Composition and Inverse Functions

Function composition chains two functions together — the output of one becomes the input of the next. Inverse functions reverse the process entirely. Together, these ideas form the foundation for solving equations, converting units, and understanding how multi-step processes can be undone.

Composition of Functions

The composition of $f$ with $g$, written $(f \circ g)(x)$, means “apply $g$ first, then apply $f$ to the result”:

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

The function $g$ is the inner function and $f$ is the outer function. Composition is not commutative — in general, $f \circ g \neq g \circ f$.

Worked Example 1: Basic Composition

Let $f(x) = 2x + 3$ and $g(x) = x^2 - 1$.

$(f \circ g)(x) = f(g(x)) = f(x^2 - 1) = 2(x^2 - 1) + 3 = 2x^2 - 2 + 3 = 2x^2 + 1$

$(g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$

Since $2x^2 + 1 \neq 4x^2 + 12x + 8$, composition order matters.

Worked Example 2: Evaluating at a Point

Using the same $f$ and $g$, find $(f \circ g)(3)$.

Method 1 (compose, then evaluate): $(f \circ g)(x) = 2x^2 + 1$, so $(f \circ g)(3) = 2(9) + 1 = 19$.

Method 2 (evaluate step by step): $g(3) = 9 - 1 = 8$, then $f(8) = 2(8) + 3 = 19$.

Both methods give the same result. Method 2 is faster for a single evaluation; Method 1 is better when you need the full composed formula.

Domain of a Composition

The domain of $(f \circ g)(x)$ is not simply the domain of $g$. You need:

1. $x$ must be in the domain of $g$ (so $g(x)$ is defined)
2. $g(x)$ must be in the domain of $f$ (so $f(g(x))$ is defined)

Worked Example 3: Domain Restrictions from Composition

Let $f(x) = \frac{1}{x - 2}$ and $g(x) = \sqrt{x}$.

Find the domain of $(f \circ g)(x) = f(\sqrt{x}) = \frac{1}{\sqrt{x} - 2}$.

Restriction 1: $g(x) = \sqrt{x}$ requires $x \geq 0$.

Restriction 2: $f$ requires its input $\neq 2$, so $\sqrt{x} \neq 2$, meaning $x \neq 4$.

Domain of $f \circ g$: $[0, 4) \cup (4, \infty)$.

Worked Example 4: Another Domain Analysis

Let $f(x) = \ln(x)$ and $g(x) = 4 - x^2$.

$(f \circ g)(x) = \ln(4 - x^2)$

Restriction from $g$: $g$ is defined for all $x$ (no restriction from $g$ alone).

Restriction from $f$: $\ln$ requires its argument positive: $4 - x^2 > 0$, so $x^2 < 4$, giving $-2 < x < 2$.

Domain of $f \circ g$: $(-2, 2)$.

Decomposing Functions

Given a complex function, we can write it as a composition of simpler functions.

Example: Express $h(x) = (3x - 7)^5$ as $f \circ g$.

Choose $g(x) = 3x - 7$ (inner) and $f(x) = x^5$ (outer). Then $f(g(x)) = (3x - 7)^5 = h(x)$.

Example: Express $k(x) = \sqrt{\ln(x)}$ as $f \circ g$.

Choose $g(x) = \ln(x)$ and $f(x) = \sqrt{x}$. Then $f(g(x)) = \sqrt{\ln(x)} = k(x)$.

Decomposition is not unique — there may be multiple valid ways to split a function.

Inverse Functions

A function $f$ has an inverse function $f^{-1}$ if and only if $f$ is one-to-one (each output comes from exactly one input). When the inverse exists:

$f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x$

The inverse “undoes” the function: if $f(a) = b$, then $f^{-1}(b) = a$.

Important notation: $f^{-1}(x)$ is the inverse function — it does not mean $\frac{1}{f(x)}$. The reciprocal would be written $(f(x))^{-1}$ or $\frac{1}{f(x)}$.

The Horizontal Line Test

A function is one-to-one if and only if every horizontal line intersects its graph at most once. If any horizontal line crosses the graph twice, the function fails the test and has no inverse (without domain restriction).

Verifying Inverses via Composition

To verify that $f$ and $g$ are inverses, check both compositions:

$f(g(x)) = x \quad \text{and} \quad g(f(x)) = x$

Both must hold on the appropriate domains.

Worked Example 5: Verification

Verify that $f(x) = 3x - 5$ and $g(x) = \frac{x + 5}{3}$ are inverses.

$f(g(x)) = f\!\left(\frac{x + 5}{3}\right) = 3 \cdot \frac{x + 5}{3} - 5 = x + 5 - 5 = x$

$g(f(x)) = g(3x - 5) = \frac{(3x - 5) + 5}{3} = \frac{3x}{3} = x$

Both compositions yield $x$, confirming the functions are inverses.

