Learn Slope Form

The ratio of rise value of the line equation to the run value of the line equation is called as slope. Slope form learning is one of the important part of algebra. In other words, it can be defined as the ratio of the change of  Y values to the change of the X value. Generally, the slope of the equation is denoted as "m". It is necessary to learn the slope form. learn slope form is mainly used in linear equation.

learn slope form of the linear equation

The slope form of the linear equation can be written as,

y = mx + b

Where, y is the line equation,

m is the slope of the equation and

b is the Y intercept value.

In point form, the line equation can be written as,

Y - Y1 = m (X - X1)

Where, X1 and y1 are the points of the line

m is the slope of the equation.

I like to share this Math Slope Formula with you all through my article.

Given two points, we could find the slope using the formula

m = (Y2 - Y1) / (X2 - X1)

Where, (X1, Y1 ) and  (X2, Y2 ) are any two points in the line.

Example problems for learn slope form

Ex:1 Find the slope of the given equation Y = 7X + 6.

Sol: In general form, the line equation is given as,

y = mx + b

where, m is the slope of the equation.

In the above equation, the slope is coefficient of x,

slope = 7

Ex:2 Find the slope of the given points (1, 3) and (3, 6).

Sol:

The given points are (1, 3) and (3, 6).

Here, X1 = 1, Y1 =3

X2 = 3, Y2 = 6

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (6 – 3) / (3 -1)

= 3 / 2

Ans:

The slope of the given equation is m= 3/2

Ex:3 Find the slope of the given points (2, 6) and (6, 14).

Sol: The given points are (2, 6) and (6, 16)

Here, X1 = 2, Y1 = 6

X2 = 6, Y2 = 16

The formula for finding the slope is given as,

m = (Y2 - Y1) / (X2 - X1)

Substitute all  values in the above equation,

m = (16 – 6) / (6 -2)

= 5 / 2

Ans:

The slope of the given equation is m = 5 / 2

Expressions With Rational Exponents

A rational number is a number represented as p/q where p and q are integers and q != 0.Exponent means  power.

The large numbers are very difficult to read and write and understand. To make them simpler we can use exponents, with the help of this many of the large numbers are converted to simpler forms. The short notation 74 stands for the product 7 x 7 x 7x 7. Here ‘7’ is called the base and ‘4’ the exponents. The number 74 is read as 7 raised to the power of 4 or simply as fourth power of 4. 74 are called the exponential form of 2401.

The above example is of a natural number as exponents.

Ex 5 ^(3/2) here 3/2 is a rational number so this is an example of rational exponents.

Rules invovled for solving rational exponents:

The Rational exponents rules are:

p,q are any real numbers except zero and m,n are positive integers.

Rule I:The base value is same under multiplication so we can directly add the power values. Then the exponent take the following form

pm  × pn = pm+n

Solving the expression 73  × 75 = 73+5 =78

Rule II:The base value is same under division so we can directly subtract the power values. Then the exponent take the following form such that m > n then

(p^m)/(p^n) = pm-n

Solving the expression  (4^4)/(4^2)= 44-2=42

Rule III:If m < n, (p^m)/(p^n) =(1)/(p^(n-m))

Solving the expression 8^3/8^2 =1/8

Rule IV:A real number to the power of power

(pm )n  = pm*n

Solving (82 )3  = 82*3=86

Rule V:The power of 0 is

p0 = 1. Anything power zero is equals to 1.

p^m/p^m   = pm-m =p0 = 1

Rule VI:p to the power of negative number is

p-m =1/p^m
Solving the expression 2-5 =  1/2^5

Rule VII:Two numbers to the same power can be written as

pm  × qm = (p*q)m

Solving the expression 32 × 82 = (3*8)2=242

Rule VIII A number to rational power is written as

p^(m/n) = root(n)(p^m)

Examples of expressions having rational exponents:

Ex : 1Solve the exponent    2x^(1/3)xx8^(5/3)xxx^(2/3)

Sol:Step 1:Given

2x^(1/3)xx8^(5/3) xx x^(2/3)

Step 2:Exponents are added

=2x^(1/3+2/3) (2^3)^(5/3)

Step 3: 8 is written as 2 power 3

Step 4 : Simplify=2x2^5

=2xx32 x

=64 x

Ex 2: Solve x^(-1/3)8^(-2/3)

x^(-1/3)8^(-2/3)

Solu: Step 1  x^(-1/3) 8^(-2/3)

Step2: 8^(-2/3) = (2^3)^(-2/3) = 2^(3 xx-2/3) = 2^-2 = 1/2^2 = 1/4

My forthcoming post is on  Define Arithmetic Mean will give you more understanding about Algebra.

