Given f(x) = 3x - 4 and g(x) = x2, work out the value of fg(2).
Worked Solution & Answer
g(2) = 4, so f(4) = 3(4) - 4 = 8 B1
Answer: 8
Functions (harder) — Notes & Worked Examples Show / Hide
Explanation:
A composite function fg(x) means apply function g first, then substitute that output into f, i.e. fg(x) = f(g(x)). Always work from the inside out, and remember that fg(x) is rarely equal to gf(x).
To find the inverse function f−1(x), write y = f(x), swap x and y (or rearrange directly to make the input variable the subject), and write your final answer in terms of x. If the original input appears twice, use the factorising rearrangement method.
Example 1: f(x) = 2x − 3 and g(x) = x2 + 1. Solve fg(x) = 15.
Solution:
Substitute g(x) into f(x):
fg(x) = 2(x2 + 1) − 3 = 2x2 − 1
Set equal to 15: 2x2 − 1 = 15 ⇒ 2x2 = 16
x2 = 8 ⇒ x = ±√8 = ±2√2
Answer: x = ±2√2
Example 2: Given h(x) = 4xx + 5, find h−1(x).
Solution:
Let y = 4xx + 5 and swap x and y: x = 4yy + 5
Multiply by (y + 5): xy + 5x = 4y
Collect y terms: 5x = 4y − xy = y(4 − x)
Divide by (4 − x): y = 5x4 − x
Answer: h−1(x) = 5x4 − x
Given f(x) = 4x - 5 and g(x) = x2, work out the value of fg(3).
Worked Solution & Answer
g(3) = 9, so f(9) = 4(9) - 5 = 31 B1
Answer: 31
f and g are functions such that
f(x) = 2x2 \quadand g(x) = 4x3
f(x) = 2x2 \quadand g(x) = 4x3
(a) Find f(-5) [1 mark]
(b) Find fg(1) [2 marks]
Worked Solution & Answer
(a)
f(–5) = 2 / (–5)² = 2/25 B1
Answer (a): 225
(b)
g(1) = 4(1)³ = 4 M1
f(4) = 2 / 4² = 2/16 = 1/8 A1
f(4) = 2 / 4² = 2/16 = 1/8 A1
Answer (b): 18
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2
27 (b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360°
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2
27 (b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360°
(a) h(x) = 3√(x) for all values of x
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2 [2 marks]
On the grid, draw the graph of the inverse function y = h-1(x) for -2 ≤ x ≤ 2 [2 marks]
(b) For all values of x
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360° [2 marks]
f(x) = \sin x
g(x) = x + 90
On the grid, draw the graph of the composite function y = fg(x) for 0° ≤ x ≤ 360° [2 marks]
Worked Solution & Answer
(a)
The inverse function is h-1(x) = x3 M1
Values: (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
Plot these points and draw a smooth cubic curve. A1
Values: (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
Plot these points and draw a smooth cubic curve. A1
Answer (a): Cubic graph y = x³ drawn through (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
(b)
fg(x) = f(g(x)) = \sin(x + 90°) = \cos x M1
Plot the cosine wave: 0° → 1, 90° → 0, 180° → -1, 270° → 0, 360° → 1. A1
Plot the cosine wave: 0° → 1, 90° → 0, 180° → -1, 270° → 0, 360° → 1. A1
Answer (b): Cosine curve drawn from (0°, 1) to (90°, 0), (180°, -1), (270°, 0), (360°, 1)
f(x) = 3x + 95 and g(x) = 6x - 1
(a) Show that gf(2) is an integer. [2 marks]
(b) Show that f-1(8) is **not** an integer. [2 marks]
Worked Solution & Answer
(a)
f(2) = (3 × 2 + 9)/5 = 15/5 = 3. gf(2) = g(3) = 6(3) – 1 = 17 M1 C1
Answer (a): f(2) = (6 + 9)/5 = 3, and g(3) = 6(3) – 1 = 17, which is an integer.
(b)
Solve f(x) = 8: (3x + 9)/5 = 8 M1
⇒ 3x + 9 = 40
⇒ 3x = 31
⇒ x = 31/3 (10 1/3), not an integer C1
⇒ 3x + 9 = 40
⇒ 3x = 31
⇒ x = 31/3 (10 1/3), not an integer C1
Answer (b): (3x + 9)/5 = 8 ⇒ 3x + 9 = 40 ⇒ 3x = 31 ⇒ x = 31/3 = 10.33..., which is not an integer.
