Module 18 of 47Lesson 1 of 4 · HL · AI 2.7 — Composite and inverse functions

Course map · Topic 2: Functions

2.7 Composite and inverse functions

HL only12 worked questions

A composite function feeds the output of one function into another, like two machines in a line. An inverse function runs a machine backwards: it takes the output and gives you back the input. Both appear in context questions (conversions, costs, areas) and in short algebra questions.

✏️ Start with the questions. Try each one on paper, then open the solution and compare.

Stuck? Open the notes ↓

✏️ Practice questions

Try each question on paper first. Then open the solution and compare every step.

Question 1●○○ Warm-up✍️ + 🖩 · Paper 1[4 marks]

Let

f(x) = 3x-2

and

g(x) = x^2+1

(a) Find $(f\circ g)(2)$ .[2]
(b) Find $(g\circ f)(2)$ .[2]

Show the full step-by-step solution

Step 1🧠 ThinkWhat does the little circle mean?

(f\circ g)(2)

means

f(g(2))

. Think of two machines. The number 2 goes into machine

g

first. Whatever comes out goes into machine

f

. Always start with the function closest to the 2.

Part (a)

Step 1✍️ By handInside first: g(2), then f of that

g(2) = 2^2+1 = 5

f(5) = 3\times5-2 = 13

Part (b)

Step 1✍️ By handNow the other order: f(2) first

f(2) = 3\times2-2 = 4

g(4) = 4^2+1 = 17

Step 2🖩 On the GDCCheck on the calculator by storing both functions

Define f(x)=3x−2›enter💻 Enter›Define g(x)=x^2+1›enter💻 Enter›f(g(2))›enter💻 Enter›g(f(2))›enter💻 Enter

Type Define (or pick it from the catalog), then the function. The calculator answers Done. After that you can type

f(g(2))

exactly as it is written.

Same answers: 13 and 17

✅ Check it The two answers are different (13 and 17). That is normal: the order of a composite matters.

Final answer (a)

13

(b)

17

Question 2●○○ Warm-up✍️ By hand · Paper 1[4 marks]

Let

f(x) = 2x+5

and

g(x) = x-3

(a) Find an expression for $(f\circ g)(x)$ .[2]
(b) Find an expression for $(g\circ f)(x)$ .[2]

Show the full step-by-step solution

Step 1🧠 ThinkPut a whole bracket into the other function

This time there is no number. We put the whole expression

g(x)

wherever we see

x

f

. Use a bracket so you do not forget to multiply everything.

Part (a)

Step 1✍️ By handReplace x in f by (x − 3)

(f\circ g)(x) = f(x-3) = 2(x-3)+5

= 2x-6+5 = 2x-1

Part (b)

Step 1✍️ By handReplace x in g by (2x + 5)

(g\circ f)(x) = g(2x+5) = (2x+5)-3

= 2x+2

✅ Check it Test with

x=4

g(4)=1

f(1)=7

, and

2(4)-1=7

. ✔

f(4)=13

g(13)=10

, and

2(4)+2=10

. ✔

Final answer (a)

(f\circ g)(x) = 2x-1

(b)

(g\circ f)(x) = 2x+2

⚠️ Common mistake Writing

2x-3+5

: the bracket was forgotten, so only the

x

was multiplied by 2, not the −3.

Question 3●○○ Warm-up✍️ By hand · Paper 1[4 marks]

Let

f(x) = 5x-7

(a) Find $f^{-1}(x)$ .[2]
(b) Hence find $f^{-1}(8)$ .[1]
(c) Verify that $f\left(f^{-1}(8)\right) = 8$ .[1]

Show the full step-by-step solution

Step 1🧠 ThinkWhat is an inverse?

f

takes a number, multiplies by 5, then subtracts 7. The inverse must undo this, in the reverse order: first add 7, then divide by 5. The swap method does this for us automatically.

Part (a)

Step 1✍️ By handWrite y = f(x), swap x and y, make y the subject

y = 5x-7

Swap:

x = 5y-7

x+7 = 5y

f^{-1}(x) = \dfrac{x+7}{5}

Part (b)

Step 1✍️ By handSubstitute 8

f^{-1}(8) = \dfrac{8+7}{5} = \dfrac{15}{5} = 3

Part (c)

Step 1✍️ By handPut the answer back into f

f(3) = 5\times3-7 = 8

✔A1

💡 Remember

f

and

f^{-1}

cancel each other out. That is why

f(f^{-1}(8))

gives back 8.

