D2.5 Compose functions algebraically and determine domains

Ontario Grade 12 Mathematics

Characteristics of Functions

Ontario Grade 12 Mathematics — Study topic D2.5

A function takes an input and follows a rule to produce an output. A composition uses the output of one function as the input of another. The order matters: one function acts on the original input, and the other acts on that result. In this lesson, you will compose functions algebraically and identify the inputs for which each composition is defined.

What you will learn

Explain what it means to compose two functions.
Write and simplify a composition algebraically.
Determine a composition’s domain by checking the restrictions of both functions.
Recognize that reversing the order of two functions can change the rule and the domain.

1. Prerequisite bridge: functions and domain restrictions

A function rule might be

f(x)=x^2+3

. The symbol

x

is the input, and

f(x)

is the output. A function’s domain is the set of inputs for which its rule gives a defined real-number output.

Some rules restrict the input. A rational expression cannot have a zero denominator. A square root needs its radicand—the expression inside the root—to be zero or positive. A logarithm needs a positive argument. These restrictions matter when functions are composed because an inner function’s output becomes an input to the outer function.

A polynomial such as

p(x)=x^2-4

allows every real number as an input. For

r(x)=\frac{1}{x-3}

, the input

3

is excluded because it makes the denominator zero. Keep the original function restrictions in view even after simplifying a composition.

\text{radicand}\geq 0

Domain means the allowed input values.
A denominator must not equal zero.
A square-root radicand must be at least zero.
A logarithm’s argument must be positive.

2. Plain language and symbolic composition

To compose functions, feed the output of one function into another. The notation

(f\circ g)(x)

means

f(g(x))

. Read it as “

f

of

g

of

x

.” The function

g

acts first; then

f

acts on the result.

For example, if

g(x)=x+2

and

f(x)=3x

, then

g

changes

x

to

x+2

. The function

f

then multiplies that result by

3

. So

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

. In the reverse order,

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

. The two compositions do not generally have the same rule.

A useful way to organize the process is input, inner function, outer function, final output. To find a composition algebraically, replace the outer function’s input with the entire inner function rule. Use parentheses around the replacement so that every part of the inner rule is included.

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

In $(f\circ g)(x)$ , $g$ acts before $f$ .
Substitute the complete inner rule into the outer rule.
Changing the order can change the composition.

3. Determine the domain of a composition

A composition is defined only when two conditions are met. First, the original input must be allowed in the inner function. Second, the inner function’s output must be allowed as an input to the outer function. Check the inner function’s domain, then apply the outer function’s restrictions to the inner output.

For a square-root outer function, require its radicand to be non-negative after substitution. For a rational outer function, require its denominator to be nonzero after substitution. For a logarithmic outer function, require its argument to be positive after substitution. Include restrictions from the inner function as well.

The domain rule for a composition can be written as a set of allowed inputs. The input must belong to the inner function’s domain, and the inner function’s output must belong to the outer function’s domain.

Simplifying can make a rule look less restricted than the original composition. Do not use a simplified expression to restore an input that was excluded by an original function rule. Keep a record of every restriction, and combine them at the end.

When a restriction is an inequality involving a rational expression, identify values that make its numerator or denominator zero. Test the intervals between those values. A value that makes the denominator zero remains excluded. \operatorname{Dom}(f° g)=\{x∈\operatorname{Dom}(g):g(x)∈\operatorname{Dom}(f)\}

Check both the inner function and the outer function’s input requirements.
Apply outer-function restrictions to the inner output.
Keep excluded values from the original function rules.

4. Apply the process in either order

In a composition problem, name the order before substituting. For each order, identify the inner rule, substitute it into the outer rule, and then check the domain. If the inner function is a square root, use its radicand restriction. If the outer function is rational, make sure its substituted denominator is not zero.

A composition’s rule and domain answer different questions. The rule gives the output for an allowed input. The domain states which inputs are allowed. Write them separately so neither is mistaken for the other.

Name the order before substituting.
Solve all restrictions created by the composition.
State the rule and domain separately.

Order of operations in a composition

Composition	First function used	Then function used
$(f\circ g)(x)$	$g$	$f$
$(g\circ f)(x)$	$f$	$g$

Worked example

Two orders, two domains

Let

f(x)=\sqrt{x+1}

and

g(x)=\frac{x-2}{x+1}

. Find both compositions and the domain of each.

