Composite Functions and Inverse Functions

Chapter 1 established what a function is and how to find its domain and range. This chapter develops two ways of combining or undoing functions: building a new function by feeding the output of one function into another (composition), and reversing a function entirely so that effect and cause swap places (the inverse function). Both ideas recur constantly for the rest of the course — composition is the algebraic engine behind the chain rule in Block D, and the inverse-function relationship underlies logarithms, inverse trigonometric functions and every "solve for the input" problem you will meet.

You need the domain/range toolkit from Chapter 1, confident algebraic manipulation, and the horizontal line test. Nothing here requires calculus.

1. Composite Functions: Definition

Given two functions $f$ and $g$, the composite function $f \circ g$ (read "$f$ of $g$", or "$f$ composed with $g$") is defined by

The instruction is to evaluate $g$ first, then feed that entire output into $f$. Composition is not multiplication — $f \circ g$ is a new function built by chaining two machines, not by multiplying $f(x)$ and $g(x)$ together.

Worked example 1. Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$. Find $(f\circ g)(x)$ and $(g \circ f)(x)$.

Key distinction. In general $f\circ g \neq g \circ f$ — composition is not commutative. The example above gives two different quadratics, which settles the question for this pair, but it is worth remembering as a general rule, not a coincidence of this example.

2. Evaluating a Composite Function at a Number

To evaluate $(f\circ g)(a)$ for a specific number $a$, two routes give the same answer: substitute $a$ into the simplified formula for $f\circ g$, or work from the inside out, computing $g(a)$ first and then applying $f$ to that result.

Worked example 2. Using $f$ and $g$ from Example 1, find $(f\circ g)(3)$ by both routes.

Inside-out: $g(3) = 2(3)-3 = 3$, then $f(3) = 3^2+1 = 10$.

Formula first: $(f\circ g)(x) = 4x^2-12x+10$, so $(f\circ g)(3) = 4(9)-36+10 = 36-36+10=10$.

Both give $10$. The inside-out method is usually faster for a single numerical evaluation; the simplified formula is better when $(f\circ g)$ must be evaluated repeatedly or analysed further (e.g. for its own domain or range).

3. The Domain of a Composite Function

The domain of $f\circ g$ is more subtle than it looks, because two conditions must both hold: $x$ must be a valid input to $g$, and the resulting output $g(x)$ must be a valid input to $f$.

Definition. $\text{domain}(f\circ g) = \{x \in \text{domain}(g) : g(x) \in \text{domain}(f)\}$.

Worked example 3. Let $f(x) = \sqrt{x}$ (domain $x\geq0$) and $g(x) = x - 5$ (domain $\mathbb{R}$). Find the domain of $(f\circ g)(x) = \sqrt{x-5}$.

Every real $x$ is a valid input to $g$, so the first condition imposes no restriction. But the output $g(x)=x-5$ must lie in the domain of $f$, i.e. $g(x) \geq 0$:

So $\text{domain}(f\circ g) = [5,\infty)$ — exactly what direct inspection of $\sqrt{x-5}$ would also give, confirming the two-condition rule is consistent with the shortcut of just finding the domain of the simplified formula. The shortcut can fail, however, when algebraic simplification removes a restriction that genuinely belongs to the composite (see the pitfalls section).

4. Decomposing a Function into Components

The reverse operation — recognising that a given function is really a composite of two simpler functions — is a skill needed constantly in Block D (the chain rule) and useful here for simplifying complicated expressions.

Worked example 4. Express $h(x) = \sqrt{3x^2+7}$ as a composite $f(g(x))$ of two simpler functions.

The "last operation performed" when evaluating $h$ at a number is taking a square root — everything else happens first, inside the root. So let the inner function be the expression under the root, and the outer function be the operation applied to it:

This decomposition is not unique — $g(x)=x^2$ and $f(x)=\sqrt{3x+7}$ also reconstruct $h$ correctly: $f(g(x))=\sqrt{3x^2+7}=h(x)$. However, not every superficially plausible split works — e.g. $g(x)=3x$ and $f(x)=\sqrt{x^2+7}$ give $f(g(x))=\sqrt{9x^2+7}\neq h(x)$ — so always verify a proposed decomposition by recomposing it.

5. The Inverse of a Function: Definition and Existence

The inverse function $f^{-1}$ of $f$ "undoes" $f$: it takes an output of $f$ back to the input that produced it.

Definition. If $f$ is one-to-one, its inverse $f^{-1}$ is the function satisfying $f^{-1}(f(x)) = x$ for every $x$ in the domain of $f$, and $f(f^{-1}(y)) = y$ for every $y$ in the range of $f$.

