Composition of functions — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Composition notation, evaluating composite functions, finding composite function rules, domain restrictions for composites, and inverse function composition properties for exponential and logarithmic function pairs, aligned with AP Precalculus CED Unit 2.

You should already know: How to find domain/range of exponential, logarithmic, and polynomial functions. How to evaluate functions at numerical and algebraic inputs. The definition of inverse functions for exponential-logarithmic pairs.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Composition of functions?

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 the AP Precalculus CED, composition of functions is a core skill within Unit 2 (Exponential and Logarithmic Functions) and appears in roughly 8-12% of Unit 2 exam questions, tested on both multiple-choice (MCQ) and free-response (FRQ) sections. Notation for composition is $(f\circ g)(x)$, read "f composed with g of x", and this is equivalent to $f(g(x))$: you first evaluate $g(x)$ to get an output, then plug that output into $f$ as the input. A common synonym is "function chaining", as you chain the output of one step to the input of the next. For 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. Unlike addition or multiplication of functions, composition is not commutative: $f\circ g$ is almost never equal to $g\circ f$, a property frequently tested on the AP exam.

2. Evaluating Composite Functions

Evaluating a composite function means finding the numerical output of $(f\circ g)(x)=f(g(x))$ for a given input value of $x$. The core rule is to always work from the inside out: evaluate the inner function (the one closest to $x$ in the notation) first, then plug that result into the outer function. This process works for all function types, but in Unit 2 you will frequently evaluate composites involving exponentials and logs, so you will often need to use inverse properties of these functions to simplify your result to a numerical value. You can also use the same inside-out process to evaluate composites at algebraic inputs, which is the first step to finding the general rule of a composite function. Common evaluation problems on the AP exam ask for the value of a composite of an exponential and a logarithmic function, leveraging the inverse identity to get a clean numerical result.

Worked Example

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

1. Recognize that $(f\circ g)(3)=f(g(3))$, so we 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 $f$ as the input: $f(g(3))=f(\ln 3)=e^{2(\ln 3)}+1$.
4. Simplify using logarithm rules and the inverse identity $e^{\ln k}=k$: $2\ln 3=\ln 3^2=\ln 9$, so $e^{\ln 9}=9$.
5. Add 1 to get the final result: $9+1=10$.

Exam tip: On multiple-choice questions, one of the most common distractors is the result of swapping the order of composition (in this example, $g(f(3))=\ln (e^6+1)\approx 6.01$, which is a common wrong answer). Always explicitly mark which function is inner before you start calculating.

3. Finding the General Rule and Domain of Composite Functions

Once you can evaluate composites at a point, you can find the general algebraic rule for $(f\circ g)(x)$ by substituting the entire expression for $g(x)$ into the outer function $f$ in place of $x$. A critical, frequently tested skill is finding the domain of the new composite function. The domain of $f\circ g$ is restricted by two rules: first, the input $x$ must be in the domain of the inner function $g$, and second, the output of $g(x)$ must be in the domain of the outer function $f$. A common mistake is simplifying the composite expression first, then finding the domain of the simplified function, which can lead to incorrectly including disallowed inputs that were restricted by the inner function. For exponential and logarithmic composites, this is especially important because logarithms have a strict domain restriction that their argument must be positive.

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 $f$ as the input: $(f\circ g)(x)=f(g(x))=\ln \left((x^2-5)+2\right)=\ln (x^2-3)$.
2. First check the domain of the inner function $g(x)$: $g(x)=x^2-5$ is a polynomial, so its domain is all real numbers, with no additional restrictions from this step.
3. Next, apply the domain restriction of the outer function $f$: the argument of the logarithm must be positive, so solve $x^2-3>0$.
4. Solve the inequality: $(x-\sqrt{3})(x+\sqrt{3})>0$, which holds when $x<-\sqrt{3}$ or $x>\sqrt{3}$.
5. Final result: $(f\circ g)(x)=\ln (x^2-3)$, domain is $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$.

Exam tip: Always find the domain before simplifying the composite expression, even if simplification removes a term that caused the restriction. For example, $\ln \left(\frac{(x-2)(x+1)}{x-2}\right)$ still excludes $x=2$ from the domain, even after canceling the $(x-2)$ term.

4. Composition of a Function and Its Inverse

A key property of inverse functions is that composing a function with its inverse gives the identity function, meaning the output equals the original input. For any one-to-one function $f$ with inverse $f^{-1}$, two 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 (Exponential and Logarithmic Functions), this is most commonly applied to the inverse pairs $f(x)=b^x$ and $f^{-1}(x)={\log }_{b}x$ (for $b>0,b\ne 1$), so the identities become $b^{{\log }_{b}x}=x$ (for $x>0$) and ${\log }_b(b^x)=x$ (for all real $x$). These identities are used constantly to simplify composite expressions involving exponentials and logs, and they are required for solving most exponential and logarithmic equations on the AP exam. Always remember to apply domain restrictions when using these identities, even after simplification.

Worked Example

Simplify the composite expression $(\ln \circ f)(x)$ where $f(x)=5e^{3x-2}$, 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 $u$, so $\ln \left(e^{3x-2}\right)=3x-2$.
4. Check the domain: the input to the outer logarithm is $5e^{3x-2}$, which is always positive (since $e^u>0$ for all real $u$, multiplied by positive 5). The inner function $f(x)$ is defined for all real $x$, so there are no additional domain restrictions.
5. Final result: $(\ln \circ f)(x)=3x-2+\ln 5$, domain is all real numbers.

