Solution to Answer questions 4 to 7 by using the given functions g and f. Hint: Drawing … - Sikademy
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Archangel Macsika

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.

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

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

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

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

Question 6

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



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

1.  20x^2+8x-12\dfrac{3}{7}

Question 7

We consider the statements provided in the different alternatives:

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

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

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

2. We prove that g is indeed injective:



g is injective.

3. The function g is NOT surjective. 


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



\dfrac{13}{56}\not\in Z^+

We conclude that g is not surjective. 

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


2. g is injective.

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Question ID: mtid-5-stid-8-sqid-2845-qpid-1402