Topic 2.7 Composition of Functions: form the composition of two functions, evaluate it, decompose a composite function, and determine the domain of a composition.

What this topic is asking

The College Board (Topic 2.7) wants you to compose two functions, written $(f \circ g)(x) = f(g(x))$: apply $g$ first, then apply $f$ to the result. You must evaluate compositions, decompose a given composite into an inner and an outer function, and find the domain of a composition, which depends on both functions.

Forming and evaluating a composition

Order is essential. $(f \circ g)(x) = f(g(x))$ applies $g$ then $f$, while $(g \circ f)(x) = g(f(x))$ applies $f$ then $g$, and these usually give different results.

Decomposing a composite

The reverse skill is decomposition: given a composite like $\sqrt{3x + 1}$, identify an inner function ($g(x) = 3x + 1$) and an outer function ($f(u) = \sqrt{u}$) so that $f(g(x))$ reproduces it. Decomposition is what lets you recognize structure, and it sets up the inverse and logarithm work later in the unit, where you peel functions apart one layer at a time.

The domain of a composition

For example, $f(x) = \frac{1}{x}$ and $g(x) = x - 2$ give $(f \circ g)(x) = \frac{1}{x - 2}$, with domain $x \neq 2$, coming from the requirement that $g(x) \neq 0$ since $0$ is excluded from $f$'s domain.

Composition also lets you evaluate a function at the output of another, which is how layered models are built. If a balloon's radius grows with time as $r(t)$, and its volume depends on radius as $V(r)$, then the composition $V(r(t))$ gives the volume directly as a function of time. The exam exploits this chaining in tables: given values of $f$ and $g$ at several inputs, you may be asked for $f(g(3))$ by first reading $g(3)$ from the table and then reading $f$ at that result. Working strictly from the inside out, one table lookup at a time, keeps the order straight and is the same discipline that makes symbolic compositions reliable.

A point that distinguishes careful work is checking both domains rather than only the surface formula. A composition like $f(g(x)) = \sqrt{x - 4}$ shows its restriction openly, but a case like $\sqrt{g(x)}$ where $g$ itself has a restricted domain hides one constraint inside another. Tracking the inner function's domain first, then the requirement that its output be acceptable to the outer function, guarantees you catch every restriction, which is exactly the layered reasoning the exam tests with composition.

Try this

Q1. If $f(x) = 2x$ and $g(x) = x + 5$, find $(g \circ f)(3)$. [1 point]

• Cue. $f(3) = 6$, then $g(6) = 6 + 5 = 11$.

Q2. Decompose $h(x) = \sqrt{x^2 + 1}$ into an inner and outer function. [1 point]

• Cue. Inner $g(x) = x^2 + 1$, outer $f(u) = \sqrt{u}$, so $h = f \circ g$.