Question #211822

Answer questions 4 to 7 by using the given functions g and f. 

Hint: Drawing graphs of f and g before answering the questions, may assist you. Keep in mind 

that g  Z+  Q and f  Z+  Z+. Please note that graphs will not be asked for in the exam.


Question 6

Which one of the following alternatives represents the image of x under g ○ f (ie g ○ f(x)))?

1. 20x2 + 8x – 12

2. 80x2 + 4 x –

3. 20x2 + 8x + 3

4. 80x2 + 4 x – 3

Question 7

Which one of the following statements regarding the function g is TRUE? 

(Remember, g  Z+  Q.)

1. g can be presented as a straight line graph.

2. g is injective.

3. g is surjective.

4. g is bijective.


1
Expert's answer
2021-07-19T18:44:08-0400

Let gg be a function from Z+Z^+ (the set of positive integers) to QQ (the set of rational numbers) defined by


(x,y)g iff y=4x37 (gZ+×Q)(x, y)\in g\ \text{iff } y=4x-\dfrac{3}{7}\ (g\sube Z^+\times Q)

and let ff be a function on Z+Z^+ defined by


(x,y)g iff y=5x2+2x3 (fZ+×Z+)(x, y)\in g\ \text{iff } y=5x^2+2x-3\ (f\sube Z^+\times Z^+)

Question 6


gf(x)=g(5x2+2x3)g\circ f(x)=g(5x^2+2x-3)

=4(5x2+2x3)37=4(5x^2+2x-3)-\dfrac{3}{7}

=20x2+8x1237=20x^2+8x-12\dfrac{3}{7}

gf:Z+Qg\circ f:Z^+\to Q defined by gf=20x2+8x1237g\circ f=20x^2+8x-12\dfrac{3}{7}


1.  20x2+8x123720x^2+8x-12\dfrac{3}{7}


Question 7

We consider the statements provided in the different alternatives:

1. gg is not defined on the set of real numbers thus gg cannot be depicted as a straight line graph.

Only positive integers can be present in the domain of g.g.

It is the case that rdered pairs such as (1,347),(2,747),(3,1147),...(1, 3\dfrac{4}{7}), (2, 7\dfrac{4}{7}), (3,11\dfrac{4}{7}),... belong to gg and these pairs can be presented as dots in a graph. 


2. We prove that gg is indeed injective:


g(u)=g(v)=>4u37=4v37g(u)=g(v)=>4u-\dfrac{3}{7}=4v-\dfrac{3}{7}

=>4(uv)=0=>u=v=>4(u-v)=0=>u=v

gg is injective.


3. The function gg is NOT surjective. 

Counterexample


y=12,12Qy=\dfrac{1}{2}, \dfrac{1}{2}\in Q

Then


y=4x37=12=>4x=1314=>x=1356y=4x-\dfrac{3}{7}=\dfrac{1}{2}=>4x=\dfrac{13}{14}=>x=\dfrac{13}{56}

1356∉Z+\dfrac{13}{56}\not\in Z^+

We conclude that gg is not surjective. 


4. Since gg is not surjective, then gg is not bijective.  



Answer

2. gg is injective.



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