Exam (elaborations) MAT1514 Assignment 1 Semester 1 | Due April 2025

Exam (elaborations)

MAT1514 Assignment 1 Semester 1 | Due April 2025




· Course

· Precalculus (MAT1514)

· Institution

· University Of South Africa (Unisa)

· Book

· Calculus




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 )

3 𝑥 2 − 4 𝑥 + 7 f(x)=3x 2 −4x+7 𝑔 ( 𝑥 )
𝑥 2 + 1 g(x)=x 2 +1

1. (g ∘ f)(x) (4 marks) This means we substitute 𝑓 ( 𝑥 ) f(x) into 𝑔 ( 𝑥 ) g(x):


𝑔(𝑓(𝑥))

𝑔 ( 3 𝑥 2 − 4 𝑥 + 7 ) g(f(x))=g(3x 2 −4x+7) Since 𝑔 ( 𝑥 )
𝑥 2 + 1 g(x)=x 2 +1, we replace 𝑥 x with 𝑓 ( 𝑥 ) f(x):


𝑔(𝑓(𝑥))

( 3 𝑥 2 − 4 𝑥 + 7 ) 2 + 1 g(f(x))=(3x 2 −4x+7) 2 +1
Expanding ( 3 𝑥 2 − 4 𝑥 + 7 ) 2 (3x 2 −4x+7) 2 using
the identity ( 𝑎 + 𝑏 + 𝑐 ) 2
𝑎 2 + 𝑏 2 + 𝑐 2 + 2 𝑎 𝑏 + 2 𝑏 𝑐 + 2 𝑎 𝑐 (a+b+c) 2 =a 2 +b 2 +c 2 +2ab+2bc+2ac: