Composition of Functions

AP Precalculus· AP Precalculus CED — Exponential and Logarithmic Functions· 14 min read

1. Introduction to Composition of Functions★★☆☆☆⏱ 3 min

Composition of functions is the process of using the output of one function as the input of a second function, combining multiple functions into a single new function. Per AP Precalculus CED, this is a core skill in Unit 2, appearing in 8-12% of Unit 2 exam questions on both multiple-choice and free-response sections.

📘 Definition

Composite Function

$(f \circ g) (x)$

A new function formed by chaining two functions: the output of the inner function $g$ becomes the input of the outer function $f$ , so $(f \circ g) (x) = f (g (x))$ . Composition is not commutative, meaning $f \circ g$ is almost never equal to $g \circ f$ .

Example:

For exponential and logarithmic pairs, composition is used to simplify expressions using inverse properties.

In AP Precalculus Unit 2, composition most frequently pairs exponentials or logarithms with polynomials, or pairs exponential and logarithmic functions with each other, especially when working with inverse functions.

2. Evaluating Composite Functions★★☆☆☆⏱ 3 min

Evaluating a composite function for a given input $x$ follows one core rule: always work from the inside out. First evaluate the inner function (the function closest to $x$ in notation), then substitute that output into the outer function as the new input.

📐 Worked Example

Given $f (x) = e^{2 x} + 1$ and $g (x) = ln x$ , find $(f \circ g) (3)$ .

1
Recognize that $(f \circ g) (3) = f (g (3))$ , so evaluate the inner function $g$ first, per the order of composition.
2
Calculate the inner function output:
$g(3) = \ln 3$
3
Substitute this output into the outer function $f$ :
$f(g(3)) = f(\ln 3) = e^{2(\ln 3)} + 1$
4
Simplify using logarithm rules and the inverse identity $e^{l n k} = k$ :
$2 \ln 3 = \ln 3^2 = \ln 9 \implies e^{\ln 9} = 9$
5
Add 1 to get the final result:
$9 + 1 = 10$

Exam tip:

Always explicitly confirm the order of composition before starting calculations to avoid falling for common distractors.

3. Finding Composite Rules and Domains★★★☆☆⏱ 4 min

To find the general algebraic rule for $(f \circ g) (x)$ , substitute the entire expression for $g (x)$ into the outer function $f$ in place of $x$ . A frequently tested skill is finding the domain of the new composite function, which must satisfy two conditions: 1) $x$ is in the domain of the inner function $g$ , and 2) the output $g (x)$ is in the domain of the outer function $f$ .

📐 Worked Example

Given $f (x) = ln (x + 2)$ and $g (x) = x^{2} - 5$ , find $(f \circ g) (x)$ and state its domain.

1
Write the composition by substituting $g (x)$ into $f$ :
$(f \circ g)(x) = f(g(x)) = \ln\left((x^2 - 5) + 2\right) = \ln(x^2 - 3)$
2
Check the domain of the inner function $g (x)$ : $g (x)$ is a polynomial, so its domain is all real numbers with no restrictions here.
3
Apply the domain restriction of the outer function $f$ : the argument of a logarithm must be positive, so solve:
$x^2 - 3 > 0$
4
Factor and solve the inequality:
$(x - \sqrt{3})(x + \sqrt{3}) > 0 \implies x < -\sqrt{3} \text{ or } x > \sqrt{3}$
5
Rule: $(f \circ g) (x) = ln (x^{2} - 3)$ , domain: $(- \infty, - 3) \cup (3, \infty)$

4. Composition of a Function and Its Inverse★★★☆☆⏱ 4 min

A key property of inverse functions is that composing a function with its inverse gives the identity function, which outputs the original input. For any one-to-one function $f$ with inverse $f^{- 1}$ , two core identities hold:

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

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

For Unit 2, this property is most commonly applied to inverse pairs of exponential and logarithmic functions: for $b > 0, b \neq = 1$ , $f (x) = b^{x}$ and $f^{- 1} (x) = lo g_{b} x$ . This gives the identities $b^{l o g_{b} x} = x$ (for $x > 0$ ) and $lo g_{b} (b^{x}) = x$ (for all real $x$ ), which are used constantly to simplify expressions and solve equations.

📐 Worked Example

Simplify the composite expression $(ln \circ f) (x)$ where $f (x) = 5 e^{3 x - 2}$ , and state any domain restrictions.

1
Write the composition explicitly:
$(\ln \circ f)(x) = \ln\left(f(x)\right) = \ln\left(5e^{3x-2}\right)$
2
Split the product using the logarithm product rule:
$\ln 5 + \ln\left(e^{3x-2}\right)$
3
Apply the inverse composition identity $ln (e^{u}) = u$ for all real $u$ :
$\ln\left(e^{3x-2}\right) = 3x - 2$
4
Check domain restrictions: The input to $ln$ is $5 e^{3 x - 2}$ , which is always positive for all real $x$ . The inner function $f (x)$ is also defined for all real $x$ , so there are no additional restrictions.
5
Final result:
$(\ln \circ f)(x) = 3x - 2 + \ln 5, \text{ domain: all real numbers}$

5. AP-Style Concept Check★★★★☆⏱ 4 min

✓ Quick check

Test your understanding with these AP-style practice questions:

If $f (x) = 2^{x}$ and $g (x) = lo g_{2} (x - 3)$ , what is the value of $(f \circ g) (11)$ ?
- A) 3
- B) 8
- C) 11
- D) 16
Reveal answer
B) 8 —
Correct: You worked inside out: $g (11) = lo g_{2} (8) = 3$ , so $f (3) = 2^{3} = 8$ . Swapping the order of composition gives ~11, a common distractor.