Step3 Simplify the variable with negative exponent

=1/(x^(1/3)) 1/4

Step 4 Simplify:=1/(4x^(1/3))

=1/(4root(3)(x))

Power Coefficient Equation

A coefficient is a multiplicative factor in a few phrase of an expression. It is typically a number, but in any case does not occupy some variables of the equation. For example in equation 7x2 − 3xy + 1.5 + y the first three stipulations correspondingly have coefficients 7, −3, and 1.5. In the third expression there are no variables, so the coefficient is the idiom itself called the power constant term or constant power coefficient of this expression

I like to share this Sample Correlation Coefficient with you all through my article.

Example Problem for the Coefficient:

Find the coefficient of the given equation:

X2 + 5y3 +2xy2 + 5yx2

Solution:

Given that X2 + 5Y3+ 2XY2 + 5YX2

Here X2+ 5Y3+ 2XY2+ 5YX2 the expression with the coefficient

The coefficient of x2 is 1

The coefficient of  5Y3 is 5

The coefficient of 2xy2 is 2

The coefficient of 5yx2 is 5.

Coefficient Of Varience

Coefficient of variation is a relative determine for standard deviation.

It is defined as 100 times the coefficient of spreading base in the lead standard difference is called coefficient power of variation.

Less C.V. designate the fewer variability or additional power.

More C.V. indicate the more variability or less power.

According to Karl Pearson who suggested this compute the equation, C.V. is the percentage dissimilarity in the mean, standard deviation being considered as the total variation in the mean.

With the help of C.V. we can find which salesman is more consistent in making sales, which batsman is more consistent in scoring runs, which student is more consistent in scoring marks, which worker is more consistent in production.

Algebra is widely used in day to day activities watch out for my forthcoming posts on word problem solver for algebra and cbse class 10 science sample papers. I am sure they will be helpful.


Coefficients Of Dispersion

Whenever we want to compare the variability of two equations which differ widely in their averages or which are measured in different units.

We do not merely compute the measures of dispersions in the equations but we calculate the coefficients of dispersion which are pure numbers autonomous of the units of measurement.

Quotient Rule for Integration

The quotient rule is a method of finding the integration of a function that is the quotient of two other functions for which derivatives exist.

By the Product Rule,

if f (x) and g(x) are differentiable functions, then
d/dx[f (x)g(x)]= f (x)g'(x) + g(x) f' (x).
Integrating on both sides of this equation,

∫[f (x)g'(x) + g(x) f '(x)]dx = f (x)g(x),
which may be rearranged to obtain

∫f (x)g'(x) dx = f (x)g(x) −∫g(x) f' (x) dx.     (A)
Letting U = f (x) and V = g(x)

then differentiating it  we get

dU = f '(x) dx and dV =g'(x) dx,

pluging these values in (A), we get

∫U dV = UV −∫V dU.                  (1).


By the Quotient Rule,

if f (x) and g(x) are differentiable functions, then

d/dx[f (x)/g(x)]= g(x) f '(x) − f (x)g'(x)/[g(x)]2 .
Integrating both sides of this equation, we get
[f (x)/g(x)]=∫g(x) f '(x) − f (x)g'(x)/[g(x)]2 dx.
That is,

f (x)/g(x)=∫f '(x)/g(x)dx -∫f (x)g'(x)/[g(x)]2 dx,
which may be rearranged to obtain

∫f '(x)g(x)dx = f (x)g(x)+∫f (x)g'(x)/[g(x)]2 dx.      (B)

Letting u = g(x) and v = f (x) and then differentiating it , we get

du = g'(x) dx and

dv = f '(x) dx,
we obtain a Quotient Rule Integration by Parts formula:

plugging values of u , v , du and dv in B we get
∫dv/u= v/u+∫(v/u²) du.                                  (2)

quotient rule for integration-Application

∫[sin(x−½)/x²] dx.
Let
u = x½, du=1/2(x-½)

v=2cos(x-½),dv = sin(x−1/2)/x³/²

Then

∫sin(x−½)/x² dx = 2 cos(x-½)/x½+∫2 cos(x-½)/x• 1/2(x-½) dx

= 2 cos(x-½)/x½− ∫2cos(x-½) ·•[−(x-³/²)/2]dx

= 2 cos(x-½)/x½− 2 sin(x-½) + C,
which may be easily verified as correct.

