Algebra of functions

Let $f$ and $g$ be real-valued functions with domains $D_f$ and $D_g$. The four combined functions are defined pointwise — you plug the same $x$ into both $f$ and $g$, get two numbers, and combine those numbers exactly as you would combine any two real numbers:

where: $f(x)$ and $g(x)$ are the values the two original functions take at the same input $x$, and the right-hand side of each equation is an ordinary arithmetic operation on two real numbers. That is the entire "rule" half of the topic. The "domain" half is a separate question, and it is answered by set intersection, not algebra:

where: $D_f \cap D_g$ is the set of inputs where both original functions are defined, and the quotient additionally strips out any input at which $g(x)$ happens to equal zero, since division by zero is never allowed. A useful side case: for a real constant $k$, $(kf)(x) = k\cdot f(x)$ has exactly the same domain as $f$, since multiplying by a constant never introduces a new restriction. Think of the construction as always happening in two separate stages that must not be blended together. Stage one is purely mechanical: substitute $f(x)$ and $g(x)$ into the arithmetic operation. Stage two is purely set-theoretic: ask "for which $x$ does this combination make sense at all," and answer that by intersecting domains (and, for a quotient, trimming zeros of the denominator) — never by simplifying the combined expression first and reading a domain off the simplified form. That single habit, kept separate and done first, prevents the most common mistake in the whole topic.

Problem. Let $f(x) = x^2 - 1$ with domain $\mathbb{R}$, and $g(x) = \sqrt{x}$ with domain $[0,\infty)$. Find $(f+g)(x)$ and $(fg)(x)$ with their domains, then evaluate $(f+g)(4)$ and $(fg)(4)$.

Step 1. Combine the rules by substitution: $(f+g)(x) = f(x) + g(x) = x^2 - 1 + \sqrt{x}$, and $(fg)(x) = f(x)\cdot g(x) = (x^2-1)\sqrt{x}$.

Step 2. Apply the domain rule $D_f \cap D_g$. Here $D_f = \mathbb{R}$ and $D_g = [0,\infty)$, so $D_f \cap D_g = [0,\infty)$. Both $f+g$ and $fg$ share this same domain — neither is a quotient, so there is no further zero-removal step.

Step 3. Evaluate $(f+g)(4)$: $(f+g)(4) = (4^2 - 1) + \sqrt{4} = 15 + 2 = \mathbf{17}$.

Step 4. Evaluate $(fg)(4)$ the same pointwise way: $(fg)(4) = (4^2-1)\cdot\sqrt{4} = 15 \times 2 = \mathbf{30}$. Both evaluations only needed $f(4)$ and $g(4)$ individually, computed first and then combined — the general symbolic rule for $(f+g)(x)$ or $(fg)(x)$ is convenient when many evaluations are needed, but is never strictly required for a single numeric evaluation.

• Cancelling a common algebraic factor in a quotient and forgetting the original exclusion still applies, exactly as $x = -3$ stays excluded from $\frac{x+3}{x^2-9}$ even though the cancelled form $\frac{1}{x-3}$ looks fine there.
• Taking the domain of $f+g$ or $fg$ to be the union $D_f \cup D_g$ instead of the intersection $D_f \cap D_g$ — the combined function needs both $f(x)$ and $g(x)$ to exist, so it can only be defined where both individually are.
• Forgetting the extra zero-removal step that applies only to a quotient $f/g$, and mistakenly applying it to a sum, difference or product too.
• Reading the domain off the simplified algebraic form of a combined function instead of computing it from the original, unsimplified $f$ and $g$.

Ask the tutor: Try asking why $D_f \cap D_g$ and not $D_f \cup D_g$ makes sense by picking a specific $x$ that's in only one domain and checking whether $(f+g)(x)$ could possibly be computed there.