Operations on functions work much like operations on numbers, but the operation is performed
on the output values of the functions. If \(f\) and \(g\) are functions, then their sum,
difference, product, and quotient are new functions.
1. Main formulas
The sum of two functions is formed by adding the two output values at the same input \(x\).
\[
\begin{aligned}
(f+g)(x) &= f(x)+g(x)
\end{aligned}
\]
The difference of two functions is formed by subtracting one output value from the other.
\[
\begin{aligned}
(f-g)(x) &= f(x)-g(x)
\end{aligned}
\]
The product of two functions is formed by multiplying the output values.
\[
\begin{aligned}
(f\cdot g)(x) &= f(x)\cdot g(x)
\end{aligned}
\]
The quotient of two functions is formed by dividing the output values, but the denominator
must not be zero.
\[
\begin{aligned}
\left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)},\\
g(x)&\ne 0.
\end{aligned}
\]
2. Domain of the result
The domain of a function operation depends on where the input functions are defined. For sums,
differences, and products, the result is defined where all functions involved are defined.
If \(D_f\) is the domain of \(f\) and \(D_g\) is the domain of \(g\), then:
\[
\begin{aligned}
D_{f+g} &= D_f\cap D_g,\\
D_{f-g} &= D_f\cap D_g,\\
D_{f\cdot g} &= D_f\cap D_g.
\end{aligned}
\]
For a quotient, one extra restriction is needed: the denominator must not equal zero.
\[
\begin{aligned}
D_{f/g}
&= \{x\in D_f\cap D_g:\ g(x)\ne 0\}.
\end{aligned}
\]
This is why quotient operations often create excluded values, holes, or vertical asymptotes.
3. Worked example: sum of functions
Let
\[
\begin{aligned}
f(x)&=x^2,\\
g(x)&=2x+1.
\end{aligned}
\]
The sum is found by adding the expressions.
\[
\begin{aligned}
(f+g)(x)
&= f(x)+g(x)\\
&= x^2+(2x+1)\\
&= x^2+2x+1.
\end{aligned}
\]
Since both \(f(x)=x^2\) and \(g(x)=2x+1\) are polynomials, both are defined for every real
number. Therefore, the domain of the sum is all real numbers.
\[
\begin{aligned}
D_{f+g} &= \mathbb{R}.
\end{aligned}
\]
4. Worked example: quotient of functions
Let
\[
\begin{aligned}
f(x)&=x^2-9,\\
g(x)&=x-3.
\end{aligned}
\]
The quotient is
\[
\begin{aligned}
\left(\frac{f}{g}\right)(x)
&= \frac{x^2-9}{x-3}\\
&= \frac{(x-3)(x+3)}{x-3}.
\end{aligned}
\]
Algebraically, the expression simplifies to \(x+3\) when \(x\ne3\), but the original quotient
still has \(x=3\) excluded because the original denominator is zero there.
\[
\begin{aligned}
x-3&\ne0,\\
x&\ne3.
\end{aligned}
\]
Therefore, the domain is
\[
\begin{aligned}
D_{f/g}
&=(-\infty,3)\cup(3,\infty).
\end{aligned}
\]
5. Polynomial shortcuts
Polynomial operations follow ordinary algebra rules. If two functions are polynomials, their
sum, difference, and product are also polynomials. The resulting domain is always
\(\mathbb{R}\).
6. Using more than two functions
Function operations can be extended to more than two functions. For example, if \(f\), \(g\),
and \(h\) are all defined at the same \(x\), then:
\[
\begin{aligned}
(f+g+h)(x)&=f(x)+g(x)+h(x),\\
(f\cdot g+h)(x)&=f(x)\cdot g(x)+h(x),\\
\left(\frac{f\cdot g}{h}\right)(x)&=\frac{f(x)\cdot g(x)}{h(x)}.
\end{aligned}
\]
In the last expression, \(h(x)\ne0\) is required because \(h(x)\) is the denominator.
7. Summary
- Write each input function clearly.
- Choose the operation: sum, difference, product, or quotient.
- Substitute the function expressions into the operation rule.
- Simplify algebraically when possible.
- Find the domain by intersecting input domains.
- For quotients, also exclude values that make the denominator zero.