Final answer (a)

f^{-1}(x) = \dfrac{x+7}{5}

(b)

3

(c)

f(3)=8

✔

Question 4●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

Let

f(x) = 2e^{x}-3

x\in\mathbb{R}

(a) Write down the range of $f$ .[1]
(b) Find $f^{-1}(x)$ , and state its domain.[3]
(c) Find the exact value of $f^{-1}(5)$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkWhat values can 2eˣ − 3 take?

e^{x}

is always positive, but it can get as close to 0 as we like. So

2e^{x}

is always above 0, and

2e^{x}-3

is always above

-3

. The graph has a horizontal asymptote

y=-3

Step 2✍️ By handRange

f(x) \gt -3

Part (b)

Step 1🧠 ThinkUndo an exponential with ln

\ln

is the inverse of

e^{x}

\ln(e^{y}) = y

. So after swapping, we get

e^{y}

by itself and then take

\ln

of both sides.

Step 2✍️ By handSwap and rearrange

y = 2e^{x}-3

Swap:

x = 2e^{y}-3

x+3 = 2e^{y}

e^{y} = \dfrac{x+3}{2}

f^{-1}(x) = \ln\left(\dfrac{x+3}{2}\right)

Domain:

x \gt -3

(the range of

f

)A1

Step 3🖩 On the GDCSee the reflection in y = x

f1(x)=2e^(x)−3›enter💻 Enter›f2(x)=ln((x+3)/2)›enter💻 Enter›f3(x)=x›enter💻 Enter

📍 Where to find it on the TI-Nspire

⌨️ ln: ctrl then eˣ on the keypad, or type ln(

On a Graphs page enter all three. The blue and red graphs are mirror images in the green line

y=x

. The red graph only exists for

x\gt -3

Part (c)

Step 1✍️ By handSubstitute 5

f^{-1}(5) = \ln\left(\dfrac{5+3}{2}\right)

= \ln 4

\;(\approx 1.39)

✅ Check it

f(\ln 4) = 2e^{\ln 4}-3 = 2\times4-3 = 5

. ✔

Final answer (a)

f(x) \gt -3

(b)

f^{-1}(x) = \ln\left(\dfrac{x+3}{2}\right)

x \gt -3

(c)

\ln 4

Question 5●●○ Standard✍️ + 🖩 · Paper 1[6 marks]

Let

f(x) = (x-2)^2+1

, for

x \ge 2

(a) Explain why the domain of $f$ has been restricted to $x\ge2$ .[1]
(b) Find $f^{-1}(x)$ and state its domain.[3]
(c) Find the coordinates of the point where the graphs of $y=f(x)$ and $y=f^{-1}(x)$ meet.[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkOne-to-one

Without a restriction,

f(1) = 2

and

f(3) = 2

. Two inputs give the same output, so the inverse would not know whether to send 2 back to 1 or to 3. Keeping only

x\ge2

(the right half of the parabola) makes each output come from just one input.

Step 2✍️ By hand

So that

f

is one-to-one, and therefore has an inverse.R1

Part (b)

Step 1✍️ By handSwap and rearrange (choose the + root)

The outputs of

f^{-1}

must be the inputs of

f

, which are

\ge 2

. So we need

+\sqrt{\ }

Swap:

x = (y-2)^2+1

x-1 = (y-2)^2

y-2 = \sqrt{x-1}

f^{-1}(x) = 2+\sqrt{x-1}

Domain

x\ge1

(range of

f

, since the vertex is

(2,1)

)A1

Part (c)

Step 1🧠 ThinkWhere do a graph and its mirror image meet?

