How to Solve Composition of Functions? (+FREE Worksheet!)
Composition of functions is one of the most powerful tools in algebra: it lets you create a new function by feeding the output of one function directly into a second function. Written as (f ∘ g)(x) or f(g(x)), function composition appears throughout Algebra 1 and is the foundation for understanding inverse functions and transformations. With the right substitution method, every composition problem becomes straightforward.
Solve Composition of Functions: what to notice and how to work it
What to notice first
Common student mistake
Key formulas and cues
A reliable path
- Identify the inputFind the x-value, expression, or inner function being used.
- Apply the ruleSubstitute with parentheses so signs and powers stay clear.
- Interpret the outputState the value, point, interval, domain, range, or inverse relationship.
Worked examples
Evaluate a function
- Replace x with 2.
- Compute 4(2) – 3.
- Simplify.
Compose functions
- Find g(3) = 6.
- Use that as the input for f.
- f(6) = 7.
Try one before moving on
Solve Composition of Functions: pop-up practice
What Is Composition of Functions?
When you compose two functions f and g, you apply one function and then immediately apply the other to the result. The notation f(g(x)) means “first apply g to x, then apply f to that result.” It is read “f of g of x.”
- Inner function: g is applied first.
- Outer function: f is applied second to g’s output.
Important: Order matters. In general, f(g(x)) ≠ g(f(x)).
How to Find a Composition of Functions
Method — Substitute the inner function into the outer
Replace every x in the outer function’s formula with the entire inner function.
Example: \(\color{blue}{f(x) = 2x + 1}\), \(\color{blue}{g(x) = x}\)\(\color{blue}{^{2} – 3}\). Find (f ∘ g)(x).
Step 1 — Write f, but replace every x with g(x):
\(\color{blue}{f(g(x)) = 2[g(x)] + 1 = 2(x^{2} – 3) + 1}\)
Step 2 — Distribute and simplify:
\(\color{blue}{= 2x^{2} – 6 + 1 = 2x^{2} – 5}\)
The reverse order: g(f(x))
Example (same f and g): Find (g ∘ f)(x).
Replace every x in g with f(x):
\(\color{blue}{g(f(x)) = [f(x)]^{2} – 3 = (2x + 1)^{2} – 3}\)
\(\color{blue}{= 4x^{2} + 4x + 1 – 3 = 4x^{2} + 4x – 2}\)
Evaluating at a specific value
You can substitute the inner function first, or work from the inside out.
Find (f ∘ g)(3): \(\color{blue}{g(3) = 9 – 3 = 6}\), then \(\color{blue}{f(6) = 2(6) + 1}\) = 13.
Step-by-Step Summary
- Identify which function is the outer and which is the inner.
- Write the outer function formula.
- Replace every variable in the outer function with the inner function expression.
- Expand and simplify (distribute, combine like terms).
- If evaluating at a number, either substitute into the simplified formula or evaluate the inner function first and feed the result into the outer.
Watch: Introduction to Composition of Functions (Video Lesson)
Khan Academy introduces function composition with intuitive examples, explaining the inside-out structure of f(g(x)):
Composition of Functions – Worked Examples
Example 1: \(\color{blue}{f(x) = 2x + 1}\), \(\color{blue}{g(x) = x}\)\(\color{blue}{^{2} – 3}\). Find (f ∘ g)(3).
Inside-out: \(\color{blue}{g(3) = (3)}\)² − \(\color{blue}{3 = 9 – 3 = 6}\)
Then: \(\color{blue}{f(6) = 2(6) + 1 = 13}\)
\(\color{blue}{(f &\#8728; g)(3) = 13}\)
Example 2: Same f and g. Find (g ∘ f)(2).
Inside-out: \(\color{blue}{f(2) = 2(2) + 1 = 5}\)
Then: \(\color{blue}{g(5) = (5)}\)² − \(\color{blue}{3 = 25 – 3 = 22}\)
\(\color{blue}{(g &\#8728; f)(2) = 22}\)
Example 3: Same f and g. Find (f ∘ g)(0).