The requirement that $f$ be one-to-one (Chapter 1, the horizontal line test) is not a technicality — it is forced by the definition of a function. If some output $y_0$ were produced by two different inputs $x_1 \neq x_2$, then $f^{-1}(y_0)$ would have to equal both $x_1$ and $x_2$ simultaneously, which is impossible for a function (exactly one output per input). This is why $f(x)=x^2$ on all of $\mathbb{R}$ has no inverse function, while $f(x)=x^2$ restricted to $x\geq0$ does.

Two further facts follow directly from the definition:

• $\text{domain}(f^{-1}) = \text{range}(f)$ and $\text{range}(f^{-1}) = \text{domain}(f)$ — inverting a function swaps the roles of domain and range exactly.
• The graphs of $f$ and $f^{-1}$ are reflections of each other in the line $y=x$, because if $(a,b)$ lies on the graph of $f$ (so $f(a)=b$), then $(b,a)$ lies on the graph of $f^{-1}$ (so $f^{-1}(b)=a$) — and reflecting a point in $y=x$ swaps its coordinates exactly.

6. Finding the Inverse of a Function Algebraically

The standard algebraic procedure has three steps: write $y=f(x)$; swap $x$ and $y$; solve the new equation for $y$.

Worked example 5 (the routine case). Find $f^{-1}(x)$ for $f(x) = 3x+7$.

So $f^{-1}(x) = \dfrac{x-7}{3}$. Check: $f(f^{-1}(x)) = 3\left(\dfrac{x-7}{3}\right)+7 = x-7+7=x$. ✓.

Worked example 6 (the edge case — domain restriction required). Find $f^{-1}(x)$ for $f(x) = x^2+4$, $x \geq 0$, and state its domain.

$f$ restricted to $x\geq0$ is one-to-one (Chapter 1), so an inverse exists on this restricted domain. Swap and solve:

Taking the positive square root when solving $x=y^2+4$ is the critical step — it is forced by the restriction $y\geq0$ carried over from the original domain of $f$, not a free choice. Omitting this restriction gives $y=\pm\sqrt{x-4}$, which is not even a function. So $f^{-1}(x) = \sqrt{x-4}$ with domain $x\geq4$ — which, consistent with the domain/range-swapping rule above, is exactly $\text{range}(f)$.

7. Verifying an Inverse

Because the swap-and-solve procedure involves irreversible-looking algebraic steps (squaring, taking reciprocals), an inverse found this way should be checked by confirming both composition identities:

Checking only one direction is not sufficient in general (though for most functions met at this level, if one direction works algebraically the other does too) — but the habit of checking matters more for catching an arithmetic slip during the solve step than for a genuine logical gap.

Common Exam Pitfalls

• Writing $f\circ g$ as $f(x)\cdot g(x)$. Composition means substitution of one whole function into another, never multiplication of the two formulas.
• Assuming composition is commutative. $f\circ g$ and $g\circ f$ are generally different functions; always compute the one actually asked for, in the stated order.
• Ignoring the inner restriction on the domain of a composite. Even if the simplified formula for $f\circ g$ looks unrestricted, a restriction coming from $\text{domain}(f)$ on the output of $g$ can still apply before any algebraic simplification occurs.
• Treating $f^{-1}(x)$ as $\dfrac{1}{f(x)}$. The $-1$ superscript on a function name denotes the inverse function, not a reciprocal — these are unrelated operations that happen to share notation.
• Forgetting that a function needs to be one-to-one before an inverse function exists. Applying the swap-and-solve procedure to a non-one-to-one function (like unrestricted $x^2$) produces a relation, not a function — check or impose one-to-one-ness first.
• Dropping the sign restriction carried from the original domain when solving for $y$ after a squaring step. This restriction is not optional — it is what makes the resulting inverse a genuine function rather than a two-valued relation.

Summary

• The composite function $(f\circ g)(x) = f(g(x))$ applies $g$ first, then $f$ to the result; composition is not commutative in general.
• $\text{domain}(f\circ g) = \{x\in\text{domain}(g) : g(x)\in\text{domain}(f)\}$ — both the input to $g$ and the output of $g$ (as an input to $f$) must be valid.
• A function has an inverse function $f^{-1}$ exactly when it is one-to-one; $f^{-1}(f(x))=x$ and $f(f^{-1}(y))=y$.
• Domain and range swap under inversion: $\text{domain}(f^{-1})=\text{range}(f)$, $\text{range}(f^{-1})=\text{domain}(f)$; the graphs of $f$ and $f^{-1}$ are reflections of each other in $y=x$.
• To find $f^{-1}$ algebraically: write $y=f(x)$, swap $x$ and $y$, solve for $y$ — carrying forward any sign restriction from the original domain when a square root appears.
• Verify an inverse by checking $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$.