IB · Mathematics: Analysis & Approaches SL & HL · Functions

Functions, composites & inverses: IB Mathematics: Analysis & Approaches questions with answers

Free Functions, composites & inverses questions for IB Mathematics: Analysis & Approaches (SL & HL), part of Functions. Each question has the answer and the marking points an examiner would look for.

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Functions, composites & inverses practice questions

Question 1Structured · 9 marksA* challenge

Let f(x) = x2 − 6x, for x ≥ k.

(a)

Find the least value of k for which f has an inverse function.

[2]
(b)

For this value of k, find f−1(x), stating its domain.

[4]
(c)

Solve f(x) = f−1(x).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)k = 3
f is one-to-one on x ≥ vertex (M1)
k = 3 A1
2
(b)f−1(x) = 3 + √x + 9, x ≥ −9
f(x) = (x − 3)2 − 9 (A1)
interchanging x and y and rearranging M1
f−1(x) = 3 + √x + 9 A1
domain x ≥ −9 A1
4
(c)x = 7
the graphs meet on y = x, so f(x) = x M1
x2 − 7x = 0 ⇒ x = 0 or x = 7 A1
x = 0 < 3 is not in the domain, so x = 7 R1
3
Question 2Structured · 6 marks

Let f(x) = 5 − x2 and g(x) = 2x, for −4 ≤ x ≤ 4.

(a)

Find the x-coordinates of the points where the graphs of f and g intersect.

[3]
(b)

Hence solve the inequality f(x) > g(x).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x = −2.19, x = 1.48
attempt to solve f(x) = g(x), e.g. graph or solver (M1)
x = −2.19, x = 1.48 A2
award A1 if one value is missing or incorrect
3
(b)−2.19 < x < 1.48
using the graph to choose the correct regions (M1)
correct critical values used as endpoints (A1)
−2.19 < x < 1.48 A1
3
Question 3Structured · 6 marks

Let f(x) = −2x + 3 and g(x) = x2 − 6, for x ∈ ℝ.

(a)

Find (f∘g)(x).

[2]
(b)

Solve (g∘f)(x) = 30.

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)−2x2 + 15
f(x2 − 6) = −2(x2 − 6) + 3 (M1)
−2x2 + 15 A1
2
(b)x = −32, x = 92
(−2x + 3)2 − 6 = 30 M1
(−2x + 3)2 = 36 (A1)
−2x + 3 = ±6 (M1)
x = −32, x = 92 A1
4
Question 4Structured · 7 marks

Let f(x) = 4x + 1x − 4, x ≠ 4.

(a)

Write down the equation of the vertical asymptote and the equation of the horizontal asymptote of the graph of f.

[2]
(b)

Find f−1(x).

[3]
(c)

Hence state, with a reason, a special property of f.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x = 4, y = 4
x = 4 A1
y = 4 A1
2
(b)f−1(x) = 4x + 1x − 4
interchanging x and y (M1)
x(y − 4) = 4y + 1 ⇒ y(x − 4) = 4x + 1 (A1)
f−1(x) = 4x + 1x − 4 A1
3
(c)f is self-inverse
f−1(x) = f(x) R1
f is self-inverse (its graph is symmetric in y = x) A1
2
Question 5Structured · 9 marksA* challenge

Let f(x) = x2 − 8x + 14, for x ≥ k.

(a)

Find the least value of k for which f has an inverse function.

[2]
(b)

For this value of k, find f−1(x), stating its domain.

[4]
(c)

Solve f(x) = f−1(x).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)k = 4
f is one-to-one on x ≥ vertex (M1)
k = 4 A1
2
(b)f−1(x) = 4 + √x + 2, x ≥ −2
f(x) = (x − 4)2 − 2 (A1)
interchanging x and y and rearranging M1
f−1(x) = 4 + √x + 2 A1
domain x ≥ −2 A1
4
(c)x = 7
the graphs meet on y = x, so f(x) = x M1
x2 − 9x + 14 = 0 ⇒ x = 2 or x = 7 A1
x = 2 < 4 is not in the domain, so x = 7 R1
3
Question 6Structured · 6 marks

Let f(x) = x3 − 4x + 1 and g(x) = e−x, for −3 ≤ x ≤ 3.

(a)

Find the x-coordinates of the points where the graphs of f and g intersect.

[3]
(b)

Hence solve the inequality f(x) < g(x).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)x = −1.37, x = 0, x = 1.88
attempt to solve f(x) = g(x), e.g. graph or solver (M1)
x = −1.37, x = 0, x = 1.88 A2
award A1 if one value is missing or incorrect
3
(b)−3 ≤ x < −1.37 or 0 < x < 1.88
using the graph to choose the correct regions (M1)
correct critical values used as endpoints (A1)
−3 ≤ x < −1.37 or 0 < x < 1.88 A1
3

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in