Study Domain and Range

The study about domain or input of a function f(x) is the total set of possible values of the independent variable in the function. The domain can also be given explicitly. The domain should be always x-values, and the range should be always y-values. The domain has not influenced by anything because it is an independent variable.

The study about range of a function is the total set of all feasible consequential values of the dependent variable of a function. The total set of all possible resulting values of the the range of function is the dependent variable of a function, after we have substituted the values in the domain.


Examples on study domain and range



Ex:1 find the domain and range of the following relation.
{(1, –4), (6, 5), (4, –1), (2, 6), (5, 3)}

Sol:

The domain should be always x-values, and the range should be always y-values. Therefore domain and range of given function is

Domain= {1, 2, 4, 5, 6}

Range= {–4, –1, 3, 5, 6}

Ex:2 State the domain and range of the following relation.
{(–4, 2), (5, 3), (–1, 1), (0, 5), (7, 5), (8, 5)}

Sol:

The domain should be always x-values, and the range should be always y-values

{(–4, 2), (5, 3), (–1, 1), (0, 5), (7, 5), (8, 5)}

domain:  {–4, –1, 5, 7, 8}

range:  {1,2,3,5}


STUDY DOMAIN AND RANGE OF TRIGONOMETRY FUNCTION


Ex:1 Let A = {1, 2, 3, 4} and B = {1, 4, 9, 16, 25}, Find the Domain and Range.

Consider the rule f: A → B: f (x) = x2 for all x Є A.

Then, each element in A has its unique image in B. So, f is a function from A to B.

f (1) = 12 = 1, f (2) = 22 = 4, f (3) = 32 = 9, f (4) = 42 = 16.

Domain(f) = {1,2,3,4} = A, co-domain(f) = {1,4,9,16,25} = B and range(f) = {1,4,9,16}.

Clearly, 25 Є B does not have its pre-image in A.

Ex:2 Let N be the set of all natural numbers. Find the domain and range for the function f(x)=2x.

Let f: N→ N: f(x) = 2x for all x fit in to N

Then, every element in N has its unique image in N.

So, f is a function from N to N.

Clearly f(1)= 2,f(2) = 4,f(3) = 6……., and so on.

Domain(f) = N, Co-domain(f) = N, Range(f) = {2, 4, 6, 8, 10}.

what is permutation math term

In math, the term permutation is the process of rearranging the given number of elements or objects. For example set {a,b,c), namely [a,b,c], [a,c,b], [b,a,c], [b,c,a], [c,a,b], and [c,b,a].

Formula for finding permutation: P(n,r) = `(n!) / ((n-r)!)` , where, n gives number of things, and r gives number of times.

Using permutation we can find how many possible ways are there to arrange the collection of objects. Permutation avoids the repletion. In this article we will discuss about the math term permutation and how to find the permutation.


Math Term Permutation – Example Problems


Example 1: How many different ways can a set of five country flags are arranged?

Solution:

P(5,5) = `(5!)/((5 - 5)!)` = `(5 * 4 * 3 * 2 * 1!) / (0!)` = 120              [0! = 1]

Therefore 120 possible ways are there to arrange a five flags.

Example 2: In how many ways 5 chocolates can be chosen from among 9 different kinds of chocolates?

Solution:

This problem involves 10 candies, taken 5 at a time.

P(10,5) = `(10!) / ((10 - 5)!)` = `(10 * 9 * 8 * 7 * 6 * 5!) / (5!)` = 30240

There are 30240 possible ways to choose 5 chocolates among 10 chocolates.

Example 3: Peter bought four movies. In how many different ways can he watch the four movies?

Solution:

P(4,4) = `(4!) / ((4 - 4)!)` = `(4 * 3 * 2 * 1) / (0!)` = 24                [0! = 1]

In 24 different ways he can watch the four movies.


Example 4: The computer password has 4 digits, if the possible digits are 2, 4, 5, 6, 7, 8, 9. How many different passwords can be made?