📐 Worked Example

In a microbiology experiment, the number of bacteria is $N (d) = 1000 e^{0.2 d}$ , where $d$ is growing days. Time is measured in hours, so $d (t) = \frac{t}{24}$ . Write the composite function for number of bacteria as a function of timer hours $t$ , then find the number of bacteria after 72 hours, rounded to the nearest whole number.

1
We need $N$ as a function of $t$ , so we compose $N$ with $d$ to get $(N \circ d) (t) = N (d (t))$ .
2
Substitute $d (t)$ into $N$ :
$(N \circ d)(t) = 1000e^{0.2\left(\frac{t}{24}\right)} = 1000e^{\frac{t}{120}}$
3
Substitute $t = 72$ and calculate:
$(N \circ d)(72) = 1000e^{\frac{72}{120}} = 1000e^{0.6} \approx 1822$
4
Final result: After 72 hours, the culture has approximately 1822 bacteria.

6. Common Pitfalls

Wrong move:

Swapping the order of composition to compute $g (f (x))$ instead of $f (g (x))$ when asked for $(f \circ g) (x)$ .

Why:

Students confuse the order of notation, forgetting the function closest to $x$ is the inner function evaluated first.

Correct move:

Always translate $(f \circ g) (x)$ explicitly to $f (g (x))$ before starting any calculation, and mark $g$ as the inner function.

Wrong move:

Simplifying the composite function first, then finding the domain from the simplified expression, ignoring restrictions from the inner function.

Why:

Simplification can cancel terms that introduced domain restrictions, leading to incorrectly including disallowed inputs.

Correct move:

Find the domain step-by-step: first find all $x$ allowed in the inner function, then filter that set to only $x$ where the inner output is allowed in the outer function.

Wrong move:

Applying the inverse composition identity $e^{l n x} = x$ to negative inputs of $ln x$ .

Why:

Students memorize the identity without remembering the domain restriction on the logarithm.

Correct move:

Before applying $b^{l o g_{b} x} = x$ , confirm the input to the logarithm is positive, and exclude any negative inputs from your result.

Wrong move:

Claiming composition is commutative, so $(f \circ g) (x) = (g \circ f) (x)$ for all functions $f, g$ .

Why:

Students confuse composition with multiplication of functions, which is commutative.

Correct move:

Always assume $f \circ g \neq = g \circ f$ unless you prove it for the specific functions given.

Wrong move:

When simplifying $e^{2 l n x}$ , writing the result as $2 x$ instead of $x^{2}$ .

Why:

Students reverse the power rule $ln x^{a} = a ln x$ , misapplying the coefficient as a multiplicative factor instead of an exponent.

Correct move:

Move the coefficient inside the logarithm as an exponent first: $a ln x = ln x^{a}$ , then apply the inverse identity, so $e^{a l n x} = x^{a}$ .

Wrong move:

For a composite $f (g (x))$ where $g (x)$ is a logarithm, forgetting that the argument of the inner logarithm must be positive in addition to any restrictions on the output of $g$ for the outer function.

Why:

Students only check the outer function's restrictions, forgetting the inner function already has a domain restriction.

Correct move:

Always check the domain of the inner function first before checking outer function restrictions.

7. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Composition Notation	$(f \circ g) (x) = f (g (x))$	$g$ is inner (evaluate first); $f$ is outer. Usually $f \circ g \neq = g \circ f$
Evaluating Composite	Work inside out	Evaluate inner $g (x)$ first, substitute result into outer $f$
Domain of Composite	Domain = ${x \in dom (g) ∣ g (x) \in dom (f)}$	Check inner domain first, then outer; do not simplify before finding domain
Inverse Composition 1	$(f \circ f^{- 1}) (x) = x$	Only valid for $x$ in the domain of $f^{- 1}$
Inverse Composition 2	$(f^{- 1} \circ f) (x) = x$	Only valid for $x$ in the domain of $f$
Exponential-Log Inverse 1	$b^{l o g_{b} x} = x$	Only for $x > 0$ , $b > 0, b \neq = 1$
Exponential-Log Inverse 2	$lo g_{b} (b^{x}) = x$	Valid for all real $x$
Simplifying Composites	$e^{a l n x} = x^{a}$	Coefficient becomes an exponent, not a multiplicative factor

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
MCQ evaluate composite exponential-log
2023 · AP Precalculus
FRQ find domain of composite

What's Next

Mastering composition of functions is a critical prerequisite for all remaining topics in AP Precalculus Unit 2, and for the entire course. You will apply the composition skills and inverse identities covered here to solve exponential and logarithmic equations, where simplifying composite expressions is required to isolate the target variable. You will also use composition regularly when building real-world models that involve multiple chained transformations, such as unit conversions or multi-step growth problems. Beyond AP Precalculus, composition is the foundational concept for the chain rule in differential calculus, a core topic in AP Calculus AB and BC. Without mastering the order of composition and domain rules here, you will struggle with these more advanced topics.