My forthcoming post is on Add Hexadecimal Numbers and cbse solved sample papers for class 9th will give you more understanding about Algebra.

quotient rule for integration -Illustration


The Quotient Rule Integration by Parts formula (2) results from applying the
standard Integration by Parts formula (1) to the integral

Let ∫dv/u

with

U = 1/u,V = v,then differentiating ,we get

dU = − 1/u² du,dV=dv
plugging these values

∫dv/u=∫U dV

= UV −∫V dU

= 1/u·( v) −∫v (− 1/u²)du
= v/u+∫v/u² du

Terminating Decimals are Rational Numbers

Rational numbers are contrast among irrational numbers like Pi and square roots and sins and logarithms of information. On rational numbers and the ending of the article you be able to click on a linkage to maintain studying about irrational numbers. A number is rational but you can note down it in a form a/b where a and b are integers, b not zero. Visibly all fractions are of that form.

Terminating decimals are rational numbers:

A rational number is several number that be able to be expressed as the section a/b of two integers, among the denominator b not equivalent to zero. While b could be equal to 1, every one integer is a rational number. The set of every rational in sequence is frequently denoted by a boldface Q stands for quotient.

The terminating decimal extension of a rational numbers forever whichever terminating follows finitely several digits or begins to replicate the similar sequence of digits over and over. Moreover, any repeating or terminating decimal represent a rational number. These statements hold true not now meant for base 10, but as well for binary, hexadecimal, or any further integer base. Terminate decimal numbers be able to simply be written in that form: for example 0.67 is 67/100, 3.40938 = 340938/100000.

A real number so as to be not rational is called irrational. Irrational numbers contain square root 2, pi, and e. The decimal extension of an irrational number continues evermore without repeating. Because locate of rational numbers is countable, and the position of real numbers is uncountable, just about each real number is irrational.

In terminating conceptual algebra, the rational numbers shape a field. This is the representative field of characteristic zero, with is the field of fractions for the ring of integers. Finite extensions of Q are called algebraic number fields, and the algebra finality of Q is the field of algebraic number.

In mathematical study, the rational numbers shape a dense separation of the real numbers. The real numbers is able to be constructed from the rational numbers by achievement, using either Cauchv sequences or Dedekind cuts.

Zero separated by any other integer equals zero, consequently zero is a rational number even though division by zero itself is indeterminate.

My forthcoming post is on Probability Permutations and sample paper for 9th class cbse will give you more understanding about Algebra.

Solving Positive Direction Coordinates

A coordinate is a number that determine the position of a point the length of a number of line or arc. A list of two, three, otherwise additional coordinates can be use to establish the position of a point on a surface, volume, otherwise higher-dimensional area.

The Positive direction represents the positive value for both the x axis and y axis. The x axis value and the y axis value should be positive to move towards the positive direction. Any of the negative change in the axis will not lead to positive direction.

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Solving positive direction examples:

Solve the following equation:

Y = x2 +2

Solution :

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when  x = 1

Y = (1)2 +2

Y = 1+2

Y = 3

The co ordinates will (1,3) which lies on the positive direction.

when we substitute x =2

We get

Y = (2)2 + 2

Y = 4 + 2

Y = 6

The co ordinates are (2,6) which lies on the positive direction.

Similarly when we substitute any positive value for the given equation the output will be positive.

Even though we give negative values, the output will be (that is y) positive. But we need to give positive values to move in the positive direction.

Solving positive direction example:

Solve the following equation:

Y = 3x +4

Solution:

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when x = 4

Y = 3(4) +4

Y = 12+4

Y = 16

The co ordinates will (4,16) which lies on the positive direction.

when we substitute x =6

We get

Y = 3(6) + 4

Y = 18 + 4

Y = 22

The co ordinates are (6,22) which lies on the positive direction.

Algebra is widely used in day to day activities watch out for my forthcoming posts on How to Find Second Derivative and Irrational Numbers Definition. I am sure they will be helpful.

Similarly when we substitute any positive value for the given equation the output will be positive.

Here when we substitute negative values for the variable x, we get negative values as output also. So we need to substitute positive values for the equation.

Poisson Distribution

If n is large, the evaluation of the binomial distribution can involve considerable computation. In such a case a simple approximation to the binomial probability could be considerable use. The approximation of binomial when n is large and p is close to zero is called the Poisson Distribution Mean.

