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GCSE · Algebra - Functions

A6 Functions: Composite and Inverse Functions

AQA 8365 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

f(x) = 2x + 3 and g(x) = x - 1.

(a)Find f(4).(1)

(b)Find g(7).(1)

(c)Find fg(2).(2)

(Total for Question 1 is 4 marks)

f(x) = 3x - 2 and g(x) = x^2.

(a)Find gf(3).(2)

(b)Find fg(-2).(2)

(Total for Question 2 is 4 marks)

f(x) = x/2 + 4.

(a)Find f^-1(x).(3)

(b)Find f^-1(10).(1)

(Total for Question 3 is 4 marks)

Explain why f(x) = x², for all real values of x, does not have an inverse function. State a suitable restricted domain for f(x) so that an inverse function does exist.

(Total for Question 4 is 2 marks)

g(x) = x² - 4, for the domain x ≥ 0.

(a)State the range of g(x) for this domain.(1)

(b)Find g^-1(x), stating its domain.(3)

(Total for Question 5 is 4 marks)

f(x) = 5x - 1 and g(x) = 2x + 3.

(a)Find fg(x), giving your answer in the form ax + b.(3)

(b)Find gf(x), giving your answer in the form ax + b.(3)

(c)State, with a reason, whether fg(x) = gf(x) for all values of x.(1)

(Total for Question 6 is 7 marks)

f(x) = (x + 2)/3.

(a)Find f^-1(x).(3)

(b)By first finding f(5), verify that f^-1(f(5)) = 5.(2)

(Total for Question 7 is 5 marks)

A taxi firm charges a fare, in pounds sterling, for a journey of m miles given by C(m) = 2m + 3. A distance of x kilometres converts to miles using the function k(x) = 0.62x.

(a)Find an expression for C(k(x)), the fare in terms of the distance x in kilometres.(2)

(b)Use your expression to find the cost of a 10 km journey.(2)

(Total for Question 8 is 4 marks)

Using C(k(x)) = 1.24x + 3 from Question 8, a customer's fare for a journey is £9.96. Find the distance travelled, in km, giving your answer to 2 decimal places.

(Total for Question 9 is 3 marks)

f(x) = 4x - 7 and g(x) = x² + 1.

(a)Find fg(x).(2)

(b)Find gf(x).(3)

(c)Solve fg(x) = 33.(3)

(Total for Question 10 is 8 marks)

f(x) = (2x - 1)/(x + 3), where x ≠ -3. Find f^-1(x), stating any value which x cannot take.

(Total for Question 11 is 4 marks)

f(x) = 3x + 2 and g(x) = x - 5.

(a)Find fg(x).(2)

(b)Find (fg)^-1(x).(3)

(c)Find f^-1(x) and g^-1(x), and hence show that g^-1(f^-1(x)) = (fg)^-1(x).(3)

(Total for Question 12 is 8 marks)

f(x) = (x + 4)/(x - 1), where x ≠ 1. Show that f is a self-inverse function.

(Total for Question 13 is 4 marks)

f(x) = x² - 3, for the domain x ≤ 0.

(a)State the range of f(x) for this domain.(1)

(b)Find f^-1(x), stating its domain.(3)

(Total for Question 14 is 4 marks)

f(x) = 2x - 1 and g(x) = 3/(x + 2), where x ≠ -2.

(a)Find fg(x), giving your answer as a single fraction.(3)

(b)State the value of x for which fg(x) is undefined.(1)

(c)Find g^-1(x).(3)

(Total for Question 15 is 7 marks)

f(x) = x³ - 2.

(a)Find f^-1(x).(3)

(b)Evaluate f^-1(25).(2)

(Total for Question 16 is 5 marks)

f(x) = ax + b, where a and b are constants. Given that f(2) = 7 and f^-1(3) = 1, find the values of a and b.

(Total for Question 17 is 5 marks)

f(x) = 2x - 3 and g(x) = x² - 2x. Solve gf(x) = 0.

(Total for Question 18 is 5 marks)

f(x) = (3x - 2)/5 and g(x) = x + 1. Determine whether there is a value of x for which fg(x) = gf(x). Justify your answer fully.

(Total for Question 19 is 4 marks)

f(x) = (2x + 1)/(x - 3), where x ≠ 3. Find the value(s) of x for which f(x) = x, giving your answers to 2 decimal places.

(Total for Question 20 is 5 marks)

f(x) = 2x - 5 and g(x) = x² + k, where k is a constant.

