Exam (elaborations) MAT1514 Assignment 1 Semester 1 | Due April 2025 • Course • Precalculus (MAT1514) • Institution • University Of South Africa (Unisa) • Book • Calculus

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:
2
f ( x )=3 x −4 x+7
2
g ( x )=x +1

1. (g ∘ f)(x) = g(f(x))

This means substituting f ( x ) into g ( x ):
2
g ( f ( x ) )= ( f ( x ) ) +1

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



Expanding ( 3 x 2 − 4 x +7 ) :
2



( 3 x 2 − 4 x +7 ) ( 3 x 2 − 4 x +7 )=9 x 4 −24 x3 + 42 x 2 −24 x 3 +16 x 2 −28 x +42 x 2 − 28 x+ 49
4 3 2
¿ 9 x − 48 x +100 x − 56 x + 49

Adding 1:

g ( f ( x ) )=9 x − 48 x + 100 x −56 x +50
4 3 2




2. (g - f)(x) = g(x) - f(x)

( x 2 +1 ) − ( 3 x 2 − 4 x +7 )
2 2
x + 1− 3 x +4 x −7
2
−2 x + 4 x −6



3. g f(x) = g(x) × f(x)

( x 2 +1 ) ( 3 x2 − 4 x+ 7 )
Expanding:

x 2 ( 3 x 2 − 4 x +7 ) +1 ( 3 x 2 −4 x+7 )
4 3 2 2
3 x −4 x +7 x +3 x − 4 x +7
4 3 2
3 x −4 x +10 x − 4 x +7



4. g⁻¹(x) (Finding the inverse of g(x))

We have:
2
y=x +1

Solving for x :
2
y −1=x