Here they meet on the mirror line

y=x

. So we solve

f(x) = x

, which is easier than

f(x)=f^{-1}(x)

Step 2✍️ By handSolve f(x) = x

(x-2)^2+1 = x \;\Rightarrow\; x^2-5x+5 = 0

x = \dfrac{5\pm\sqrt{5}}{2}

; only

x = \dfrac{5+\sqrt{5}}{2} = 3.618\ldots

\ge 2

Point

(3.62,\ 3.62)

Step 3🖩 On the GDCSee it on a Graphs page

f1(x)=(x−2)^2+1 | x≥2›enter💻 Enter›f2(x)=2+√(x−1)›enter💻 Enter›f3(x)=x›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

📍 Where to find it on the TI-Nspire

⌨️ ≤ and ≥: type <= or >=; the TI turns them into ≤ and ≥

⌨️ the “such that” bar |: type | (Shift + \ on the keyboard)

⌨️ square root: ctrl then x² on the keypad, or type sqrt(

The “|” (such that) symbol restricts the domain; find it with ctrl =. Then use Intersection on f1 and f3.

Final answer (a) So that

f

is one-to-one. (b)

f^{-1}(x) = 2+\sqrt{x-1}

x\ge1

(c)

(3.62,\ 3.62)

— exactly

\left(\dfrac{5+\sqrt5}{2},\ \dfrac{5+\sqrt5}{2}\right)

⚠️ Common mistake Writing

f^{-1}(x) = 2\pm\sqrt{x-1}

. The

-

root gives values below 2, which are not in the domain of

f

Question 6●●○ Standard🖩 GDC · Paper 2[3 marks]

Let

f(x) = x^3+2x-1

. The function

f

is one-to-one, so

f^{-1}

exists. Find

f^{-1}(5)

Show the full step-by-step solution

Step 1🧠 ThinkYou do not need a formula for the inverse

Rearranging

y = x^3+2x-1

for

x

is too hard. But

f^{-1}(5)

simply asks: which input gives an output of 5? So we solve

f(x) = 5

on the GDC.

Step 2✍️ By handWrite the equation

f^{-1}(5) = a \iff f(a) = 5 \iff a^3+2a-1 = 5

Step 3🖩 On the GDCSolve with nSolve on a Calculator page

nSolve(x^3+2x−1=5,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

nSolve(is inmenu💻 click menu on keypad›3: Algebra›1: Numerical Solve

nSolve is in 3: Algebra › 1: Numerical Solve, or just type it.

x = 1.45616…

Step 4🖩 On the GDCOr: intersection with y = 5 on a Graphs page

f1(x)=x^3+2x−1›enter💻 Enter›f2(x)=5›enter💻 Enter›menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Final answer

f^{-1}(5) \approx 1.46

⚠️ Common mistake Calculating

f(5) = 134

. That is the function, not its inverse.

Question 7●●○ Standard✍️ By hand · Paper 1[4 marks]

Let

f(x) = \dfrac{4}{x}

x\neq0

(a) Find $(f\circ f)(x)$ .[2]
(b) Hence write down $f^{-1}(x)$ , and describe the symmetry of the graph of $y=f(x)$ .[2]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkPut f inside itself

(f\circ f)(x) = f(f(x))

. Wherever

f

has an

x

, we write

\dfrac{4}{x}

Step 2✍️ By handSimplify a fraction over a fraction

f\left(\dfrac{4}{x}\right) = \dfrac{4}{\;\dfrac{4}{x}\;}

= 4\times\dfrac{x}{4}

= x

Part (b)

Step 1🧠 ThinkDoing f twice gets you back to the start

f(f(x)) = x

means

f

undoes itself. So the inverse of

f

f

itself — we call it self-inverse. Because

f^{-1}

is the reflection of

f

y=x

, the graph must be its own mirror image in

y=x

Step 2✍️ By hand

f^{-1}(x) = \dfrac{4}{x}

x\neq0

The graph is symmetrical about the line

y=x

.A1

✅ Check it

f(1) = 4

and

f(4) = 1

: the points

(1,4)

and

(4,1)

are mirror images in

y=x

. ✔

Final answer (a)

(f\circ f)(x) = x

(b)

f^{-1}(x) = \dfrac{4}{x}

(self-inverse); the graph is symmetrical in

y=x

Question 8●●○ Standard✍️ By hand · Paper 1[6 marks]

Kagiso changes Botswana pula into South African rand at a bureau in Gaborone. The bureau converts

p

pula into rand using

R(p) = 1.32p

. It then takes a fixed fee of 50 rand, using

F(r) = r-50

, where

r

is an amount in rand.

