Functions in Year 4: composite, inverse, graphs

End of September 2026, Year 4 at the liceo classico (the Italian high school focused on Latin, Greek and the humanities): after the revision of conic sections the chapter on functions is closing — domain, composition, inverse, graph transformations. The test falls in five to eight weeks, so between the end of October and mid-November. With 2 hours of maths a week there is no time to practise in class: whatever you do not fix at home arrives at the test as it is.

The level here is that of the classical high school, not the science one: no technicalities that the classical track does not need, but the ideas — composition, inverse, transformations — have to be understood well, because they are the basis of calculus and logarithms that will arrive in the following years.

What a function is, in practice

A function $f$ from a set $A$ to a set $B$ is a rule that associates with every element $x$ of $A$ one and only one element $y=f(x)$ of $B$. The key phrase is “only one”: if for a certain $x$ two different values of $y$ correspond, that is not a function, it is just a relation.

In the Cartesian plane this is read with the vertical line test: if a vertical line $x=k$ meets the graph at more than one point, that graph does not represent a function. The circle $x^2+y^2=r^2$, for example, is not the graph of a function: a vertical line passing through the inside cuts it at two points.

The domain $A$ is the set of values of $x$ for which the rule makes sense. I will not go over it in full here — I have already covered it in detail, with all the conditions case by case, in domain and symmetries of a function. The two conditions that come back most often in the test on composite and inverse functions are those on fractions and irrational expressions:

When the function you are working with is a composition of several pieces, each piece generates its own condition and all of them have to be put in a system, never as a union — this is the point on which most marks are lost, and it is the reason why it is worth rereading that post before the test.

Composite function: the order matters

Given two functions $f$ and $g$, the composite function $f\circ g$ is defined by

It is read from right to left: first you apply $g$, then you apply $f$ to the result. It is a detail that seems obvious written like this, but it is the source of the commonest mistake in the chapter.

$f(g(x))$ is not equal to $g(f(x))$, almost never. Let us check it with a simple example: $f(x)=x+1$ and $g(x)=x^2$.

They are two different functions: composition is not commutative. When the question writes $f\circ g$ you have to start from $g$, substitute its result in place of $x$ in $f$, and that is all — without swapping the order “because it seems more convenient”.

The domain of the composite function

Here is the second delicate point. The domain of $f(g(x))$ is not simply the domain of $g$: an extra condition is needed.

That is: $x$ must lie in the domain of $g$ (otherwise $g(x)$ does not even exist), and the value $g(x)$ you obtain must lie in the domain of $f$ (otherwise $f$ does not accept it).

Example: $f(x)=\sqrt{x}$ (domain $x\ge 0$) and $g(x)=x-3$ (domain $\mathbb{R}$). The composite is

The domain of $g$ is all of $\mathbb{R}$, but it is also necessary that $g(x)=x-3$ be $\ge 0$, that is $x\ge 3$. The domain of the composite is $x\ge 3$, much smaller than the domain of $g$ alone. Whoever stops at the domain of $g$ and ignores the condition on $f$ gets the domain wrong almost every time.

Injective function and inverse function

A function $f$ admits an inverse only if it is injective (one-to-one): different values of $x$ must correspond to different values of $y$. In a formula, $f$ is injective if

Graphically it is checked with the horizontal line test: if a horizontal line $y=k$ meets the graph at more than one point, $f$ is not injective on that domain and has no inverse (on the whole domain).

The case that confuses most is $f(x)=x^2$: taken on the whole of $\mathbb{R}$ it is not injective, because $f(2)=f(-2)=4$ — a horizontal line at $y=4$ cuts the graph at two points. It becomes injective, and therefore invertible, only if you restrict the domain, for example to $x\ge 0$: there every positive $y$ has only one $x$ that produces it.

How to find the inverse

The procedure is always the same, three steps:

1. Write $y=f(x)$.
2. Swap $x$ and $y$: you get $x=f(y)$.
3. Solve for $y$: that $y$ is $f^{-1}(x)$.

Example: $f(x)=2x+3$.

So $f^{-1}(x)=\frac{x-3}{2}$.

With $x^2$ restricted to $x\ge 0$:

So $f^{-1}(x)=\sqrt{x}$, with domain $x\ge 0$ (which coincides with the range of the restricted $f$).

Graph of the inverse: symmetry about $y=x$

The graph of $f^{-1}$ is obtained from that of $f$ by reflection in the line $y=x$. It is a direct consequence of the swap $x\leftrightarrow y$ in the procedure: every point $(a,b)$ of the graph of $f$ becomes the point $(b,a)$ on the graph of $f^{-1}$, and $(a,b)\to (b,a)$ is exactly the reflection in $y=x$.

Two practical consequences to remember for the test:

• domain and range swap: $\text{dom}(f^{-1})=\text{range}(f)$ and vice versa;
• if you can draw $f$, you can already draw $f^{-1}$ without doing any calculation: reflect the graph in the bisector of the first and third quadrants.

Graph transformations

Knowing a few basic graphs (line, parabola, hyperbola $y=1/x$, absolute value) and the transformation rules lets you sketch by eye much more complex functions, without doing a full study. The main rules:

Translations.