(a)Find fg(x) in terms of x and k.(2)

(b)Given that fg(3) = 21, find the value of k.(3)

(c)Hence find gf(x), giving your answer in the form ax² + bx + c.(3)

(Total for Question 21 is 8 marks)

Mark scheme · A6 Functions: Composite and Inverse Functions

Question 1

(a) B1 f(4) = 11 cao
(a) Answer: 11
(b) B1 g(7) = 6 cao
(b) Answer: 6
(c) M1 g(2) = 1 found or used
(c) A1 fg(2) = 5 cao
(c) Answer: 5

Question 2

(a) M1 f(3) = 7 found
(a) A1 gf(3) = 49 cao
(a) Answer: 49
(b) M1 g(-2) = 4 found
(b) A1 fg(-2) = 10 cao
(b) Answer: 10

Question 3

(a) M1 let y = x/2 + 4 and rearrange to isolate x, e.g. x/2 = y - 4
(a) M1 x = 2(y - 4) or equivalent correct rearrangement
(a) A1 f^-1(x) = 2x - 8 oe (e.g. 2(x - 4))
(a) Answer: f^-1(x) = 2x - 8
(b) B1 f^-1(10) = 12 ft from part (a)
(b) Answer: 12

Question 4

B1 correct reason, e.g. f(x) is many-to-one since f(2) = f(-2) = 4, so f fails the horizontal line test / two x-values give the same output
B1 a valid restricted domain stated, oe, e.g. x ≥ 0 (or x ≤ 0)
Answer: f is many-to-one (e.g. f(2) = f(-2) = 4); restrict domain to x ≥ 0 (or x ≤ 0)

Question 5

(a) B1 g(x) ≥ -4 oe
(a) Answer: g(x) ≥ -4
(b) M1 y = x² - 4 rearranged to x² = y + 4
(b) A1 g^-1(x) = √x + 4 (positive root taken since x ≥ 0)
(b) B1 domain stated as x ≥ -4 oe, ft from part (a)
(b) Answer: g^-1(x) = √x + 4, domain x ≥ -4

Question 6

(a) M1 substitute g(x) into f, e.g. 5(2x + 3) - 1
(a) M1 expand correctly, e.g. 10x + 15 - 1
(a) A1 fg(x) = 10x + 14 cao
(a) Answer: fg(x) = 10x + 14
(b) M1 substitute f(x) into g, e.g. 2(5x - 1) + 3
(b) M1 expand correctly, e.g. 10x - 2 + 3
(b) A1 gf(x) = 10x + 1 cao
(b) Answer: gf(x) = 10x + 1
(c) B1 No, because the constant terms differ (14 ≠ 1), ft from (a) and (b)
(c) Answer: No, since 10x + 14 ≠ 10x + 1

Question 7

(a) M1 let y = (x+2)/3 and multiply both sides by 3, i.e. 3y = x + 2
(a) M1 rearrange to x = 3y - 2
(a) A1 f^-1(x) = 3x - 2 cao
(a) Answer: f^-1(x) = 3x - 2
(b) M1 f(5) = 7/3 (or awrt 2.33) found
(b) A1 f^-1(7/3) = 3(7/3) - 2 = 5, shown correctly cso
(b) Answer: f^-1(f(5)) = 5, verified

Question 8

(a) M1 substitute k(x) into C, i.e. 2(0.62x) + 3
(a) A1 C(k(x)) = 1.24x + 3 oe
(a) Answer: C(k(x)) = 1.24x + 3
(b) M1 substitute x = 10 into 1.24x + 3, ft from (a)
(b) A1 £15.40 cao
(b) Answer: £15.40

Question 9

M1 set up the equation 1.24x + 3 = 9.96
M1 rearrange to 1.24x = 6.96
A1 awrt 5.61 (km)
Answer: x = awrt 5.61 km

Question 10

(a) M1 substitute g(x) into f, e.g. 4(x²+1) - 7
(a) A1 fg(x) = 4x² - 3 cao
(a) Answer: fg(x) = 4x² - 3
(b) M1 substitute f(x) into g, e.g. (4x-7)² + 1
(b) M1 expand (4x-7)² correctly to 16x² - 56x + 49
(b) A1 gf(x) = 16x² - 56x + 50 cao
(b) Answer: gf(x) = 16x² - 56x + 50
(c) M1 set up 4x² - 3 = 33, ft from (a)
(c) M1 rearrange to x² = 9
(c) A1 x = 3 or x = -3 (both values required) cao
(c) Answer: x = 3 or x = -3

Question 11

M1 let y = (2x-1)/(x+3) and cross-multiply: y(x+3) = 2x - 1
M1 collect x terms on one side, e.g. yx - 2x = -1 - 3y, i.e. x(y-2) = -1-3y
A1 f^-1(x) = (3x+1)/(2-x) oe
B1 x ≠ 2 stated
Answer: f^-1(x) = (3x+1)/(2-x), x ≠ 2

