Exercise 5.6.28. For each of the following descriptions of a function f, find a suh S of the
domain of f such that the restriction fIs is a bijection onto the range of f. (a)f : N N, wheref(n)-n
+ 1 if n is odd and/(n) =n/2 if n is even (b)f : R Z, wheref(x)- . Here lxl denotes the floor\' or.
(c)f : R R, where,f(x) =x3-3x2 , (d)f : R R, where,f(x) =x4-4x2 (e)f : R R, wheref(x)=xe-x. ,
find a subset 7 The floor function from R to Z assigns to each x R the greatest integer less than or
equal to x. It is denoted by Lx. For example, L = 3 and L-r=-4 ie, | | = 3
Solution
A bijective function is a one to one function that is form every x value there will be exactly one y
value.
In simopler terms each element in the domain set of a function when plugged in the function
gives a distinct value of the function
that is a part of the range of the function.
in our problem we need to find the subset of a domain for which the function is bijective
a> f : N-->N , f(n) = n+1
the domain and the range are all natural numbers
that is n E [0,1,2,3,4,.....N]
if n is odd
f(n)= n+1 ,this is a straight line and this function will be bijective for all odd natural numbers that
is n E (2k+1) will be the
domain in which f(n) will be bijective ,as for all odd values of n we\'ll get distinct f values.
n E [1 , 2k+1] , here k = 1,2,3........k just the integral vaslues
when n is even
f(n) = n/2 , this again is a straight line so the domain in whicj f(n) is bijective will be all even
natural, numbers\\that is
n E 2k ,k E {1,2,3,4...N}
b> f(x) = [x]
x : Z-->Z
here f(x) represents a step function
and a step function have the same range for all the x values within a particular step domain
so f(x) = [x] , cannot be a one to one function as we do not get distinct f(x) values for different x
values
c> f(x) = x^3-3x^2
f(x)= x^2(x-3)
this function will be a curve with zeros at x = 0 and 3
and it will not be a bijective or one to one function over the entire real number line.
d>
f(x) = x^4-4x^2
this function will be a curve with zeros at x = 0 and 2 and -2
and it will not be a bijective or one to one function over the entire real number line..
Exercise 5.6.28. For each of the following descriptions of a function.pdf
1. Exercise 5.6.28. For each of the following descriptions of a function f, find a suh S of the
domain of f such that the restriction fIs is a bijection onto the range of f. (a)f : N N, wheref(n)-n
+ 1 if n is odd and/(n) =n/2 if n is even (b)f : R Z, wheref(x)- . Here lxl denotes the floor' or.
(c)f : R R, where,f(x) =x3-3x2 , (d)f : R R, where,f(x) =x4-4x2 (e)f : R R, wheref(x)=xe-x. ,
find a subset 7 The floor function from R to Z assigns to each x R the greatest integer less than or
equal to x. It is denoted by Lx. For example, L = 3 and L-r=-4 ie, | | = 3
Solution
A bijective function is a one to one function that is form every x value there will be exactly one y
value.
In simopler terms each element in the domain set of a function when plugged in the function
gives a distinct value of the function
that is a part of the range of the function.
in our problem we need to find the subset of a domain for which the function is bijective
a> f : N-->N , f(n) = n+1
the domain and the range are all natural numbers
that is n E [0,1,2,3,4,.....N]
if n is odd
f(n)= n+1 ,this is a straight line and this function will be bijective for all odd natural numbers that
is n E (2k+1) will be the
domain in which f(n) will be bijective ,as for all odd values of n we'll get distinct f values.
n E [1 , 2k+1] , here k = 1,2,3........k just the integral vaslues
when n is even
f(n) = n/2 , this again is a straight line so the domain in whicj f(n) is bijective will be all even
natural, numbersthat is
n E 2k ,k E {1,2,3,4...N}
b> f(x) = [x]
x : Z-->Z
here f(x) represents a step function
and a step function have the same range for all the x values within a particular step domain
so f(x) = [x] , cannot be a one to one function as we do not get distinct f(x) values for different x
values
c> f(x) = x^3-3x^2
f(x)= x^2(x-3)
2. this function will be a curve with zeros at x = 0 and 3
and it will not be a bijective or one to one function over the entire real number line.
d>
f(x) = x^4-4x^2
this function will be a curve with zeros at x = 0 and 2 and -2
and it will not be a bijective or one to one function over the entire real number line.
![Exercise 5.6.28. For each of the following descriptions of a function f, find a suh S of the
domain of f such that the restriction fIs is a bijection onto the range of f. (a)f : N N, wheref(n)-n
+ 1 if n is odd and/(n) =n/2 if n is even (b)f : R Z, wheref(x)- . Here lxl denotes the floor' or.
(c)f : R R, where,f(x) =x3-3x2 , (d)f : R R, where,f(x) =x4-4x2 (e)f : R R, wheref(x)=xe-x. ,
find a subset 7 The floor function from R to Z assigns to each x R the greatest integer less than or
equal to x. It is denoted by Lx. For example, L = 3 and L-r=-4 ie, | | = 3
Solution
A bijective function is a one to one function that is form every x value there will be exactly one y
value.
In simopler terms each element in the domain set of a function when plugged in the function
gives a distinct value of the function
that is a part of the range of the function.
in our problem we need to find the subset of a domain for which the function is bijective
a> f : N-->N , f(n) = n+1
the domain and the range are all natural numbers
that is n E [0,1,2,3,4,.....N]
if n is odd
f(n)= n+1 ,this is a straight line and this function will be bijective for all odd natural numbers that
is n E (2k+1) will be the
domain in which f(n) will be bijective ,as for all odd values of n we'll get distinct f values.
n E [1 , 2k+1] , here k = 1,2,3........k just the integral vaslues
when n is even
f(n) = n/2 , this again is a straight line so the domain in whicj f(n) is bijective will be all even
natural, numbersthat is
n E 2k ,k E {1,2,3,4...N}
b> f(x) = [x]
x : Z-->Z
here f(x) represents a step function
and a step function have the same range for all the x values within a particular step domain
so f(x) = [x] , cannot be a one to one function as we do not get distinct f(x) values for different x
values
c> f(x) = x^3-3x^2
f(x)= x^2(x-3)](https://image.slidesharecdn.com/exercise5-230709050800-092e4701/85/Exercise-5-6-28-For-each-of-the-following-descriptions-of-a-function-pdf-1-320.jpg)