f(x) = 9 - √(x) where x ≥ 0
g(x) = 4x2
g(x) = 4x2
(a) Find f(9) [1 mark]
(b) Solve fg(x) < 0
Show clear algebraic working. [3 marks]
Show clear algebraic working. [3 marks]
Worked Solution & Answer
(a)
9 – √9 = 6 B1
Answer (a): 6
(b)
fg(x) = 9 – √(4x²) = 9 – 2x < 0 (since x ≥ 0) M1
⇒ 2x > 9 M1
⇒ x > 4.5 A1
⇒ 2x > 9 M1
⇒ x > 4.5 A1
Answer (b): x > 4.5
For all values of x
f(x) = (x + 1)2 and g(x) = 2(x - 1)
f(x) = (x + 1)2 and g(x) = 2(x - 1)
(a) Show that gf(x) = 2x(x + 2) [2 marks]
(b) Find g-1(7) [2 marks]
Worked Solution & Answer
(a)
gf(x) = 2[(x + 1)² – 1] = 2[x² + 2x + 1 – 1] = 2x(x + 2) M1 C1
Answer (a): gf(x) = 2((x + 1)² - 1) = 2(x² + 2x + 1 - 1) = 2(x² + 2x) = 2x(x + 2)
(b)
2(x – 1) = 7 M1
⇒ x – 1 = 3.5
⇒ x = 4.5 A1
⇒ x – 1 = 3.5
⇒ x = 4.5 A1
Answer (b): 4.5
The functions f and g are given by
f(x) = 12x + 1 and g(x) = 5 - 3x
f(x) = 12x + 1 and g(x) = 5 - 3x
(a) Find f(-3) [1 mark]
(b) Find fg(1) [2 marks]
(c) Find g-1(4) [2 marks]
Worked Solution & Answer
(a)
12/(–3 + 1) = 12/(–2) = –6 B1
Answer (a): –6
(b)
g(1) = 5 – 3(1) = 2 M1
⇒ f(2) = 12/(2 + 1) = 4 A1
⇒ f(2) = 12/(2 + 1) = 4 A1
Answer (b): 4
(c)
Solve 5 – 3x = 4 M1
⇒ 3x = 1
⇒ x = 1/3 A1
⇒ 3x = 1
⇒ x = 1/3 A1
Answer (c): 13
The functions f(x) and g(x) are defined, for x ≥ 0, by
f(x) = x2, g(x) = x + 65
f(x) = x2, g(x) = x + 65
(a) Find the value of gf(8). [2 marks]
(b) Find an expression for g-1(x). [2 marks]
(b) The diagram shows a sketch of the graph of y = h(x).
The point A has coordinates (-2, -3) and the point B has coordinates (2, 1). [2 marks]
The point A has coordinates (-2, -3) and the point B has coordinates (2, 1). [2 marks]
On the same diagram, sketch the graph of y = h(x) + 4.
You must indicate the coordinates of the new positions of the points A and B.
You must indicate the coordinates of the new positions of the points A and B.
Worked Solution & Answer
(a)
f(8) = 82 = 64. M1
gf(8) = g(64) = 64 + 65 = 705 = 14. A1
gf(8) = g(64) = 64 + 65 = 705 = 14. A1
Answer (a): 14
(b)
Let y = x + 65 ⇒ 5y = x + 6 ⇒ x = 5y - 6 M1
So g-1(x) = 5x - 6. A1
So g-1(x) = 5x - 6. A1
Answer (b): 5x - 6
(b)
The transformation y = h(x) + 4 is a vertical translation by vector (04). M1
A' = (-2, -3 + 4) = (-2, 1).
B' = (2, 1 + 4) = (2, 5). A1
A' = (-2, -3 + 4) = (-2, 1).
B' = (2, 1 + 4) = (2, 5). A1
Answer (b): Graph translated 4 units vertically up. New A: (-2, 1), new B: (2, 5)
(a) The functions f(x) and g(x) are defined by
f(x) = \sin x°, g(x) = x - 90.
Find the value of fg(120).
(b) The diagram shows a sketch of the graph of y = h(x).
(c) The function k(x) is defined by
k(x) = x3 - 23.
Solve k-1(x) = 5.
f(x) = \sin x°, g(x) = x - 90.
Find the value of fg(120).
(b) The diagram shows a sketch of the graph of y = h(x).
(c) The function k(x) is defined by
k(x) = x3 - 23.
Solve k-1(x) = 5.
(a) The functions f(x) and g(x) are defined by
f(x) = \sin x°, g(x) = x - 90.
Find the value of fg(120). [2 marks]
f(x) = \sin x°, g(x) = x - 90.
Find the value of fg(120). [2 marks]
(b) The diagram shows a sketch of the graph of y = h(x). [2 marks]
On the same diagram, sketch the graph of y = -h(x).
Write down the coordinates of the maximum and minimum points of y = -h(x).
Write down the coordinates of the maximum and minimum points of y = -h(x).
(c) The function k(x) is defined by
k(x) = x3 - 23.
Solve k-1(x) = 5. [4 marks]
k(x) = x3 - 23.
Solve k-1(x) = 5. [4 marks]
x =
Worked Solution & Answer
(a)
g(120) = 120 - 90 = 30 M1
fg(120) = f(30) = \sin 30° = 0.5 A1
fg(120) = f(30) = \sin 30° = 0.5 A1
Answer (a): 0.5 (or 12)
(b)
The transformation y = -h(x) reflects the graph in the x-axis. M1
• The minimum point (-2, -1) becomes a maximum at (-2, 1).
• The maximum point (-6, 0.5) becomes a minimum at (-6, -0.5). A1
• The minimum point (-2, -1) becomes a maximum at (-2, 1).
• The maximum point (-6, 0.5) becomes a minimum at (-6, -0.5). A1
Answer (b): Maximum: (-2, 1); Minimum: (-6, -0.5)
(c)
If k-1(x) = 5, then x = k(5). M1
k(5) = 53 - 23 = 125 - 23 = 102 M1
(Alternatively: y = x3 - 23 ⇒ x = 3√(y + 23) ⇒ k^{-1}(x) = 3√(x + 23) M1
3√(x + 23) = 5 ⇒ x + 23 = 53 = 125 ⇒ x = 102) A1
k(5) = 53 - 23 = 125 - 23 = 102 M1
(Alternatively: y = x3 - 23 ⇒ x = 3√(y + 23) ⇒ k^{-1}(x) = 3√(x + 23) M1
3√(x + 23) = 5 ⇒ x + 23 = 53 = 125 ⇒ x = 102) A1
Answer (c): x = x = 102