(a) Find $(F\circ R)(p)$ and explain what it represents.[2]
(b) Find the number of rand Kagiso receives for P 2 000.[1]
(c) Find $(F\circ R)^{-1}(x)$ .[2]
(d) Hence find how many pula Kagiso must change to receive exactly 3 000 rand.[1]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkFollow the money

First the pula is converted (

R

), then the fee is taken from the rand (

F

). “

R

first, then

F

” is written

F\circ R

Step 2✍️ By hand

(F\circ R)(p) = F(1.32p) = 1.32p-50

The number of rand received, after the fee, for

p

pula.A1

Part (b)

Step 1✍️ By hand

1.32\times2000-50 = 2590

randA1

Part (c)

Step 1🧠 ThinkThe inverse works backwards: rand → pula

Let

h(p) = 1.32p-50

. Its inverse takes the rand received and tells you the pula you needed.

Step 2✍️ By hand

Swap:

x = 1.32y-50

x+50 = 1.32y

(F\circ R)^{-1}(x) = \dfrac{x+50}{1.32}

Part (d)

Step 1✍️ By hand

\dfrac{3000+50}{1.32} = 2310.606\ldots \approx \text{P }2\,310.61

✅ Check it

1.32\times2310.61-50 = 3000.0

rand. ✔

Final answer (a)

1.32p-50

: rand received after the fee for

p

pula. (b) R 2 590 (c)

\dfrac{x+50}{1.32}

(d)

\text{P }2\,310.61

Question 9●●● Exam-level✍️ + 🖩 · Paper 2[7 marks]

A circular patch of algae is growing on the surface of Gaborone Dam. Its radius, in metres,

t

days after it is first seen is

r(t) = 2+0.5t

. The area of a circle of radius

r

A(r) = \pi r^2

(a) Find $(A\circ r)(t)$ .[2]
(b) Find the area of the patch after 10 days.[2]
(c) Find the time it takes for the area to reach $300\text{ m}^2$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTime → radius → area

r

turns time into a radius.

A

turns a radius into an area. Feeding

r(t)

into

A

gives area directly from time.

Step 2✍️ By hand

(A\circ r)(t) = A(2+0.5t)

= \pi(2+0.5t)^2

Part (b)

Step 1✍️ By hand

r(10) = 2+5 = 7

A = \pi\times7^2 = 49\pi = 153.938\ldots \approx 154\text{ m}^2

Part (c)

Step 1🧠 ThinkNow we know the area and want the time

Solve

\pi(2+0.5t)^2 = 300

. You can do it by hand by undoing each step, or let the GDC solve it.

Step 2✍️ By handBy hand: undo each step

(2+0.5t)^2 = \dfrac{300}{\pi}

2+0.5t = \sqrt{\dfrac{300}{\pi}} = 9.77205\ldots

t = \dfrac{9.77205-2}{0.5} = 15.544\ldots

Step 3🖩 On the GDCOr on the GDC

nSolve(π(2+0.5t)^2=300,t)›enter💻 Enter

📍 Where to find it on the TI-Nspire

nSolve(is inmenu💻 click menu on keypad›3: Algebra›1: Numerical Solve

Step 4✍️ By hand

t \approx 15.5

daysA1

Final answer (a)

\pi(2+0.5t)^2

(b)

154\text{ m}^2

(c)

15.5

days

Question 10●●● Exam-level✍️ + 🖩 · Paper 1[8 marks]

Let

f(x) = x^2-4x+7

, for

x\ge2

, and

g(x) = x+1

x\in\mathbb{R}

(a) Show that $f(x)$ can be written as $(x-2)^2+3$ .[1]
(b) Hence write down the range of $f$ .[1]
(c) Find $f^{-1}(x)$ , stating its domain.[3]
(d) Find $(f^{-1}\circ g)(10)$ , giving your answer in the form $a+b\sqrt{2}$ .[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By handExpand the given form (this is the safest way to “show”)

(x-2)^2+3 = x^2-4x+4+3 = x^2-4x+7

Part (b)

Step 1🧠 ThinkSmallest value

(x-2)^2

is never negative, and it is 0 when

x=2

. So the smallest value of

f

0+3=3

, and it grows from there.