Exam tip: When simplifying $e^{\ln x^2}$, the result is $|x|$, not just $x$. $\ln x^2$ is defined for all non-zero $x$, and $e^{\ln x^2}=x^2$, so the simplified form requires the absolute value to retain the correct domain for negative inputs.

5. Common Pitfalls (and how to avoid them)

• 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 students to incorrectly include 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 $e^{\ln x}=x$ to negative inputs of $\ln x$. Why: Students memorize the identity without remembering the domain restriction on the logarithm. Correct move: Before applying $b^{{\log }_{b}x}=x$, 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 $f\circ g\ne g\circ f$ unless you prove it for the specific functions given.
• Wrong move: When simplifying $e^{2\ln x}$, writing the result as $2x$ 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 $e^{a\ln x}=x^a$.
• 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.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

If $f(x)=2^x$ and $g(x)={\log }_2(x-3)$, what is the value of $(f\circ g)(11)$? A) $3$ B) $8$ C) $11$ D) $16$

Worked Solution: First, translate the notation: $(f\circ g)(11)=f(g(11))$, so we evaluate the inner function $g$ first. Calculate $g(11)={\log }_2(11-3)={\log }_{2}8=3$. Next, substitute this output into the outer function $f$: $f(g(11))=f(3)=2^3=8$. The most common error is swapping the order of composition, which gives $g(f(11))\approx 11$, a common distractor. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\sqrt{x+4}$ and $g(x)=\ln (2x-1)$. (a) Write the rule for $(f\circ g)(x)$ and find its domain. (b) Write the rule for $(g\circ f)(x)$ and find its domain. (c) Is $(f\circ g)(x)=(g\circ f)(x)$ for this pair of functions? Justify your answer.

Worked Solution: (a) Substitute $g(x)$ into $f$ to get $(f\circ g)(x)=f(g(x))=\sqrt{\ln (2x-1)+4}$. First, the inner function $g(x)$ requires $2x-1>0⟹x>\frac{1}{2}$. The outer function $f$ requires its argument to be non-negative: $\ln (2x-1)+4\ge 0⟹\ln (2x-1)\ge -4⟹2x-1\ge e^{-4}⟹x\ge \frac{1+e^{-4}}{2}$. Final result: Rule: $(f\circ g)(x)=\sqrt{\ln (2x-1)+4}$, domain: $\left[\frac{1+e^{-4}}{2},\infty \right)$.

(b) Substitute $f(x)$ into $g$ to get $(g\circ f)(x)=g(f(x))=\ln \left(2\sqrt{x+4}-1\right)$. First, the inner function $f(x)$ requires $x+4\ge 0⟹x\ge -4$. The outer function $g$ requires $2\sqrt{x+4}-1>0⟹\sqrt{x+4}>\frac{1}{2}⟹x+4>\frac{1}{4}⟹x>-\frac{15}{4}$. Combining restrictions gives domain $\left(-\frac{15}{4},\infty \right)$. Final result: Rule: $(g\circ f)(x)=\ln \left(2\sqrt{x+4}-1\right)$, domain: $\left(-\frac{15}{4},\infty \right)$.

(c) No, $(f\circ g)(x)\ne (g\circ f)(x)$. Test $x=1$, which is in the domain of both functions: $(f\circ g)(1)=\sqrt{\ln (2(1)-1)+4}=\sqrt{0+4}=2$, while $(g\circ f)(1)=\ln \left(2\sqrt{5}-1\right)\approx 1.46$, which is not equal to 2. Since the functions have different outputs at a shared input, they are not equal.

Question 3 (Application / Real-World Style)

In a microbiology experiment, the number of bacteria in a temperature-controlled culture is given by $N(d)=1000e^{0.2d}$, where $d$ is the number of days the culture has been growing. The experiment timer measures time in hours, so the number of growing days as a function of timer hours $t$ is $d(t)=\frac{t}{24}$. Write the composite function that gives number of bacteria as a function of timer hours $t$, then calculate the number of bacteria after 72 hours on the timer. Round your final answer to the nearest whole number.

Worked Solution: We need $N$ as a function of $t$, so we compose $N$ with $d$ to get $(N\circ d)(t)=N(d(t))$. Substitute $d(t)$ into $N$: $(N\circ d)(t)=1000e^{0.2\left(\frac{t}{24}\right)}=1000e^{\frac{t}{120}}$. For $t=72$, substitute to get $(N\circ d)(72)=1000e^{\frac{72}{120}}=1000e^{0.6}\approx 1000(1.8221)=1822$. In context, this means after 72 hours (3 full days) of growth, the culture contains approximately 1822 bacteria.

7. Quick Reference Cheatsheet

8. What's Next

Mastering composition of functions is a critical prerequisite for the remaining topics in AP Precalculus Unit 2, and for the entire course. Next, you will apply composition and the inverse function identities we 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 models for real-world scenarios that involve multiple chained transformations, such as unit conversions or multi-step growth. Beyond Unit 2, 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.

Solving exponential equations Solving logarithmic equations Inverse functions Modeling with exponential functions