The typical mistake is the sign in the horizontal translation: $f(x-3)$ shifts to the right, not to the left, because to get the same value of $y$ as before you now need an $x$ that is 3 units larger.

Reflections.

Absolute value. These are two different transformations that are often confused:

flips upwards only the part of the graph that lies below the $x$-axis, leaving the part above unchanged — because the modulus changes sign only for negative values.

deletes the part of the graph with $x<0$ and replaces that zone with the mirror image of the part with $x\ge 0$, because $f(|x|)$ for negative $x$ equals $f$ evaluated at its positive opposite — the resulting graph is always symmetric about the $y$-axis.

The typical mistake: domain of fractions and irrational expressions, not the composition itself

The point on which the test on functions does the most damage is not the composition rule itself, but the domain when fractions and irrational expressions combine with each other or inside a composition. Two examples that sum up the problem:

Here the condition is not $x-2\ge 0$: under a root that sits in the denominator the inequality must be strict, because if $x-2=0$ the denominator becomes zero anyway. Correct condition: $x-2>0⟹x>2$.

Here you need to study the sign of a fraction, not of a single factor: $\frac{2x-4}{x+1}\ge 0$ has to be solved by studying the sign of numerator and denominator separately (with the denominator in any case $\ne 0$), not by “solving” the inequality as if it were a product of whole numbers. It is exactly the type of compound condition that the post on the domain covers in full, with the system of conditions compared step by step.

Three worked examples

1. Composite function, domain included

Given $f(x)=\frac{1}{x}$ and $g(x)=x+2$, find $f(g(x))$, $g(f(x))$ and their respective domains.

They are two different functions with different domains: direct confirmation that composition is not commutative, both in the expression and in the domain.

2. Inverse with a restricted domain

Find the inverse of $f(x)=(x-1{)}^2$ restricted to $x\ge 1$, and check the result.

I set $y=(x-1{)}^2$ and swap:

Since the restriction is $x\ge 1$, that is $x-1\ge 0$, the absolute value is removed by choosing the positive sign:

So $f^{-1}(x)=1+\sqrt{x}$, domain $x\ge 0$. Check: $f(f^{-1}(x))=f(1+\sqrt{x})=(1+\sqrt{x}-1{)}^2=(\sqrt{x}{)}^2=x$. It works.

3. Transformations in sequence

Start from the graph of $f(x)=x^2$ and describe the graph of $y=-f(x+1)+3=-(x+1{)}^2+3$.

It is best to read the transformations from the inside out, in the order in which they act on the variable:

1. $x\to x+1$: shifts the graph 1 unit to the left. Vertex at $(-1,0)$.
2. The minus sign in front of $f$: reflects in the $x$-axis. Vertex still at $(-1,0)$, but the parabola now opens downwards.
3. $+3$: shifts upwards by 3. Final vertex at $(-1,3)$.

The final graph is a parabola with concavity downwards, vertex $(-1,3)$, obtained without expanding the expression: just reading the transformations one at a time, in the right order.

Frequently asked questions

Does every function have an inverse? No. It has to be injective on the domain considered. If it is not (like $x^2$ on the whole of $\mathbb{R}$), either you restrict the domain or the inverse does not exist.

How do I tell at a glance from the graph whether a function is injective? Horizontal line test: if every horizontal line cuts the graph at most once, it is injective.

Why is the graph of the inverse symmetric about $y=x$ and not about the $y$-axis? Because finding the inverse means swapping $x$ and $y$, and the swap $x\leftrightarrow y$ is geometrically exactly the reflection in the bisector $y=x$, not in a coordinate axis.

How I organise things with those who have 2 hours a week

With only two weekly hours of maths the time to practise in class is almost zero: the teacher explains, and the consolidation part — the one where you really learn to tell $f(g(x))$ from $g(f(x))$, or to read a transformation without expanding the calculations — stays at home. In online lessons I work with targeted exercises on the two points that cost the most marks at the classical high school: the domain of composite fractional/irrational functions, and reading transformations in the correct sequence. It is the same reasoning scheme needed afterwards for logarithms and exponentials, so the time invested now is not used up with this test.

To get an idea of where you stand you can try the mock test on functions: it is calibrated on the Year 3 syllabus of the science high school, which covers the same content — domain, composite, inverse, transformations — that the classical high school does in Year 4, so it remains a good testing ground even if your class is different. The correct answer to every question is free, the guided solution unlocks with the first lesson.

The domain alone, with all the conditions case by case, is covered in domain and symmetries of a function; for the rest of the Year 4 syllabus at the classical high school (and how it fits with the physics of the same year) there is Year 4: the whole maths syllabus in one formula sheet and, for the physics of the same class, Year 4 physics at the classical high school: energy and circular motion. The general picture of the subject is on the maths page; if you want a hand organising your revision before the test, start with the free introductory call. The most used inverse of the Year 4 syllabus is the logarithm as the inverse of the exponential: I wrote about it in exponentials and logarithms in Year 4.