Some of these questions are from the following textbook, Discrete Mathematics withApplications, by Susanna S. Epp. 1 Answer the following questions:Define F: NN by the rule F(n) = 2-3n, for all natural numbers n E N. i. Is Fafunction? Why or why not?&ihhe *** 4r152191i. If we modify the codomain of F to be Z (the set of all integers), is F a function? If f is afunction, is it an injective (i.e., one-to-one) function? Is it a surjective function? Brieflyexplain why.PEWAYI44**# 4pDefine G: RR by the rule G(x) = 23x for all real numbers x. Is G surjective?Explain your answer./0. a #,CS Scanned withCamScanner 5) It's possible to compose two functions to get a composite function. For example,let f R R, and f(x) 2x 1 (i.e., function f maps a real number x to a real numbergiven by 2x +1;), and let g: RR, and g(x) = x ^2 (i.e., function g maps any realnumber to its square). Then function (g f) is defined as follows: (gg(f(x)), i.e., function (g f) maps x to g(f(x)) (i.e., x is first mapped to a real numberf(x), and then f(x) is in turn mapped to number g(f(x)) using rule of function g.So (g f)(x) g(2x 1) (2x 1)^2.i. Is (g f)(x) a function? Why?f)(x)ii. Calculate (gf)(1), (g. f)(0.2).Iii. What is function (f g)(x)= f(g(x))? Calculate (fg)(1), (fe)(0.1)

Question
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Step 1

4(i)Consider the given mapping F: N→N defined by

F(n)=2-3n , for all natural numbers n belongs to N

for, F to be a function, for each n there will be exactly one value of F(n),

for example: n=1 gives F(n)=-1, n=2 gives F(n)=-4 but -1 and -4 are not natural numbers,

so, for each n there does not exist value of F(n) on the defined codomain.

Hence, F(n) is not a function.

Step 2

(ii)Now, when the codomain if F is modified to be Z (the set of all integers) , for each n there will be exactly one value of F(n) on defined codomain.

Hence, F(n) is a function.

So, here F:N→Z defined by F(n)=2-3n

Now, check injectivity:

Let n1 and n2 belongs to N such that

F( n1)= F( n2)

2-3n1=2-3n2

n1= n2

Thus, F is one-to-one function.

Now, check surjectivity:

It is known that a function f is surjective if for every y there exist an x such that f(x)=y,

Now, here, let y=F(n)=0 from Z, then

2-3n=0

n=2/3, which is not a natural number.

So, for every y there does not exist an x such that f(x)=y,

Hence, F is not surjective.

Step 3

(iii)

Now, Consider the given mapping G: R→R defined by

G(x)=2-3x ,

Now, let y=G(x) from R,then

y=2-3x

3x=2-y

x=(2-y)/3

Now, G(x)=G((2-y)/3)

=2-3[(2-y)/3]

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