Solution:

This problem involves 7 digits, 4 digits at a time.

P(7,4) = `(7!) / ((7 - 4)!)` = `(7 * 6 * 5 * 4 * 3!) / (3!)` = 840

Therefore, 840 different passwords can be made.


Math Term Permutation – Practice Problems


Problem 1:  How many different ways can a set of seven country flags are arranged?

Problem 2: In how many ways can 6 candies chosen from among 9 different colors of candies?

Problem 3: Anita bought six movies. In how many different ways can she watch the six movies?

Answer: 1) 5040 2) 60480 3) 720

Learning High School Algebra

Algebra is a branch of mathematics that deals with  the study of  rules of operations and relations. Diophantus is regarded as the father of algebra. Basic algebraic concepts include variables and constants, expressions, terms, polynomials, equations and algebraic structures. So much is known about the subject that study of this takes a life time. Algebra offers lot of food for thought for students and mathematics lovers will have lot of fun solving problems in algebra. The beauty of the subject is that it prompts the student to think indicatively to solve problems and helps to develop analytical and logical skills for the student. Study of mathematics at any level will cover some topics in algebra.

High school algebra covers topics such as polynomials, algebraic expressions and identities, equtaions and factorization.

Let us learn some solved problems in the above topics of high school algebra.


Learning Polynomials and Factorization in high school algebra:


At high school level, polynomials and factorization are important topics covered in algebra.

POLYNOMIALS:

An Algebraic expression of the form axn is called Monomial in x. For example, 7x3   .The sum of two monomials are called a Binomial and the sum of three monomials are called Trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is Trinomial. The sum of a finite number of monomials in x is called a polynomial in x.

Example:

Find the sum of 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Solution:

Using the associative and distributive properties of real numbers, we obtain

(2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1) = 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.  (answer)

FACTORIZATION :

The process of writing polynomial as a product of two or more simpler polynomials is called Factorization.

The way of writing a polynomial as a product of two or more simpler polynomials is called factorization. The process of factorization is also known as the resolution into factors.

Example 1: Factorize x2 – 2xy – x + 2y.

Solution:

x2 – 2xy – x + 2y = (x2 – 2xy) – (x – 2y)

= x(x – 2y) + (–1) (x – 2y)

= (x – 2y) [x + (–1)]

= (x – 2y) (x – 1).   (answer)


Example 2: Solve 9x2 = x.

Solution:

9x2 = x

9x2 − x = 0

x (9x − 1) = 0        ab = 0

x = 0      or    x =

What if we attempt the same problem using the another method?

9x2 = x

9x = 1 divide by x

x =

Every quadratic equations has 2 roots. Dividing the quadratic equation by x removes the root x = 0.

However, dividing by a constant does not impact the roots.

Learning Algebraic Identites in high school algebra:

Algebraic Identities:

Algebraic identities is important in High school algebra . Algebraic identities are algebraic equations satisfied by any value of the variables.
The algebraic identity of x + 0 = x tells us that anything (x) added to zero equals the original "anything," no matter what rate that "anything" (x) may be. Like normal algebra, Boolean algebra has its own unique identities base on the bivalent states of Boolean variables.

Example 4: Solve for variables x,y and z:

x + y + 2z = 2  ------> (1)

3x - y + 3z = 4  ------>(2)

2x + y + 4z = 6  ------>(3)

Solution:

Solve (1) and (2),

x + y + 2z = 2 ----> (1)

3x - y + 3z = 4 -----> (2)

add the above two equations.

we get 4x + 5z = 6    ------> (4)

solve (1) and (3)

(3) * 2 ---->        4x + 4y +8z = 12

(4)     ----->        4x + 0y +5z = 6
(-)           (-)      (-)

3z = 6

z = 2

substitute z = 2 in (4) eqn

4x + 5(2) = 6

4x = -4

x = -1

substitute x = -1, z = 2 in (1) eqn.

(-1) + y + 2(2) = 2

y -3 = 2

y = 5.

x  = -1, y = 5, z = 2.