MAT1514 Assignment 1 (COMPLETE ANSWERS) 2025 - DUE 2025; 100% TRUSTED Complete, trusted solutions and explanations

MAT1514 Assignment 1 (COMPLETE ANSWERS) 2025


Question 1 Given: 𝑓 ( 𝑥 )

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

( 𝑔 ∘ 𝑓 ) ( 𝑥 ) (g∘f)(x) This represents 𝑔 ( 𝑓 ( 𝑥 ) ) g(f(x)), meaning we substitute 𝑓 ( 𝑥 ) f(x)
into 𝑔 ( 𝑥 ) g(x):


𝑔(𝑓(𝑥))

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


(3𝑥2−4𝑥+7)(3𝑥2−4𝑥+7)
9 𝑥 4 − 24 𝑥 3 + 49 𝑥 2 − 56 𝑥 + 49 (3x 2 −4x+7)(3x 2 −4x+7)=9x 4 −24x 3 +49x 2 −56x+49
Adding 1:

9 𝑥 4 − 24 𝑥 3 + 49 𝑥 2 − 56 𝑥 + 50 9x 4 −24x 3 +49x 2 −56x+50 2. ( 𝑔 − 𝑓 ) ( 𝑥 ) (g−f)(x) This
means:


𝑔(𝑥)−𝑓(𝑥)

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

𝑥 2 + 1 − 3 𝑥 2 + 4 𝑥 − 7 =x 2 +1−3x 2 +4x−7
− 2 𝑥 2 + 4 𝑥 − 6 =−2x 2 +4x−6 3. 𝑔 𝑓 ( 𝑥 ) gf(x) This means 𝑔 ( 𝑥 ) × 𝑓 ( 𝑥 ) g(x)×f(x):

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


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

3 𝑥 4 − 4 𝑥 3 + 7 𝑥 2 + 3 𝑥 2 − 4 𝑥 + 7 =3x 4 −4x 3 +7x 2
+3x 2 −4x+7

3 𝑥 4 − 4 𝑥 3 + 10 𝑥 2 − 4 𝑥 + 7 =3x 4 −4x 3 +10x 2
−4x+7 4. 𝑔 − 1 ( 𝑥 ) g −1 (x) To find the inverse of 𝑔 ( 𝑥
)

𝑥 2 + 1 g(x)=x 2 +1, we set 𝑦
𝑥 2 + 1 y=x 2 +1:


𝑦
𝑥 2 + 1 y=x 2 +1 Solving for 𝑥 x:


𝑥2

𝑦 − 1 x 2 =y−1 𝑥
± 𝑦 − 1 x=± y−1

Since 𝑔 ( 𝑥 ) g(x) is not one-to-one, its inverse is: