MathsISC Class 12 · Relations and Functions

Composite and Invertible Functions: fog, gof and the Inverse f⁻¹

Updated 15 October 2026 · Fact-checked

The composite gof applies f first and then g, so (gof)(x) = g(f(x)); it needs the range of f inside the domain of g. A function is invertible exactly when it is one-one and onto. To find the inverse, solve y = f(x) for x and then rename y as x.

f acts first, then g, so g∘f jumps from A to C; the inverses undo the steps in reverse order, which works only if f is one-one and onto.

Illustration drawn for QuestionBuddy.

Understand Composite and Invertible Functions

Composite functions. If $f : A \to B$ and $g : B \to C$ , the composite $g \circ f : A \to C$ is defined by $(g \circ f) (x) = g (f (x))$ . Read it from the inside: $f$ acts first, then $g$ . It exists only when every output of $f$ is an allowed input of $g$ , that is, the range of $f$ lies inside the domain of $g$ .

Order matters. With $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , $(g \circ f) (x) = (2 x + 1)^{2}$ but $(f \circ g) (x) = 2 x^{2} + 1$ . So in general $g \circ f \neq = f \circ g$ . Composition is, however, associative: $h \circ (g \circ f) = (h \circ g) \circ f$ . If $f$ and $g$ are both one-one, so is $g \circ f$ ; if both are onto, so is $g \circ f$ .

Invertible functions. $f : A \to B$ is invertible if there is a function $g : B \to A$ with $g \circ f = I_{A}$ and $f \circ g = I_{B}$ , where $I$ is the identity function. Then $g$ is written $f^{- 1}$ , and it undoes $f$ : $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (y)) = y$ . The key theorem is: ** $f$ is invertible if and only if it is one-one and onto.** One-one guarantees that $f^{- 1}$ never has to choose between two answers; onto guarantees that $f^{- 1}$ is defined at every point of $B$ . The inverse, when it exists, is unique. For a composite, $(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$ , like taking off socks and shoes in the reverse order.

Finding the inverse. Put $y = f (x)$ , solve for $x$ in terms of $y$ , and you have $f^{- 1} (y)$ . Then replace $y$ by $x$ to write $f^{- 1} (x)$ , and state its domain, which is the co-domain of $f$ . The CISCE syllabus restricts this to algebraic functions, so the solving step is always algebra: cross-multiplying, completing the square, or taking a root whose sign is fixed by the domain.

Domain restrictions make functions invertible. $x^{2}$ is not invertible on $R$ , but on $[0, \infty)$ onto $[0, \infty)$ its inverse is $x$ . This is why ISC questions give carefully chosen domains such as $R - {\frac{2}{3}}$ and co-domains such as $R - {\frac{2}{3}}$ for rational functions.

Graphs. The point $(a, b)$ lies on the graph of $f$ exactly when $(b, a)$ lies on the graph of $f^{- 1}$ . So the graph of $f^{- 1}$ is the mirror image of the graph of $f$ in the line $y = x$ , and the domain and range swap. When the syllabus asks you to sketch a function and its inverse, draw $y = x$ as a dashed line first, plot two or three points of $f$ , reflect them, and join.

A function that is its own inverse satisfies $f (f (x)) = x$ ; its graph is symmetric about $y = x$ . Rational functions such as $\frac{a x + b}{c x - a}$ have this property, which ISC questions often ask you to prove.

Composite and Invertible Functions: formulas to remember

Composite function: $(g \circ f) (x) = g (f (x))$; f acts first. Needs Range(f) ⊆ Domain(g).
Order matters: $g \circ f \neq = f \circ g in general$; Always check by a numerical value if unsure.
Associativity: $h \circ (g \circ f) = (h \circ g) \circ f$; Brackets can be moved; order cannot.
Condition for an inverse: $f invertible ⟺ f one-one and onto$; Prove both before finding the inverse unless invertibility is given.
Defining property: $f^{- 1} (f (x)) = x, f (f^{- 1} (y)) = y$; Use it to check your inverse.
Inverse of a composite: $(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$; Reverse the order, like socks and shoes.
Graph of the inverse: $(a, b) \in f ⟺ (b, a) \in f^{- 1}$; Reflect the graph of f in the line y = x.

How to solve Composite and Invertible Functions questions

Most ISC questions here are "find fog/gof" (1 to 2 marks) or "show f is invertible and find f⁻¹" (2 to 5 marks). Use these steps.

1For a composite, write the outer function with a blank: $g () = ()^{2}$ . Put the whole inner expression into the blank and simplify.
2Check the composite is defined: is the range of the inner function inside the domain of the outer one?
3For an inverse, first confirm that $f$ is one-one and onto (or note that the question says it is invertible).
4Write $y = f (x)$ and solve for $x$ . Keep track of excluded values and of the sign of any square root, using the domain.
5Write $f^{- 1} (y) = \dots$ , then replace $y$ by $x$ to get $f^{- 1} (x)$ , and state its domain (the co-domain of $f$ ).
6Check one value: pick a simple $x$ , compute $f (x)$ , and confirm that $f^{- 1}$ takes it back to $x$ .
7For "show $f$ is its own inverse", compute $f (f (x))$ in one fraction and simplify to $x$ .

Composite and Invertible Functions in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Order of a composite function

Suppose an online shop applies a Rs 100 coupon, $f (x) = x - 100$ , and a tax of 18 per cent, $g (x) = 1.18 x$ , to a basket worth Rs $x$ (with $x > 100$ ).

Find $(g \circ f) (x)$ and $(f \circ g) (x)$ and say in words what each one means. For a basket of Rs 1000, which order is cheaper for the customer, and by how much?

Try it first, then show the solutionSolution

$(g \circ f) (x) = g (x - 100) = 1.18 (x - 100) = 1.18 x - 118$ : coupon first, then tax on the reduced price.
$(f \circ g) (x) = f (1.18 x) = 1.18 x - 100$ : tax first, then coupon off the taxed amount.
At $x = 1000$ : $g \circ f$ gives $1.18 \times 900 = 1062$ ; $f \circ g$ gives $1180 - 100 = 1080$ .
The difference is $1080 - 1062 = 18$ rupees, which is the tax on the Rs 100 coupon.

Answer: Coupon first ( $g \circ f$ ) costs Rs 1062; tax first ( $f \circ g$ ) costs Rs 1080. Coupon first is cheaper by Rs 18.

Memory hook: At the checkout, the order of coupon and tax changes the bill: gof is not fog.

Inverse of a function as decoding

Imagine two friends who hide a whole number $x$ in their messages by sending $f (x) = 7 x + 4$ . To make it harder to crack, they sometimes encode twice, sending $(f \circ f) (x)$ .

Find the decoding function $f^{- 1}$ and decode the number 151. Find $(f \circ f) (x)$ and decode a twice-encoded 1110.

Try it first, then show the solutionSolution

$y = 7 x + 4 \Rightarrow x = \frac{y - 4}{7}$ , so $f^{- 1} (x) = \frac{x - 4}{7}$ .
$f^{- 1} (151) = \frac{147}{7} = 21$ . Check: $7 \times 21 + 4 = 151$ .
$(f \circ f) (x) = 7 (7 x + 4) + 4 = 49 x + 32$ .
Decode 1110: $49 x + 32 = 1110 \Rightarrow 49 x = 1078 \Rightarrow x = 22$ . (Equivalently, apply $f^{- 1}$ twice: $f^{- 1} (1110) = 158$ , $f^{- 1} (158) = 22$ .)

Answer: $f^{- 1} (x) = \frac{x - 4}{7}$ ; 151 decodes to 21; $(f \circ f) (x) = 49 x + 32$ , and 1110 decodes to 22.

Memory hook: Encoding is f, decoding is f⁻¹: a code only works if it is a bijection.

Tricks to remember Composite and Invertible Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

Inside first
In g(f(x)), the function nearest x acts first. Read gof from right to left.
Use it when: Avoiding the common swap of fog and gof.
Swap and solve
Inverse = solve y = f(x) for x, then swap the letters. Domain and range swap too.
Use it when: Every "find f⁻¹(x)" question.
Socks and shoes
You put on socks then shoes, and take off shoes then socks: (gof)⁻¹ = f⁻¹ o g⁻¹.
Use it when: Inverse of a composite.
Mirror y = x
The graph of the inverse is the graph of f seen in a mirror placed along y = x.
Use it when: Sketching a function and its inverse.

Composite and Invertible Functions: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Inverse of a rational function · 2 marks · Short answerBoard level

The function $f : R - {\frac{1}{2}} \to R - {0}$ defined by $f (x) = \frac{3}{2 x - 1}$ is invertible. Find $f^{- 1} (x)$ .

Try it first, then show the solutionSolution

Let $y = \frac{3}{2 x - 1}$ , with $y \neq = 0$ . Then $2 x - 1 = \frac{3}{y}$ . (½ mark)
$2 x = \frac{3}{y} + 1 = \frac{3 + y}{y} \Rightarrow x = \frac{y + 3}{2 y}$ . (½ mark)
So $f^{- 1} (y) = \frac{y + 3}{2 y}$ . Replacing $y$ by $x$ : $f^{- 1} (x) = \frac{x + 3}{2 x}$ , $x \neq = 0$ . (1 mark)
Check: $f (1) = 3$ and $f^{- 1} (3) = \frac{6}{6} = 1$ .

Answer: $f^{- 1} (x) = \frac{x + 3}{2 x}$ , $x \in R - {0}$ . A question of this type was asked in ISC 2024; the examiners noted that candidates left the answer as $f^{- 1} (y)$ instead of replacing $y$ by $x$ .

Example 2: Composites in both orders · 1 mark · MCQBoard level

If $f (x) = 2 x + 1$ and $g (x) = x^{2}$ for $x \in R$ , then $(g \circ f) (2) - (f \circ g) (2)$ equals:
(a) 0
(b) 16
(c) 25
(d) 34

Try it first, then show the solutionSolution

$(g \circ f) (2) = g (5) = 25$ .
$(f \circ g) (2) = f (4) = 9$ .
$25 - 9 = 16$ .

Answer: (b) 16.

Example 3: A function that is its own inverse · 3 marks · Short answerHarder board level

Let $f : R - {\frac{2}{3}} \to R - {\frac{2}{3}}$ be $f (x) = \frac{4 x + 3}{6 x - 4}$ . Show that $(f \circ f) (x) = x$ and hence find $f^{- 1}$ .

Try it first, then show the solutionSolution

$(f \circ f) (x) = \frac{4 ( \frac{4 x + 3}{6 x - 4} ) + 3}{6 ( \frac{4 x + 3}{6 x - 4} ) - 4}$ . (½ mark)
Multiply numerator and denominator by $6 x - 4$ : numerator $= 16 x + 12 + 18 x - 12 = 34 x$ . (1 mark)
Denominator $= 24 x + 18 - 24 x + 16 = 34$ . So $(f \circ f) (x) = \frac{34 x}{34} = x$ . (1 mark)
Since $f \circ f = I$ , $f$ is invertible and $f^{- 1} = f$ . (½ mark)

Answer: $(f \circ f) (x) = x$ , so $f^{- 1} (x) = \frac{4 x + 3}{6 x - 4}$ .

Example 4: Restricting the domain to get an inverse · 3 marks · Short answerHarder board level

Show that $f : [0, \infty) \to [4, \infty)$ , $f (x) = x^{2} + 4$ , is invertible and find $f^{- 1}$ .

Try it first, then show the solutionSolution

One-one: $x_{1}^{2} + 4 = x_{2}^{2} + 4 \Rightarrow x_{1}^{2} = x_{2}^{2} \Rightarrow x_{1} = x_{2}$ , because both are non-negative. (1 mark)
Onto: for $y \geq 4$ , $x = y - 4$ is real and non-negative, and $f (x) = (y - 4) + 4 = y$ . (1 mark)
Hence $f$ is invertible with $f^{- 1} (y) = y - 4$ , that is, $f^{- 1} (x) = x - 4$ for $x \geq 4$ . (1 mark)

Answer: $f^{- 1} (x) = x - 4$ , $x \in [4, \infty)$ . The positive root is chosen because the domain of $f$ is $[0, \infty)$ .

Example 5: Showing gof and fog differ · 2 marks · Short answerHarder board level

For $f (x) = ∣ x ∣$ and $g (x) = 2 x - 3$ on $R$ , find $(g \circ f) (x)$ and $(f \circ g) (x)$ , and show they are not equal.

Try it first, then show the solutionSolution

$(g \circ f) (x) = g (∣ x ∣) = 2∣ x ∣ - 3$ ; $(f \circ g) (x) = f (2 x - 3) = ∣2 x - 3∣$ . (1 mark)
At $x = - 1$ : $(g \circ f) (- 1) = 2 - 3 = - 1$ but $(f \circ g) (- 1) = ∣ - 5∣ = 5$ . So $g \circ f \neq = f \circ g$ . (1 mark)

Answer: $(g \circ f) (x) = 2∣ x ∣ - 3$ , $(f \circ g) (x) = ∣2 x - 3∣$ ; they differ, for example at $x = - 1$ .

Example 6: Inverse by completing the square · 5 marks · Long answerHOTS / competency

Let $f : N \to S$ , $f (x) = 4 x^{2} + 12 x + 15$ , where $S$ is the range of $f$ . Show that $f$ is invertible, find $f^{- 1}$ , and use it to find the natural numbers whose images are 31 and 87.

Try it first, then show the solutionSolution

Complete the square: $f (x) = (2 x + 3)^{2} + 6$ . (1 mark)
One-one: $(2 x_{1} + 3)^{2} = (2 x_{2} + 3)^{2}$ . Both $2 x_{1} + 3$ and $2 x_{2} + 3$ are positive for natural numbers, so $2 x_{1} + 3 = 2 x_{2} + 3$ and $x_{1} = x_{2}$ . (1 mark)
Onto: $S$ is defined as the range, so every element of $S$ is an image. Hence $f : N \to S$ is bijective and invertible. (1 mark)
Inverse: $y = (2 x + 3)^{2} + 6 \Rightarrow 2 x + 3 = y - 6$ (positive root) $\Rightarrow x = \frac{y - 6 - 3}{2}$ . So $f^{- 1} (y) = \frac{y - 6 - 3}{2}$ , $y \in S$ . (1 mark)
$f^{- 1} (31) = \frac{5 - 3}{2} = 1$ and $f^{- 1} (87) = \frac{9 - 3}{2} = 3$ . Check: $f (1) = 31$ , $f (3) = 36 + 36 + 15 = 87$ . (1 mark)

Answer: $f^{- 1} (y) = \frac{y - 6 - 3}{2}$ ; the pre-images of 31 and 87 are 1 and 3.

Where students lose marks in Composite and Invertible Functions

The slips examiners see most often, and what to write instead.