Definition:

A random variable is said to follows Poisson distribution if it assumes only non-negative values and its probability mass function is given by

P(x,lambda ) =P(X=x) = elambda lambda x  /  X!    

x=0, 1, 2…

O            

otherwise

lambda is known as parameter of Poisson distribution .X~p() denotes it.

Characteristic function of Poisson distribution

phi x ( t )  =E[ei t x ]

=sum_(x=0)^oo ei t x   e-lambda lambda x / x!

=  e-lambda sum_(x=0)^oo ei t x lambda x  / x!

=  e-lambda [1+(lambda eit ) + (lambda eit)2 / 2! +(lambda ei t)3  /3!  + ............. ]

=e-lambda elambda eit

=  elambda (ei t -1)

Additive property of Poisson distribution:

Independent Poisson variate is also a Poisson variate xi (i=1,2,......n). xi follows Poisson with parameter lambda i.

xi ~ P(lambda i )    (i=1,2.....n)

then sum_(i=1)^n    xi ~ P(sum_(i=1)^n lambda i )

Proof:

Mxi(t) = elambda i(e^t -1)

Mx1+x2+x3+.....Xn(t) = Mx1 (t) Mx2 (t). . . Mxn (t)

= elambda 1(e^t-1). elambda 2(e^t-1) ..........elambda n(e^t-1)

=e(lambda 1 +lambda 2 +lambda 3+.......lambda n )(et-1)

=esum_(i=1)^n lambda i(et-1)

M.G.F of Poisson distribution:

Mx ( t ) = E [ etx ]

= sum_(x=0)^oo etx   e-lambda lambda x  / x!

= e-lambda sum_(x=0)^oo etxlambda x / x!

= e-lambda [ 1+ et lambda +  (lambda et )2 /2! + (  (lambda et )3  / 3! +...... ]

=e-lambda elambda

= elambda (et -1)

Applications of Poisson distribution

The following are some instance where the distribution is applicable

Number of deaths from a disease
Number of suicide reported in a particular city.
Number of defective materials in packing manufactured by good concern.
Number of printing mistakes at each page of the book.
Number of air accidents in some unit of the time.

Needs Assessment Definition

Assessment is an important and essential part of teaching. If teachers are to ask themselves whether what they are teaching and how they are teaching has the desired outcomes, they will needs to assess what children are able to do according to a set of criteria.

The criteria, which indicate what children, should be able to do and think in mathematics by the end of the foundation phase.

Teachers need to ask themselves with assessment:

The following are two very important questions that teachers have to ask when teaching.

Is what I am doing helping children to develop a desire to learn mathematics?

Is what I am doing teaching children to become numerate?

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Why we needs assessment?

Some reasons for assessing learners are so that we can measure whether a lesson has been effective for each learner (whether we have taught the lesson effectively enough). We assess learners to see what each one can do, for example, which learners are able to calculate change, or which learners are able to solve relevant word problem. We assess learners to see which of the children are ready for a new challenge and which must still practice what has already been taught. We assess learners so that we can plan further lessons that suit the needs of the children.

Types of assessment:

There are two main ways in which to assess children

Formative assessment

Summative assessment.

Formative assessment is assessing a learner while the learner is forming the new knowledge.

Example for formative assessment:

An example of formative assessment would be sitting with a learner while he or she is doing a task (say using a number line to count in groups), watching how the child goes about the task and asking the child to explain how and what he or she is doing. In this way, you find out what strategies the child is using and developing and what strategies you should be helping the child with; you are getting direct and instant feedback on hoe the child is coping and you are able to respond to the situation immediately through re-teaching and explaining again, asking anther learner to help, or planning another lesson on that needs for the next day.

Summative assessment is assessing a learner at the end of the lesson, section, topic, quarter or year as a summing up of what the learner knows. Therefore, tests and exams are summative versions of assessment.

When both formative and summative assessments are used, that is continuous assessment. In an outcomes-based education system, continuous assessment is used. The teacher studies the learning outcomes required of the learners and then plans lessons to teach to achieve these outcomes. During the lessons, the teacher observers what children are doing and saying and how children are doing a task. The teacher asks for explanations from the children as to what and how they are doing a take. The teacher helps those learners who are confused and continually monitors which learners are gaining control of the skills and concepts. Once a child can do something independently within the number range for that learner, a teacher can say that that child has learnt what was intended by the lesson and so can record that performance as a desired learning outcome for that child.