Compose $f$ after $g$
In $(f\circ g)(x)$ , use the full rule for $g$ as the input to $f$ . The square root requires its radicand to be non-negative. The inner function $g$ is also undefined when its denominator is zero.
$(f\circ g)(x)=\sqrt{\frac{x-2}{x+1}+1}$
Simplify and restrict
Combine the terms inside the root. The denominator restriction excludes $x=-1$ . For the square root, solve the resulting rational inequality. Its critical values are $-1$ and $\frac12$ . Testing the intervals shows the expression is non-negative when $x<-1$ and when $x\geq\frac12$ .
$(f\circ g)(x)=\sqrt{\frac{2x-1}{x+1}}$
Compose $g$ after $f$
Now $f$ acts first. Its square root requires $x+1\geq0$ . Substitute its output into the numerator and denominator of $g$ . The new denominator is a square root plus $1$ , which is at least $1$ for every allowed input, so it is never zero.
$(g\circ f)(x)=\frac{\sqrt{x+1}-2}{\sqrt{x+1}+1}$

Answer: For

(f\circ g)(x)

, the rule is

\sqrt{\frac{2x-1}{x+1}}

and the domain is

(-\infty,-1)\cup[\frac12,\infty)

. For

(g\circ f)(x)

, the rule is

\frac{\sqrt{x+1}-2}{\sqrt{x+1}+1}

and the domain is

[-1,\infty)

.

Check: For the first composition,

x=-1

is excluded by

g

, and

x=\frac12

is included because the radicand is zero. For the second, the square root allows

x=-1

, and the denominator is positive for every allowed input.

Common mistakes and how to avoid them

Treating

(f\circ g)(x)

as though

f

acts first because it is written first.

Correction: Read the composition as

f(g(x))

. The inside function

g

acts first.

Checking only the visible denominator or root in the simplified composition.

Correction: Check the restrictions of both original functions. Keep any excluded input even if algebraic simplification hides it.

Using a strict inequality for a square-root radicand.

Correction: A square-root radicand may equal zero, so use a non-negative condition and include valid boundary values.

Assuming the two orders have the same rule or domain.

Correction: Form each order separately. The inner function changes both the substitution and the restrictions.

Lesson summary

A composition substitutes one function rule into another.
In $(f\circ g)(x)$ , $g$ acts first and $f$ acts second.
The domain includes inputs allowed in the inner function whose outputs are also allowed in the outer function.
Keep denominator, square-root, logarithm, and other original function restrictions when simplifying.

Check your understanding

Question 1

If

f(x)=x^2+1

and

g(x)=x-3

, which expression is (f° g)(x)?

$(x-3)^2+1$
$(x^2+1)-3$
$x^2-3+1$
$(x-3)+1$

Show answer and explanation

(x-3)^2+1

In

f(g(x))

, substitute the entire rule

x-3

into the input of

f

. This gives

(x-3)^2+1

.

Question 2

Let

f(x)=\sqrt{x}

and

g(x)=x+4

. What is the domain of (f° g)(x)?

$[-4,\infty)$
$(0,\infty)$
$(-\infty,-4]$
$(-\infty,\infty)$

Show answer and explanation

[-4,\infty)

The composition is

\sqrt{x+4}

. Its radicand must satisfy

x+4\geq0

, so

x\geq-4

.

Question 3

Let

f(x)=\frac{1}{x}

and

g(x)=x-2

. Which input is excluded from the domain of (f° g)(x)?

$x=2$
$x=-2$
$x=0$
No input is excluded

Show answer and explanation

x=2

The composition is

\frac{1}{x-2}

. Its denominator is zero at

x=2

, so that input is excluded.

Key terms

Composition: A function formed by using the output of one function as the input of another.
Domain: The set of inputs for which a function is defined.
Inner function: The function applied first in a composition; it appears inside the other function’s input.
Radicand: The expression inside a radical, such as $x+1$ inside $\sqrt{x+1}$ .

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation D2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.5 · Compose functions algebraically and determine domains

Ontario Grade 12 Mathematics

Ontario Grade 12 Mathematics — Study topic D2.5

What you will learn

1. Prerequisite bridge: functions and domain restrictions

2. Plain language and symbolic composition

3. Determine the domain of a composition

4. Apply the process in either order

Order of operations in a composition

Two orders, two domains

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MHF4U

About this lesson and its review