Mistake: Leaving the answer as f⁻¹(y)
The question asks for f⁻¹(x); stopping at f⁻¹(y) leaves the last step undone.
Wrong working
$x = \frac{y + 3}{2 y}$ , so $f^{- 1} (y) = \frac{y + 3}{2 y}$ .
Right working
$f^{- 1} (y) = \frac{y + 3}{2 y}$ , hence $f^{- 1} (x) = \frac{x + 3}{2 x}$ , $x \neq = 0$ .
Do this: Always finish by replacing y with x and stating the domain.
Usually costs: 1 mark (noted by the ISC 2024 examiners).
Mistake: Composing in the wrong order
gof means f first; students often substitute g into f instead.
Wrong working
$(g \circ f) (x) = f (g (x)) = 2 x^{2} + 1$ .
Right working
$(g \circ f) (x) = g (f (x)) = (2 x + 1)^{2}$ .
Do this: Write $g (f (x))$ in words: "g of (f of x)", and fill the inner bracket first.
Usually costs: The whole part.
Mistake: Taking both signs of a square root
An inverse must be a function, so only one root can be kept; the domain decides which.
Do this: Write "positive root, since x ≥ 0" (or the domain condition) beside the step.
Usually costs: 1 mark.
Mistake: Finding an inverse without checking invertibility
If f is not one-one and onto, the algebra can still produce a formula that is not an inverse.
Do this: Unless the question states that f is invertible, prove one-one and onto first.
Usually costs: 1 to 2 marks in a 3- or 5-mark question.

Exam tips

State the domain of $f^{- 1}$ ; it is the co-domain of $f$ , and it often carries the last half-mark.
In "show that $f \circ f = I$ " questions, multiply the top and bottom of the big fraction by the inner denominator at once; it keeps the algebra to two lines.
If a question asks for $(g \circ f)^{- 1}$ , you may either invert the composite directly or use $f^{- 1} \circ g^{- 1}$ ; check one numerical value either way.
For a sketch of $f$ and $f^{- 1}$ , label the line $y = x$ and mark at least two pairs of reflected points.
Keep inverse functions and reciprocals apart: $f^{- 1} (x)$ is not $\frac{1}{f ( x )}$ .

