MAT1514 Assignment 1 (COMPLETE ANSWERS) Semester 1 2025 DUE APRIL 2025; 100% trusted, comprehensive and complete reliable solution with clear explanation.

,MAT1514 Assignment 1 (COMPLETE ANSWERS)
Semester 1 2025 DUE APRIL 2025; 100% trusted,
comprehensive and complete reliable solution with clear
explanation.
ALL QUESTIONS ANSWERED CORRECTLY


NOTE:This assignment covers Chapters 1, 2, and 3 of the
prescribed book and contributes to your year mark.
Question 1 Given f (x) = 3x2 − 4x + 7 and g(x) = x2 + 1.
Find and simplify the following: 1. (g f )(x) (4) 2. (g − f )(x) (2)
3. g f (x) (4) 4. g−1(x) (3) [13 marks]

Given Functions:
 𝑓(𝑥) = 3𝑥2 − 4𝑥 + 7𝑓(𝑥) = 3𝑥^2 − 4𝑥 + 7𝑓(𝑥) =
3𝑥2 − 4𝑥 + 7
 𝑔(𝑥) = 𝑥2 + 1𝑔(𝑥) = 𝑥^2 + 1𝑔(𝑥) = 𝑥2 + 1
𝟏. (𝒈 ∘ 𝒇)(𝒙)(𝒈 \𝒄𝒊𝒓𝒄 𝒇)(𝒙)(𝒈 ∘ 𝒇)(𝒙) (𝟒 𝒎𝒂𝒓𝒌𝒔)
𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 (𝑔 ∘ 𝑓)(𝑥)(𝑔 \𝑐𝑖𝑟𝑐 𝑓)(𝑥)(𝑔
∘ 𝑓)(𝑥) 𝑚𝑒𝑎𝑛𝑠 𝑤𝑒 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑓(𝑥)𝑓(𝑥)𝑓(𝑥) 𝑖𝑛𝑡𝑜 𝑔(𝑥)𝑔(𝑥)𝑔(𝑥). 𝑆𝑜, 𝑤
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(3𝑥2 − 4𝑥 + 7)(𝑔 \𝑐𝑖𝑟𝑐 𝑓)(𝑥)
= 𝑔(𝑓(𝑥)) = 𝑔(3𝑥^2 − 4𝑥 + 7)(𝑔 ∘ 𝑓)(𝑥)
= 𝑔(𝑓(𝑥)) = 𝑔(3𝑥2 − 4𝑥 + 7)
𝑁𝑜𝑤, 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑓(𝑥)𝑓(𝑥)𝑓(𝑥) 𝑖𝑛𝑡𝑜 𝑔(𝑥) = 𝑥2 + 1𝑔(𝑥)
= 𝑥^2 + 1𝑔(𝑥) = 𝑥2 + 1:

, 𝑔(3𝑥2 − 4𝑥 + 7)
= (3𝑥2 − 4𝑥 + 7)2 + 1𝑔(3𝑥^2 − 4𝑥 + 7)
= (3𝑥^2 − 4𝑥 + 7)^2 + 1𝑔(3𝑥2 − 4𝑥 + 7)
= (3𝑥2 − 4𝑥 + 7)2 + 1
𝐹𝑖𝑟𝑠𝑡, 𝑒𝑥𝑝𝑎𝑛𝑑 (3𝑥2 − 4𝑥 + 7)2(3𝑥^2 − 4𝑥 + 7)^2(3𝑥2
− 4𝑥 + 7)2:
(3𝑥2 − 4𝑥 + 7)2
= (3𝑥2)2 − 2(3𝑥2)(4𝑥) + (4𝑥)2 + 2(3𝑥2)(7)
− 2(4𝑥)(7) + 72(3𝑥^2 − 4𝑥 + 7)^2
= (3𝑥^2)^2 − 2(3𝑥^2)(4𝑥) + (4𝑥)^2
+ 2(3𝑥^2)(7) − 2(4𝑥)(7) + 7^2(3𝑥2 − 4𝑥
+ 7)2
= (3𝑥2)2 − 2(3𝑥2)(4𝑥) + (4𝑥)2 + 2(3𝑥2)(7)
− 2(4𝑥)(7) + 72
= 9𝑥4 − 24𝑥3 + 16𝑥2 + 42𝑥2 − 56𝑥 + 49
= 9𝑥^4 − 24𝑥^3 + 16𝑥^2 + 42𝑥^2 − 56𝑥
+ 49 = 9𝑥4 − 24𝑥3 + 16𝑥2 + 42𝑥2 − 56𝑥 + 49
= 9𝑥4 − 24𝑥3 + 58𝑥2 − 56𝑥 + 49
= 9𝑥^4 − 24𝑥^3 + 58𝑥^2 − 56𝑥 + 49
= 9𝑥4 − 24𝑥3 + 58𝑥2 − 56𝑥 + 49
𝑁𝑜𝑤, 𝑎𝑑𝑑 1:
𝑔(3𝑥2 − 4𝑥 + 7)
= 9𝑥4 − 24𝑥3 + 58𝑥2 − 56𝑥 + 50𝑔(3𝑥^2 − 4𝑥
+ 7)
= 9𝑥^4 − 24𝑥^3 + 58𝑥^2 − 56𝑥 + 50𝑔(3𝑥2
− 4𝑥 + 7) = 9𝑥4 − 24𝑥3 + 58𝑥2 − 56𝑥 + 50