Learn Horizontal Line Test

Definition of function:

The mathematical concept of a function expresses the intuitive idea that one quantity (the argument of the function, also known as the input) completely determines another quantity (the value, or the output). A function assigns a unique value to each input of a specified type. The argument and the value may be real numbers, but they can also be elements from any given sets: the domain and the codomain of the function. An example of a function with the real numbers as both its domain and codomain is the function f(x) = 2x, which assigns to every real number the real number with twice its value. In this case, it is written that f(5) = 10.

A relation f:A`->B ` is said to be a function if every element in domain(A) has an unique image in codomain(B).

Definition of One-to-One Function:

A function is said to be One-to-One if no two different elements in domain have same image in codomain.The definition of one-to-one function can be written algebraically as follows:

Let a  and b any elements of domain.

A function f(x) is said to be one-to-one

1.if a is not equal to b then f(a) is not equal to f(b)

OR contra positive of the above

2.if f(a)=f(b) then a=b

Horizontal Line Test::

The horizontal line test is used to determine if a function is one-to-one.The lines used for this test are parallel to x axis.

If the function is one-to-one, then it can be visualized as one whose graph is never intersected by any horizontal line more than once.



If and only if  f is onto, any horizontal line will intersect the graph at least at one point (when the horizontal line is in the codomain).



If f is bijective, any line horizontal or vertical will intersect the graph at exactly one point.


Graph of one-to-one function:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at only one point so it is an one-to-one function.

Graph of a function which is not one-to-one:

If a line is drawn parallel to x axis (horizontal line) to this curve  then it will cut the curve at more than  one point so it is not an one-to-one function.

Algebra is widely used in day to day activities watch out for my forthcoming posts on answers to math problems for free and cbse 9th class science book. I am sure they will be helpful.

Meaning of Correlation Coefficient

Two variables are related in such a way that:

(i) if there is an increase in one accompanied by an accompanied by a decrease in the other, Then the variables are said to be correlation The value of  the correlation will be in the interval [1, -1].

If the value of the correlation is positive then it is direct and if it is negative, then it is inverse.

If  the value of correlation is 1, it is said to be perfect positive correlation. If it is -1, it is said to be perfect negative correlation. If  the correlation is zero, then there is no correlation.

The formula to find the correlation is [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]

where dx = x – barx ,  dy = y – bary

Now let us see few problems on correlation.

Example problems on meaning of correlation coefficient:

Ex 1: Find the correlation between two following set of data:

Cor_Tab1

Soln: barX = 180 / 9 = 20,

barY = 360 / 9 = 40.

Cor_SolTab1

Therefore the correlation =  [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]] = 193 / [sqrt [120 xx346]] = 0.94

This value shows that there is a very high relationship between x and y.

More example problem on meaning of correlation coefficient:

Ex 2: Find the correlation between the following set of data.

Cor_Tab2

Soln: barX = 36 / 6 = 6,  barY = 60 / 6 = 10

Cor_SolTab2

Therefore the correlation is   [sum dx dy] /[ sqrt [sum d_x^2 . sum d_y^ 2]]  = -67 / [sqrt [50 xx 106]]

= -0.92

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This value shows that there is a very low relationship between x and y.

By now the meaning of correlation coefficient will be more clearer. I believe that these examples would have helped you to do problems on correlation coefficient.

Average Velocity

Before going to the concept of average velocity,you should know about the concept of velocity.So, here is the introduction to velocity.

Velocity - the rate of change of position of an object. Velocity is a vector physical quantity because it requires both magnitude and direction to define. Speed is an absolute scalar value (magnitude) of velocity. Velocity is measured in meter/second (m/s).

In mathematics, velocity is the ratio of  displacement and  during the time taken to the interval. As it is associated to how fast waves travel between layers of the earth, it also tells us of how compact they are in between.

For example, 10 meter per second is a scalar value of velocity, but 10 m/s east is a vector. The rate of change in the velocity is called as acceleration.

Average velocity:

The Average velocity topic is dealt under mathematics. Students get to learn how to calculate the average velocity when the rest of the values are given. Students get to learn to practice various word problems on the concept of average velocity and also they get ample support from the online tutors regarding their homework problems and examination preparation.They can get all the answers for homework problems.

The “Rate of change(derivative) of position at which an object ” is known as velocity . It is a vector physical quantity; both magnitude and direction are essential to define it. The scalar complete value (magnitude) of velocity is speed, a quantity that is measured in meters per second (m/s or ms-1) when using the SI (metric) system.