\(\color{blue}{g(0) = 0 – 3}\) = −3; \(\color{blue}{f(-3) = 2(-3) + 1}\) = −5
\(\color{blue}{(f &\#8728; g)(0) = -5}\)
Example 4: Same f and g. Find (g ∘ f)(0).
\(\color{blue}{f(0) = 1}\); \(\color{blue}{g(1) = 1 – 3}\) = −2
\(\color{blue}{(g &\#8728; f)(0) = -2}\)
More Practice: Composite Functions Video
The Organic Chemistry Tutor covers composite functions with additional examples and tips for avoiding common mistakes:
Exercises for Composition of Functions
Let \(\color{blue}{f(x) = 2x + 1}\) and \(\color{blue}{g(x) = x}\)\(\color{blue}{^{2} – 3}\). Evaluate each composition.
- (f ∘ g)(x) in simplified form
- (g ∘ f)(x) in simplified form
- (f ∘ g)(−2)
- (g ∘ f)(−1)
- (f ∘ f)(1)
Answers
- \(\color{blue}{(f &\#8728; g)(x) = 2(x^{2} – 3) + 1}\) = \(\color{blue}{2x^{2} – 5}\)
- \(\color{blue}{(g &\#8728; f)(x) = (2x + 1)}\)\(\color{blue}{^{2} – 3}\) = 4x² + \(\color{blue}{4x – 2}\)
- \(\color{blue}{g(-2) = 4 – 3 = 1}\); \(\color{blue}{f(1) = 3}\). Answer: 3
- f(−1) = −1; \(\color{blue}{g(-1) = 1 – 3}\) = −2. Answer: −2
- \(\color{blue}{f(1) = 3}\); \(\color{blue}{f(3) = 7}\). Answer: 7
Free Composition of Functions Worksheet
Ready to practice on your own? Download our free Composition of Functions worksheet below, work through each problem at your own pace, and then check your answers. If a few give you trouble, scroll back up to the worked examples and try again, steady practice is the surest way to master Composition of Functions before a quiz or test.
Download Combining Functions Worksheet
Frequently Asked Questions
What is composition of two functions?
Composition of two functions means plugging one function into another, so the output of the inner function becomes the input of the outer one. It is written (f ∘ g)(x) = f(g(x)). For example, if f(x) = x² and g(x) = x + 3, then f(g(x)) = (x + 3)², and f(g(2)) = (2 + 3)² = 25.
How do you solve composition functions?
Work from the inside out. Evaluate or write the inner function first, then substitute that whole result, in parentheses, into the outer function and simplify. Example: f(x) = 3x − 1 and g(x) = x² + 2. Then f(g(x)) = 3(x² + 2) − 1 = 3x² + 5. Parentheses keep the signs and powers correct.
How do you evaluate composite functions using graphs?
Read the inner function first. To find f(g(1)), locate x = 1 on the graph of g and read its y-value. Suppose g(1) = 4. Then use 4 as the new input: find x = 4 on the graph of f and read that y-value. That number is f(g(1)). Keep track of which graph belongs to which function.
Are composite functions commutative?
No, in general f(g(x)) is not the same as g(f(x)), so order matters. With f(x) = x + 1 and g(x) = 2x, f(g(x)) = 2x + 1 but g(f(x)) = 2(x + 1) = 2x + 2. The two orders match only in special cases, such as a function and its inverse, where both compositions give x.
What grade is composite functions?
Composite functions usually appear in Algebra 1 or Algebra 2, often around grades 9 to 11, and come back in precalculus and calculus. In calculus, the chain rule is all about differentiating composite functions, so getting comfortable with f(g(x)) in algebra pays off later.
How do you find the domain of a composition of functions?
The domain of f(g(x)) includes every x in the domain of g whose output g(x) is also in the domain of f. Check both conditions. Example: f(x) = √x and g(x) = x − 4. Then f(g(x)) = √(x − 4), and the expression under the root must be at least 0, so the domain is x ≥ 4.
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