Step 2✍️ By hand

f(x)\ge3

Part (c)

Step 1✍️ By hand

Swap:

x = (y-2)^2+3

y-2 = \sqrt{x-3}

(+ root, since

y\ge2

)M1

f^{-1}(x) = 2+\sqrt{x-3}

Domain

x\ge3

Part (d)

Step 1🧠 Thinkg first, then the inverse

(f^{-1}\circ g)(10) = f^{-1}(g(10))

. Work out

g(10)

first.

Step 2✍️ By hand

g(10) = 11

f^{-1}(11) = 2+\sqrt{11-3} = 2+\sqrt{8}

= 2+2\sqrt{2}

\;(\approx 4.83)

Step 3🖩 On the GDCCheck the decimal value

2 + √8 = 2 + 2√2 ✔

💡 Simplifying surds

\sqrt{8} = \sqrt{4\times2} = 2\sqrt2

Final answer (a) shown (b)

f(x)\ge3

(c)

f^{-1}(x) = 2+\sqrt{x-3}

x\ge3

(d)

2+2\sqrt2

Question 11●●● Exam-level✍️ By hand · Paper 2[8 marks]

A taxi company in Gaborone charges a fare of

F(d) = 8+4.5d

pula for a journey of

d

km. During a promotion, customers get 10% off,

D(x) = 0.9x

, and also a P 5 voucher,

S(x) = x-5

, where

x

is an amount in pula.

(a) Find the fare for a 12 km journey with no promotion.[1]
(b) Find $(S\circ D)(62)$ and $(D\circ S)(62)$ , and explain in words what each one means.[4]
(c) The company always applies the cheaper order. Mpho has P 80. Find the greatest distance she can travel.[3]

Show the full step-by-step solution

Part (a)

Step 1✍️ By hand

F(12) = 8+4.5\times12 = \text{P }62

Part (b)

Step 1🧠 ThinkWhich comes first: the discount or the voucher?

(S\circ D)(62)

: the 10% discount

D

is used first, then the voucher

S

(D\circ S)(62)

: the voucher first, then 10% off what is left.

Step 2✍️ By hand

(S\circ D)(62) = S(55.8) = 55.8-5 = \text{P }50.80

(D\circ S)(62) = D(57) = 0.9\times57 = \text{P }51.30

S\circ D

: 10% off first, then P 5 off.R1

D\circ S

: P 5 off first, then 10% off.R1

💡 Remember Taking 10% off a bigger number saves more money — so discount first is cheaper. Same pieces, different order, different answer.

Part (c)

Step 1🧠 ThinkWrite the whole cost as one function, then work backwards

The cheaper order is

S\circ D\circ F

: fare, then 10% off, then P 5 off. Set this equal to 80 and solve for

d

Step 2✍️ By hand

0.9(8+4.5d)-5 = 80

0.9(8+4.5d) = 85 \;\Rightarrow\; 8+4.5d = 94.444\ldots

d = \dfrac{94.444\ldots-8}{4.5} = 19.209\ldots

Greatest distance

\approx 19.2

kmA1

✅ Check it

F(19.2) = 94.4

;

0.9\times94.4 = 84.96

;

84.96-5 = 79.96 \le 80

. ✔

Final answer (a) P 62 (b)

(S\circ D)(62) = \text{P }50.80

(discount then voucher);

(D\circ S)(62) = \text{P }51.30

(voucher then discount) (c)

19.2

Question 12●●● Exam-level✍️ By hand · Paper 1[6 marks]

The function

f

is defined by

f(x) = ax+b

. It is known that

f(1) = 5

and

f^{-1}(11) = 3

(a) Find the value of $a$ and the value of $b$ .[3]
(b) Let $g(x) = x^2$ . Solve $(g\circ f)(x) = 64$ .[3]

Show the full step-by-step solution

Part (a)

Step 1🧠 ThinkTurn the inverse fact into a normal fact

f^{-1}(11) = 3

means “the input that gives 11 is 3”. So

f(3) = 11

. Now we have two ordinary equations.

Step 2✍️ By hand

f(3) = 11

a+b = 5

and

3a+b = 11

Subtract:

2a = 6

a = 3,\ b = 2

A1A1

Part (b)

Step 1🧠 Thinkf first, then square

(g\circ f)(x) = g(3x+2) = (3x+2)^2

. A square equal to 64 means the bracket is 8 or −8.

