Algebra of Real Functions — Sum, Difference, Product and Quotient

Given two real functions $f$ and $g$, each with its own domain, new functions can be built from them by combining their output values pointwise, at every $x$ common to both domains.

Sum. $(f+g)(x) = f(x) + g(x)$, defined for every $x$ in $D_f \cap D_g$ (the intersection of the two domains) — since both $f(x)$ and $g(x)$ must individually be defined before they can be added.

Difference. $(f-g)(x) = f(x) - g(x)$, again defined on $D_f \cap D_g$.

Product. $(fg)(x) = f(x)\cdot g(x)$, defined on $D_f \cap D_g$.

Quotient. $\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, defined on $D_f \cap D_g$ with the further restriction that $g(x) \ne 0$ — so the domain of $f/g$ is $\{x \in D_f \cap D_g : g(x) \ne 0\}$, which can be strictly smaller than $D_f \cap D_g$ itself.

Scalar multiple. For a fixed real number $k$, $(kf)(x) = k\cdot f(x)$, defined on $D_f$ (the domain of $f$ alone).

In every one of these operations, the new function's domain is worked out first, from the two original domains, before its rule is used — a very common error is to simplify the algebraic expression for $f/g$ (cancelling a common factor, say) and forget that a point excluded by $g(x) = 0$ stays excluded even if the simplified formula looks defined there.

Worked illustration. Let $f(x) = x^2 + 1$ and $g(x) = 2x - 3$, both with domain $\mathbb{R}$. Then $(f+g)(x) = x^2+2x-2$, $(f-g)(x) = x^2-2x+4$, $(fg)(x) = (x^2+1)(2x-3) = 2x^3-3x^2+2x-3$, all defined on $\mathbb{R}$; but $\left(\dfrac{f}{g}\right)(x) = \dfrac{x^2+1}{2x-3}$ is defined only on $\mathbb{R} - \left\{\dfrac{3}{2}\right\}$, since $g\left(\dfrac{3}{2}\right)=0$. …