Composition and Inverse Functions

Pass outputs through composed functions in the correct order, including domain restrictions on intermediate values. Determine when an inverse can uniquely recover an input and distinguish inversion from reciprocation.

Function composition joins rules by using one function's output as another function's input. An inverse function reverses a rule without losing which input produced which output. Both ideas require attention to order, domains, and information that may have been discarded.

Composition Passes an Output Forward

Let

f(x)=2x

and

g(x)=x+3.

The composition $g\circ f$ means apply $f$ first and then apply $g$ :

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

$g$ composed with $f$ , evaluated at $x$ , means apply $f$ to $x$ first and then apply $g$ to that result.

At $x=4$ , the intermediate value is $f(4)=8$ , and the final output is $g(8)=11$ .

Composition is read from the inside outward. In $g(f(x))$ , the inner expression $f(x)$ is evaluated first. The notation $g\circ f$ follows the same order even though $g$ is written on the left.

Order matters. Reversing the functions gives:

(f\circ g)(x)=f(g(x))=f(x+3)=2x+6,

which is generally different from $2x+3$ .

Exercise: Evaluate a composition

Let $f(x)=x^2$ and $g(x)=x+1$ . What is $(f\circ g)(3)$ ?

Your answer

Compute it first, then check your number.

HintWork from the inside outward

First compute $g(3)$ . Use that output as the input to $f$ .

SolutionFollow the composition order

Apply $g$ first:

g(3)=3+1=4.

Then apply $f$ :

f(4)=4^2=16.

Therefore $(f\circ g)(3)=16$ . Reversing the order would evaluate $g(f(3))=g(9)=10$ , a different result.

The Intermediate Output Must Be Allowed

The composition $g\circ f$ is defined only for inputs $x$ for which $f(x)$ is defined and the resulting value $f(x)$ belongs to the domain of $g$ .

For example, let:

f(x)=x-2, \qquad g(u)=\sqrt{u}.

Then:

(g\circ f)(x)=\sqrt{x-2}.

Over the real numbers, the intermediate value must satisfy $x-2\ge0$ . Therefore the composition has domain $x\ge2$ , even though $f(x)=x-2$ itself accepts every real input.

Exercise: Find the domain of a composition

Let $f(x)=x+5$ and $g(u)=\sqrt{u}$ . What is the smallest real input allowed for $(g\circ f)(x)$ ?

Your answer

Compute it first, then check your number.

HintApply the outer function's domain

The outer square root requires $x+5\ge0$ .

SolutionRestrict the intermediate value

The composition is:

(g\circ f)(x)=\sqrt{x+5}.

Its real-valued domain requires:

x+5\ge0 \quad\Longrightarrow\quad x\ge-5.

The smallest allowed input is $-5$ .

Long compositions are often easier to inspect when intermediate values are named. If $h=f(x)$ and $y=g(h)$ , then $y=g(f(x))$ . The same pattern describes unit conversion, coordinate transformations, signal processing stages, and layers in a neural network.

Composition is associative when all expressions are defined:

h\circ(g\circ f)=(h\circ g)\circ f.

The grouping may change, but the order $f$ then $g$ then $h$ does not.

An Inverse Reverses a Function

Suppose $f$ maps an input $x$ to an output $y$ . An inverse function $f^{-1}$ maps that output back to the original input. On the relevant domains, both directions must hold:

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

and

f(f^{-1}(y))=y.

The notation $f^{-1}$ means inverse function. It does not mean the reciprocal $1/f(x)$ .

For:

f(x)=3x+2,

write $y=3x+2$ and solve for $x$ :

y-2=3x, \qquad x=\frac{y-2}{3}.

Therefore:

f^{-1}(y)=\frac{y-2}{3}.

The inverse undoes the forward operations in reverse order: subtract $2$ and then divide by $3$ .

Exercise: Evaluate an inverse

For $f(x)=5x-1$ , what is $f^{-1}(19)$ ?

Your answer

Compute it first, then check your number.

HintReverse the two operations

The forward rule multiplies by $5$ and then subtracts $1$ . Undo them in reverse order.

SolutionRecover the original input

Solving $y=5x-1$ for $x$ gives:

f^{-1}(y)=\frac{y+1}{5}.

Therefore:

f^{-1}(19)=\frac{20}{5}=4.

Checking with the forward rule gives $f(4)=5(4)-1=19$ .

Exercise: Distinguish an inverse from a reciprocal

If $f(x)=2x$ , which rule is its inverse?

Choose the inverse rule

f⁻¹(y) = y / 2

f⁻¹(y) = 1 / (2y)

f⁻¹(y) = -2y

Select one choice, then check.

HintTest the round trip

Apply the candidate rule to $f(x)=2x$ and see whether it returns $x$ .

SolutionUndo multiplication

The forward function multiplies by $2$ , so the inverse divides by $2$ :

f^{-1}(y)=\frac{y}{2}.

The round trip confirms it:

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

The reciprocal $1/f(x)$ is a different function and does not undo $f$ .

An Inverse Requires Unique Recovery

For an inverse to recover an input, different allowed inputs must not produce the same output. Such a function is called one-to-one, or injective.

The function $f(x)=x^2$ is not one-to-one on all real numbers because both $3$ and $-3$ map to $9$ . From the output $9$ alone, an inverse cannot know which input to return. On a graph, this failure appears when a horizontal line crosses the graph more than once.

If the domain is restricted to $x\ge0$ , every output in $[0,\infty)$ has one allowed input, and the inverse is:

f^{-1}(y)=\sqrt{y}, \qquad y\ge0.

The inverse swaps the original function's domain and range. Its graph is the reflection of the original graph across the line $y=x$ .

Exercise: Identify a lost input distinction

Why does $f(x)=|x|$ have no inverse on all real numbers?

Choose the reason

Positive and negative inputs can produce the same output.

Absolute value is undefined for negative inputs.

The function can output zero.

Select one choice, then check.

HintTest opposite inputs

Evaluate the function at $2$ and $-2$ .

SolutionShow the collision

Both inputs produce the same output:

|2|=2, \qquad |-2|=2.

An inverse receiving $2$ would not know which original input to return. Restricting the domain to $x\ge0$ or to $x\le0$ would remove the collision and permit an inverse on the corresponding range.

Trace Forward, Then Reverse

For a composition, work from the innermost function outward and check that each intermediate value belongs to the next function's domain. For an inverse, ask whether the output uniquely identifies its input, reverse the operations in the opposite order, and verify both round trips. If a stage merges several inputs into one output, later stages cannot recover the lost distinction.