥ symbols can as well be found beneath the [Test] menu. When you try to place in 1≤x<3 like we right, the calculator will interpret that as 1 ≤ 3 and forever return true.

Be sure to utilize some kind of decimal setting or dot mode whenever you are viewing a piecewise function. Or else, you might get some weird outcomes.

Finding the Domain:Implied Domain:

This is the set of all real numbers where the expression is defined.

a) Start off with all the real numbers.

b) Eliminate any values of the independent variable that cause division by zero

c) Eliminate any values of the independent variable that result in taking the square root of a negative number. The square root can be substituted with any even (4th , 6th and so on) root.

You do not require to state restrictions that result from the implied domain. In another words, if there is an (x-2) in the denominator, then you do not need to state that x can’t be 2.

Other Exclusions:At times, we will need to exclude the other values.

A) Problem Constraints: These are utilized in applications. Here are some illustrations:

a) When x is the length of the side of a triangle, then x cannot be negative.

b) When x is the position on a 12' teeter totter at which the fulcrum should be positioned, then x should be between 0 and 12.

c) When x is the number of people in a room, then x cannot be negative.

B) Stated Restrictions: Sometimes the problem will merely come out and state you cannot use a specific value, or that you can employ only specific values. These are not in the implied domain; therefore they necessitate to be stated.

If you ever simplify the function, and a value that was in the implied domain is no longer in the implied domain, then it requires becoming a stated restriction. Illustration, divide (x

^{2}-4) by (x-2). The implied domain is x can’t equal to 2 as that would cause division by zero. Though, if you factor the numerator as (x-2)(x+2), then (x-2) in the numerator divides with (x-2) in the denominator and you are left with just (x+2). The fact that x can’t be 2 is no longer implied by the simple x+2 in the numerator, therefore you should now state that x can’t be 2.Combining Domains:Whenever you have a function that is a composition of some pieces, then the values in domain must be able to be used in all portions of the function. Let us state the numerator consists a square root of x in it (and hence x should be non-negative) and the denominator consists of an x-2 in it (and hence x can’t be 2). Whenever you join such domains, you get all the values of x which are non-negative apart from 2.

In another words, the domain for function is the intersection of all the domains of individual parts.

Now, let us state the numerator consists a square root of x in it (and hence x should be non-negative) and the denominator consists an x+2 in it (and hence x can’t be -2). Whenever you join those domains, you get all the values of x which are non-negative apart from -2. Well, you do not need to state that x can’t be -2 as you have already state that x is non-negative. Therefore, in this condition, the domain would be all non-negative values of x.

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