Practice questions on Composite and Invertible Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $f : R \to R$ is $f (x) = 5 x - 7$ , then $f^{- 1} (x)$ is:
(a) $\frac{x - 7}{5}$
(b) $\frac{x + 7}{5}$
(c) $5 x + 7$
(d) $\frac{1}{5 x - 7}$
Hint
Solve $y = 5 x - 7$ for $x$ .
Answer
(b) $\frac{x + 7}{5}$ .
Q2 · 1 mark · Very short answerBoard level
If $f (x) = x + 2$ and $g (x) = x^{2}$ , find $(g \circ f) (3)$ and $(f \circ g) (3)$ .
Hint
Work from the inside out.
Answer
$(g \circ f) (3) = g (5) = 25$ ; $(f \circ g) (3) = f (9) = 11$ .
Q3 · 3 marks · Short answerHarder board level
Show that $f : R - {- 1} \to R - {1}$ , $f (x) = \frac{x}{x + 1}$ , is invertible and find $f^{- 1} (x)$ .
Hint
Show one-one by cross-multiplying; for onto, solve for $x$ and check it is not $- 1$ .
Answer
One-one: $\frac{x _{1}}{x _{1} + 1} = \frac{x _{2}}{x _{2} + 1} \Rightarrow x_{1} = x_{2}$ . Onto: $y = \frac{x}{x + 1} \Rightarrow x = \frac{y}{1 - y}$ for $y \neq = 1$ , and this is never $- 1$ (that would need $y = - 1 + y$ ). So $f^{- 1} (x) = \frac{x}{1 - x}$ , $x \neq = 1$ .
Q4 · 3 marks · Short answerHarder board level
Let $f (x) = \frac{3 x + 2}{5 x - 3}$ , $x \neq = \frac{3}{5}$ . Show that $(f \circ f) (x) = x$ for all $x \neq = \frac{3}{5}$ . What is $f^{- 1}$ ?
Hint
Clear the inner denominator $5 x - 3$ .
Answer
Numerator: $3 (3 x + 2) + 2 (5 x - 3) = 19 x$ . Denominator: $5 (3 x + 2) - 3 (5 x - 3) = 19$ . So $(f \circ f) (x) = x$ and $f^{- 1} = f$ .
Q5 · 2 marks · Short answerHarder board level
The graph of $f (x) = 2 x + 4$ meets the graph of its inverse at a point. Find the point.
Hint
For a strictly increasing function, the graphs of f and f⁻¹ can only meet on the line y = x.
Answer
$f^{- 1} (x) = \frac{x - 4}{2}$ . On $y = x$ : $2 x + 4 = x \Rightarrow x = - 4$ . Check: $f^{- 1} (- 4) = - 4$ . The point is $(- 4, - 4)$ .
Q6 · 3 marks · Short answerHOTS / competency
Find all linear functions $f (x) = a x + b$ such that $(f \circ f) (x) = 4 x + 9$ for all real $x$ .
Hint
Expand $a (a x + b) + b$ and compare coefficients.
Answer
$(f \circ f) (x) = a^{2} x + ab + b$ . So $a^{2} = 4$ and $b (a + 1) = 9$ . $a = 2$ gives $b = 3$ ; $a = - 2$ gives $b = - 9$ . So $f (x) = 2 x + 3$ or $f (x) = - 2 x - 9$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a cab company charges $f (d) = 25 + 12 d$ rupees for a ride of $d$ km, and during rain applies a surge $g (F) = 1.5 F$ to the fare $F$ .
(i) Find $(g \circ f) (d)$ . (1 mark)
(ii) Find the rainy-day fare for a 10 km ride. (1 mark)
(iii) Find $(g \circ f)^{- 1}$ and hence the distance of a rainy-day ride that costs Rs 307.50. (1 mark)
(iv) Find $(f \circ g) (d) = 25 + 18 d$ and explain why it does not describe the surge correctly. (1 mark)
Hint
The surge multiplies the whole fare, so it must act last.
Answer
(i) $1.5 (25 + 12 d) = 37.5 + 18 d$ . (ii) $37.5 + 180 =$ Rs 217.50. (iii) $(g \circ f)^{- 1} (y) = \frac{y - 37.5}{18}$ ; $\frac{270}{18} = 15$ km. (iv) $f \circ g$ multiplies the distance by 1.5 (in effect raising only the per-km rate to Rs 18) and leaves the Rs 25 base charge unsurged, so it gives Rs 205 for 10 km instead of Rs 217.50.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does $(g \circ f) (x)$ mean?Show answer
$g (f (x))$ : apply $f$ first, then $g$ .
2.State the condition for $f : A \to B$ to be invertible.Show answer
$f$ must be one-one and onto (bijective).
3.Find $f^{- 1} (x)$ for $f (x) = \frac{x - 2}{3}$ .Show answer
$f^{- 1} (x) = 3 x + 2$ .
4.Write $(g \circ f)^{- 1}$ in terms of $f^{- 1}$ and $g^{- 1}$ .Show answer
$(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}$ .
5.How is the graph of $f^{- 1}$ related to that of $f$ ?Show answer
It is the reflection of the graph of $f$ in the line $y = x$ .
6.If $f (x) = x^{3}$ , what is $(f \circ f^{- 1}) (8)$ ?Show answer
8, since $f \circ f^{- 1}$ is the identity.

Composite and Invertible Functions: frequently asked questions

What is the difference between f⁻¹(x) and 1/f(x)?

f⁻¹ is the inverse function, which undoes f. 1/f(x) is the reciprocal of the value of f. For f(x) = 2x, the inverse is x/2, while the reciprocal is 1/(2x).

Why must a function be one-one and onto to have an inverse?

If two inputs shared an output, the inverse would have to send that output back to two places, which a function cannot do. If some element of the co-domain had no pre-image, the inverse would have nowhere to send it. One-one and onto remove both problems.

Is gof always defined when fog is defined?

No. fog needs the range of g inside the domain of f, while gof needs the range of f inside the domain of g. One can hold without the other.

Which functions are included for composites and inverses in the ISC 2027 syllabus?

The CISCE syllabus restricts composite and invertible functions to algebraic functions, so the questions use polynomials, rational functions and roots rather than exponential or logarithmic functions.

How can I check my inverse quickly?

Pick a simple value of x, work out f(x), and put the result into your f⁻¹. You should get back the value you started with. Doing this for one value catches most sign and algebra errors.