Step 2✍️ By hand

(3x+2)^2 = 64

3x+2 = 8

3x+2 = -8

x = 2

x = -\dfrac{10}{3}

A1A1

✅ Check it

f(2) = 8,\ 8^2 = 64

✔

f\left(-\dfrac{10}{3}\right) = -8,\ (-8)^2 = 64

✔

Final answer (a)

a=3,\ b=2

(b)

x = 2

x = -\dfrac{10}{3}

⚠️ Common mistake Forgetting the negative square root and losing the second solution.

🆘 Stuck on a question? Open the notes, GDC toolkit and common mistakes

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

🎯 What you must know

Composite: $(f\circ g)(x) = f(g(x))$ . Work from the inside out: do $g$ first, then put the answer into $f$ .
Order matters: usually $(f\circ g)(x) \neq (g\circ f)(x)$ .
Inverse $f^{-1}$ undoes $f$ : if $f(a)=b$ then $f^{-1}(b)=a$ . So $(f\circ f^{-1})(x) = (f^{-1}\circ f)(x) = x$ , the identity function.
To find $f^{-1}(x)$ : write $y=f(x)$ , swap $x$ and $y$ , make $y$ the subject. (Or make $x$ the subject first, then swap.)
Only a one-to-one function has an inverse. A function like $x^2$ must have its domain restricted (e.g. $x\ge0$ ) first.
Domain of $f^{-1}$ = range of $f$ ; range of $f^{-1}$ = domain of $f$ .
The graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$ .

Composite function

(f\circ g)(x) = f(g(x))

🧠 Learn this one

Inverse and identity

(f\circ f^{-1})(x) = (f^{-1}\circ f)(x) = x

🧠 Learn this one

🖩 Your GDC toolkit for this subtopic

Follow the steps on your own TI-Nspire CX II. The screens show what you should see.

Store a function on a Calculator page (then use f(g(2)) directly)

Define f(x)=3x−2›enter💻 Enter

Type Define f(x)=3x−2.
Press enter.

Find

f^{-1}(k)

without the formula: solve

f(x)=k

nSolve(f(x)=5,x)›enter💻 Enter

📍 Where to find it on the TI-Nspire

nSolve(is inmenu💻 click menu on keypad›3: Algebra›1: Numerical Solve

Type nSolve(f(x)=5,x).
Press enter.

x = 1.45616…

Worked example: Question 6 — Solve with nSolve on a Calculator page

nSolve is in 3: Algebra › 1: Numerical Solve, or just type it.

Graph

f

f^{-1}

and

y=x

to see the reflection (Graphs page)

f1(x)=…›enter💻 Enter›f3(x)=x›enter💻 Enter

Type f1(x)=….
Press enter.
Type f3(x)=x.
Press enter.

Worked example: Question 4 — See the reflection in y = x

On a Graphs page enter all three. The blue and red graphs are mirror images in the green line

y=x

. The red graph only exists for

x\gt -3

Find where two graphs meet

menu💻 click menu on keypad›6: Analyze Graph›4: Intersection

Start on a Graphs page (ctrl doc adds one).
Press menu.
Choose 6: Analyze Graph.
Choose 4: Intersection.
Move the pointer to the LEFT of the crossing point and press click, then to the RIGHT and press click. The intersection point and its coordinates appear.

The menu path

Worked example: Question 5 — See it on a Graphs page

The “|” (such that) symbol restricts the domain; find it with ctrl =. Then use Intersection on f1 and f3.

⚠️ Where students lose marks

Doing the composite in the wrong order: $f(g(x))$ means $g$ first, even though $f$ is written first.
Confusing $f^{-1}(x)$ with $\dfrac{1}{f(x)}$ . The −1 means “inverse”, not “one over”.
Forgetting to state the domain of $f^{-1}$ — it is the range of $f$ , and it is often worth a mark.
Using the wrong sign of a square root: if the domain of $f$ is $x\ge2$ , the inverse must give values $\ge2$ , so take the + root.
Writing $f^{-1}(x)$ in terms of $y$ . The final answer must be in terms of $x$ .

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · AI 2.7 — Composite and inverse functions

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

2.7 Composite and inverse functions

✏️ Practice questions

🎯 What you must know

🖩 Your GDC toolkit for this subtopic

⚠️ Where students lose marks