Let f be the function defined by f(x) = 2x + 3. For the function f in this context, find f(0).
(Total for Question 1 is 1 mark)
2
The function g is defined by g(x) = x2 - 4. For the function g in this question, find g(3).
(Total for Question 2 is 1 mark)
3
Let f(x) = 3x - 1 and h(x) = 5. For these functions, work out f(h(2)).
(Total for Question 3 is 2 marks)
4
For functions f(x) = x + 2 and g(x) = 2x, find and simplify g(f(x)). State the composite as an algebraic expression in x for the functions named.
(Total for Question 4 is 2 marks)
5
With the same f(x) = x + 2 and g(x) = 2x as in Question 4, find and simplify f(g(x)).
(Total for Question 5 is 2 marks)
6
Let f(x) = 4x - 5 and g(x) = x2. Work out fg(2) and gf(2), where fg(x) means f(g(x)) and gf(x) means g(f(x)).
(Total for Question 6 is 3 marks)
7
Define f(x) = 1/(x + 1) for x ≠ -1. For the function f in this context, find f(f(1)) and show your working.
(Total for Question 7 is 3 marks)
8
Let f(x) = 2x + 1 and g(x) = x - 3. Show that the composite fg(x) = 2x - 5, where fg(x) denotes f(g(x)).
(Total for Question 8 is 3 marks)
9
The function f is defined by f(x) = 3x + 2. Find the inverse function f-1(x) by swapping variables and rearranging, and state f-1(x) as an algebraic expression.
(Total for Question 9 is 4 marks)
10
Find the inverse of the function g(x) = (x - 4)/2 by swapping variables and rearranging, for the function g named here. Give g-1(x) in simplest form.
(Total for Question 10 is 3 marks)
11
Let f(x) = 2x + 1 and g(x) = (x - 1)/2. By inspection or composition, show whether g is the inverse of f by calculating f(g(x)) and g(f(x)). Name the conclusion for the functions in this question.
(Total for Question 11 is 3 marks)
12
Let f(x) = x2 + 1, defined for x ≥ 0 only. Find f-1(x) for this restricted function by swapping variables and rearranging, and state the domain for f-1 in this context.
(Total for Question 12 is 3 marks)
13
Given f(x) = 2x + 3 and g(x) = x2, solve the equation f(g(x)) = 11 for x, where f(g(x)) denotes f composed with g, and show working.
(Total for Question 13 is 3 marks)
14
Let f(x) = (x + 2)/3. Solve f(x) = 5 for x by rearrangement, showing a correct method and final value for the function named here.
(Total for Question 14 is 2 marks)
15
The function f is defined by f(x) = x/2 - 7. Find f-1(x) by swapping variables and rearranging, and give f-1(3).
(Total for Question 15 is 3 marks)
16
Let f(x) = x - 4 and g(x) = x/5. Find and simplify fg(x) and gf(x) for the functions named, and state whether fg = gf as functions.
(Total for Question 16 is 2 marks)
Mark scheme · IG.M4 Functions: Composite and Inverse Functions
Question 1
B1 f(0) = 3 cao
Answer: 3
Question 2
B1 g(3) = 5 cao
Answer: 5
Question 3
M1 evaluates h(2) = 5 and substitutes into f, or shows f(5) seen
A1 f(h(2)) = 14 cao
Answer: 14
Question 4
M1 substitutes f(x) into g as g(f(x)) = 2(f(x)) or 2(x + 2)
A1 g(f(x)) = 2x + 4 cao
Answer: g(f(x)) = 2x + 4
Question 5
M1 substitutes g(x) into f as f(g(x)) = g(x) + 2 or x*2 + 2
A1 f(g(x)) = 2x + 2 cao
Answer: f(g(x)) = 2x + 2
Question 6
M1 evaluates g(2) = 4 and substitutes into f, or shows f(g(2)) process
M1 evaluates f(2) = 3 and substitutes into g, or shows g(f(2)) process
A1 fg(2) = 11 and gf(2) = 9 cao
Answer: fg(2) = 11; gf(2) = 9
Question 7
M1 finds f(1) = 1/2 or equivalent
M1 substitutes f(1) into f correctly, showing f(1/2) = 1/(1/2 + 1)
A1 f(f(1)) = 2/3 cao
Answer: 2/3
Question 8
M1 substitutes g(x) into f: f(g(x)) = 2(g(x)) + 1
M1 substitutes g(x) = x - 3 giving 2(x - 3) + 1
A1 simplifies to fg(x) = 2x - 5 cao
Answer: fg(x) = 2x - 5
Question 9
M1 starts with y = 3x + 2 and swaps x and y or writes x = 3y + 2
M1 rearranges correctly: x - 2 = 3y or y = (x - 2)/3 seen
A1 f-1(x) = (x - 2)/3 cao
B1 states domain restriction or checks composition eg f(f-1(x)) = x, oe
Answer: f-1(x) = (x - 2)/3
Question 10
M1 starts with y = (x - 4)/2 and swaps x and y to x = (y - 4)/2 or equivalent
M1 rearranges to y = 2x + 4
A1 g-1(x) = 2x + 4 cao
Answer: g-1(x) = 2x + 4
Question 11
M1 computes f(g(x)) = 2((x - 1)/2) + 1 and simplifies to x
M1 computes g(f(x)) = (2x + 1 - 1)/2 and simplifies to x
A1 concludes g is the inverse of f since both compositions give x cao
Answer: Both f(g(x)) = x and g(f(x)) = x, so g is the inverse of f
Question 12
M1 starts with y = x2 + 1 and swaps x and y to x = y2 + 1 or writes x = y2 + 1
M1 rearranges to y = √x - 1 and states the principal (non-negative) root
A1 f-1(x) = √x - 1 with domain x ≥ 1 cao
Answer: f-1(x) = √x - 1, domain x ≥ 1
Question 13
M1 writes f(g(x)) = 2x2 + 3 and sets equal to 11
M1 solves 2x2 + 3 = 11 to get x2 = 4
A1 x = 2 or x = -2 cao
Answer: x = 2 or x = -2
Question 14
M1 multiplies both sides by 3 or shows rearrangement 3f(x) = x + 2
A1 x = 13 cao
Answer: x = 13
Question 15
M1 writes y = x/2 - 7 and swaps to x = y/2 - 7 or equivalent