The average velocity v of an object affecting through a displacement  (?x) during a time interval (?t) is described by the formula.Here is the Formula for average velocity:

V = Final displacement / total time taken

=  ?x / ?t

Students can get more detailed explanation on the topic on the Physics help page.

Finding average velocity

Example 1: If an object is thrown from the ground at an initial velocity of 10 meter per second, after t seconds the height of the object “h” in meter is given by,   h = t2 + 10t.  Find the time taken to the object to reach a height of 200 meter.

Solution:

Given, Height, h = 200 meter

h = t2 + 10t --> (1)

t =? (At 200 meter)
Plug h = 200 in equation (1)

200 = t2 + 10t

t2 + 10t – 200 = 0

By solving the quadratic equation,

t2 + 20t -10t -200 = 0

t(t+20) – 10(t+20) = 0

(t+20) (t -10) = 0

t = -20 or t = 10

We have two values for time “t”. Since time cannot be negative, any problem that gives a negative answer for one of its answers is always false, so we just go for the positive value.
So the final answer is 10 seconds.

So, the object reaches the height (200 meters) in 10 seconds.

Students can get more solved examples and problems to practice upon on the mastering physics page.

Calculate average velocity

Here is one more example of how to Calculate average velocity.

Example :  A particle moving along the x axis is located at 17.2 m at 1.79 s and at 4.48 m at 4.28 s. what is the average velocity in the particular time interval?

Solution:

Distance traveled by particle beside x-axis = Final location  - Initial location

= 17.20 - 4.48

= 12.72 meters

Time taken to cover up this distance = Final time - Initial time

= 4.28 - 1.79

= 2.49 seconds

Average velocity    = (Distance traveled)/(Time taken)

= 12.72 / 2.48

= 5.1084 (approximately)

Average velocity of the particle is    5.1084 m/s

Bijective Function Examples

In mathematics, a bijection, or a bijective function, is a function f from a set X to a set Y with the property that, for every y in Y, there is exactly one x in X such that f(x) = y and no unmapped element exists in either X or Y.

A one - one onto function is said to be bijective or a one-to-one correspondence.

A few examples of bijective function is given below which helps you for learning bijective function.

(Source: Wikipedia)

Examples of bijective function:

Example 1:

Show that the function f : R → R : f(x) = 3 - 4x  is one-one onto and hence bijective .

Solution:

We have

f(x1) = f(x2)

3 - 4x1 = 3 - 4x2

x1 = x2

Therefore, the function f is one-one.

Now, let y = 3 - 4x. Then, x = (3 - y)/4

Thus, for each y ε R (codomain of f), there exists x = (3 - y)/4 ε R

such that f(x) = f((3 - y)/4)

= {3 - 4 (3 - y)/4 }

= 3 - (3 - y)

= y

This shows that every element in codomain of f has its pre-image in dom(f).

Therefore, the function f is onto.

Hence, the given function is bijective.

Example 2:

Let A = R - {3} and B = R - {1}. Let f : A → B : f(x) = (x - 2)/(x - 3) for all values of x ε A.

Show that f is one-one and onto.

Solution:

f is one-one, since

f(x1) = f(x2)

(x1 - 2)/(x1 - 3) = (x2 - 2)/(x2 - 3)

(x1 - 2)(x2 - 3) = (x2 - 2)(x1 - 3)

x1x2 - 3x1 - 2x2 + 6 = x1x2 - 2x1 - 3x2 + 6

x1 = x2

Let y ε B such that y =  (x - 2)/(x - 3) .

Then, (x - 3)y = (x - 2)

x = (3y - 2)/(y - 1)

Clearly, x is defined when y ≠ 1.

Also, x = 3 will give us 1 = 0, which is false.

Therefore,

x ≠ 3.

And, f(x) = ((3y - 2)/(y - 1) - 2)/((3y - 2)/(y - 1)- 3) = y

Thus, for each y ε B, there exists x ε A such that f(x) = y.

Therefore, f is onto.

Hence, the given function is one-one onto.

These examples of bijective function help you to solve the following practice problems.

My forthcoming post is on icse board question papers and Nonlinear Partial Differential Equations will give you more understanding about Algebra.

Practice problems of bijective function:

Following examples of bijective function is given for your practice which helps you to learn more about bijective function.

1) Show that the function f : R → R : f(x) = x3  is one-one and onto.

2) Let R0 be the set of all non zero real numbers. Show that f : R0 → R0 : f(x) = 1/x is a one-one onto function.