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

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

1. (g ∘ f)(x) [4 marks]

This means we substitute f ( x ) into g ( x ):

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

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



Expanding:
2
( 3 x 2 − 4 x +7 ) =9 x 4 −24 x 3 +49 x 2 − 56 x+ 49
Thus:

, g ( f ( x ) )=9 x − 24 x + 49 x −56 x +50
4 3 2




2. (g − f)(x) [2 marks]

( g − f ) ( x ) =g ( x ) − f ( x )

Substituting:

( 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) [4 marks]

This represents the product of g ( x ) and f ( x ) :

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

Expanding:
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 ) [3 marks]

To find the inverse function g− 1 ( x ), we set y=g ( x ):
2
y=x +1

Solving for x :
2
x = y −1

x=± √ y − 1

Thus, the inverse function is:

g ( x )=± √ x −1
−1