Question 12

(a) M1 substitute g(x) into f, e.g. 3(x-5) + 2
(a) A1 fg(x) = 3x - 13 cao
(a) Answer: fg(x) = 3x - 13
(b) M1 let y = 3x - 13, ft from (a)
(b) M1 rearrange to x = (y+13)/3
(b) A1 (fg)^-1(x) = (x+13)/3 cao
(b) Answer: (fg)^-1(x) = (x+13)/3
(c) M1 f^-1(x) = (x-2)/3 and g^-1(x) = x+5 both found
(c) M1 substitute f^-1(x) into g^-1, e.g. (x-2)/3 + 5
(c) A1 simplifies to (x+13)/3, matching (fg)^-1(x) from (b) cso
(c) Answer: g^-1(f^-1(x)) = (x+13)/3 = (fg)^-1(x), as required

Question 13

M1 let y = (x+4)/(x-1) and cross-multiply: y(x-1) = x + 4
M1 collect x terms, e.g. yx - x = y + 4, i.e. x(y-1) = y+4
A1 f^-1(x) = (x+4)/(x-1) obtained correctly
A1 conclusion that f^-1(x) = f(x), so f is self-inverse cso
Answer: f^-1(x) = (x+4)/(x-1) = f(x), so f is self-inverse

Question 14

(a) B1 f(x) ≥ -3 oe
(a) Answer: f(x) ≥ -3
(b) M1 y = x² - 3 rearranged to x² = y + 3
(b) A1 f^-1(x) = -√x+3 (negative root taken since x ≤ 0)
(b) B1 domain x ≥ -3 stated, ft from part (a)
(b) Answer: f^-1(x) = -√x+3, domain x ≥ -3

Question 15

(a) M1 substitute g(x) into f, e.g. 2(3/(x+2)) - 1
(a) M1 write over a common denominator, e.g. (6 - (x+2))/(x+2)
(a) A1 fg(x) = (4-x)/(x+2) oe
(a) Answer: fg(x) = (4-x)/(x+2)
(b) B1 x = -2
(b) Answer: x = -2
(c) M1 let y = 3/(x+2) and cross-multiply: y(x+2) = 3
(c) M1 rearrange to x = (3-2y)/y
(c) A1 g^-1(x) = (3-2x)/x, x ≠ 0 cao
(c) Answer: g^-1(x) = (3-2x)/x

Question 16

(a) M1 y = x³ - 2 rearranged to x³ = y + 2
(a) M1 take cube roots, x = cbrt(y+2)
(a) A1 f^-1(x) = cbrt(x+2) cao
(a) Answer: f^-1(x) = cbrt(x+2)
(b) M1 25 + 2 = 27 found, ft from (a)
(b) A1 f^-1(25) = 3 cao
(b) Answer: 3

Question 17

M1 interpret f^-1(3) = 1 as f(1) = 3
M1 form two equations: 2a + b = 7 and a + b = 3
M1 solve simultaneously, e.g. subtract to find a
A1 a = 4 cao
A1 b = -1 cao
Answer: a = 4, b = -1

Question 18

M1 substitute f(x) into g, e.g. (2x-3)² - 2(2x-3)
M1 expand and simplify to 4x² - 16x + 15 (or unsimplified equivalent)
M1 correct method to solve the quadratic, e.g. factorise (2x-3)(2x-5) = 0 or use the quadratic formula
A1 x = 1.5 cao
A1 x = 2.5 cao
Answer: x = 1.5 or x = 2.5

Question 19

M1 fg(x) = (3(x+1)-2)/5 = (3x+1)/5 found
M1 gf(x) = (3x-2)/5 + 1 = (3x+3)/5 found
M1 attempt to equate fg(x) and gf(x) and simplify, e.g. 3x+1 = 3x+3
A1 correct conclusion: no such x exists, since this simplifies to 1 = 3, a contradiction cso
Answer: No value of x exists; fg(x) = gf(x) leads to the contradiction 1 = 3

Question 20

M1 set (2x+1)/(x-3) = x and cross-multiply: 2x+1 = x(x-3)
M1 rearrange to x² - 5x - 1 = 0
M1 correct use of the quadratic formula with a=1, b=-5, c=-1
A1 x = awrt 5.19
A1 x = awrt -0.19
Answer: x = 5.19 or x = -0.19 (2 dp)

Question 21

(a) M1 substitute g(x) into f, e.g. 2(x²+k) - 5
(a) A1 fg(x) = 2x² + 2k - 5 oe
(a) Answer: fg(x) = 2x² + 2k - 5
(b) M1 substitute x=3 into fg(x), ft from (a), i.e. 2(9) + 2k - 5
(b) M1 simplify to 13 + 2k = 21
(b) A1 k = 4 cao
(b) Answer: k = 4
(c) M1 substitute f(x) into g using k=4 (ft from (b)), i.e. (2x-5)² + 4
(c) M1 expand (2x-5)² correctly to 4x² - 20x + 25
(c) A1 gf(x) = 4x² - 20x + 29 cao (ft)
(c) Answer: gf(x) = 4x² - 20x + 29

Functions: Composite and Inverse Functions - Worksheets, Questions and Revision

A6 Functions: Composite and Inverse Functions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21