Finding Inverses Algebraically

Method: Replace $f(x)$ with $y$, swap $x$ and $y$, then solve for $y$.

Worked Example 6: Linear Function

Find $f^{-1}$ for $f(x) = \frac{2x + 1}{3}$.

Write $y = \frac{2x + 1}{3}$. Swap: $x = \frac{2y + 1}{3}$. Solve:

$3x = 2y + 1 \implies 2y = 3x - 1 \implies y = \frac{3x - 1}{2}$

$f^{-1}(x) = \frac{3x - 1}{2}$

Worked Example 7: Rational Function

Find $f^{-1}$ for $f(x) = \frac{x + 3}{x - 1}$, $x \neq 1$.

Swap: $x = \frac{y + 3}{y - 1}$. Solve:

$x(y - 1) = y + 3 \implies xy - x = y + 3 \implies xy - y = x + 3 \implies y(x - 1) = x + 3$

$y = \frac{x + 3}{x - 1}$

Remarkably, $f^{-1}(x) = \frac{x + 3}{x - 1} = f(x)$. This function is its own inverse — it is called an involution.

Worked Example 8: Radical Function

Find $f^{-1}$ for $f(x) = \sqrt{x - 2} + 5$, where $x \geq 2$.

Swap: $x = \sqrt{y - 2} + 5$. Solve:

$x - 5 = \sqrt{y - 2} \implies (x - 5)^2 = y - 2 \implies y = (x - 5)^2 + 2$

$f^{-1}(x) = (x - 5)^2 + 2, \quad x \geq 5$

The domain of $f^{-1}$ is the range of $f$: since $\sqrt{x - 2} \geq 0$, $f(x) \geq 5$.

Restricting Domains to Create Inverses

Many common functions are not one-to-one on their entire domain. By restricting the domain, we can create a one-to-one function that has an inverse.

Worked Example 9: Quadratic with Restricted Domain

$f(x) = x^2$ is not one-to-one on $(-\infty, \infty)$ because, for instance, $f(3) = f(-3) = 9$.

Restrict to $x \geq 0$: $f(x) = x^2$, $x \geq 0$.

Now $f$ is one-to-one (the right half of the parabola passes the horizontal line test).

Find the inverse: $x = y^2$, $y \geq 0$, so $y = \sqrt{x}$.

$f^{-1}(x) = \sqrt{x}, \quad x \geq 0$

We could also restrict to $x \leq 0$, giving $f^{-1}(x) = -\sqrt{x}$.

Graphing Inverses

The graph of $f^{-1}$ is the reflection of the graph of $f$ over the line $y = x$. Every point $(a, b)$ on $f$ corresponds to $(b, a)$ on $f^{-1}$.

This means:

• The domain of $f$ becomes the range of $f^{-1}$ (and vice versa)
• Intercepts reflect: if $f$ has $y$-intercept $(0, b)$, then $f^{-1}$ has $x$-intercept $(b, 0)$

Real-World Application: Nursing Dosage Conversions

In nursing, drug concentrations often require converting between units. If the conversion function is:

$C(x) = 0.001x + 2.5$

where $x$ is the dosage in micrograms and $C$ is the blood concentration in mg/L, the inverse tells us what dosage produces a target concentration:

$x = \frac{C - 2.5}{0.001} = 1000(C - 2.5)$

$C^{-1}(y) = 1000(y - 2.5) = 1000y - 2500$

For a target concentration of 3.2 mg/L: $C^{-1}(3.2) = 1000(3.2) - 2500 = 700$ micrograms.

Real-World Application: Temperature Conversion

The Fahrenheit-to-Celsius conversion is $C = f(F) = \frac{5}{9}(F - 32)$. The inverse converts Celsius back to Fahrenheit:

$F = f^{-1}(C) = \frac{9}{5}C + 32$

Verify: $f(f^{-1}(C)) = \frac{5}{9}\!\left(\frac{9}{5}C + 32 - 32\right) = \frac{5}{9} \cdot \frac{9}{5}C = C$.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Composition $(f \circ g)(x) = f(g(x))$ applies $g$ first, then $f$ — order matters
• The domain of $f \circ g$ requires $x$ in the domain of $g$ and $g(x)$ in the domain of $f$
• Inverse functions satisfy $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$
• To find an inverse algebraically: write $y = f(x)$, swap $x$ and $y$, solve for $y$
• A function must be one-to-one (pass the horizontal line test) to have an inverse
• Restricting the domain can make a non-one-to-one function invertible
• The graph of $f^{-1}$ is the reflection of $f$ over the line $y = x$

Return to College Algebra for more topics in this section.