Adding, Multiplying, and Dividing Functions, foundations for calculus lesson

A shop’s profit is its revenue minus its costs, and both depend on how many items it sells. To find the profit at $40$ items you take the revenue at $40$ and the cost at $40$ and subtract. That is all it means to combine two functions. The care goes into one question, which inputs both functions can actually handle.

Key terms

Pointwise operation

Definition A pointwise operation builds a new function from two functions $f$ and $g$ by taking one input $x$ , evaluating both $f(x)$ and $g(x)$ at that same $x$ , and combining the two output numbers by an arithmetic operation. The new function is defined only at inputs where both outputs exist and the operation makes sense.

In plain words Feed the same number into both functions, then add, subtract, multiply or divide the two answers. You never mix outputs from different inputs.

Example Let $f(x) = x^2$ and $g(x) = 3x + 1$ . At $x = 2$ the outputs are $f(2) = 4$ and $g(2) = 7$ , so the pointwise sum at $2$ is $4 + 7 = 11$ .

Common domain

Definition The common domain of $f$ and $g$ is the intersection of their domains,

\operatorname{dom} f \cap \operatorname{dom} g,

the set of inputs at which both $f$ and $g$ are defined. It is the domain of $f + g$ , $f - g$ and $fg$ .

In plain words An input is allowed only if both functions accept it. Find each domain separately, then keep the overlap.

Example The function $f(x) = \sqrt{x - 1}$ needs $x \ge 1$ , and $g(x) = \frac{1}{x - 3}$ needs $x \ne 3$ . Their common domain is

[1, 3) \cup (3, \infty).

Sum of functions

Definition The sum of $f$ and $g$ is the function $f + g$ defined by

\begin{gathered} (f + g)(x) = f(x) + g(x), \\[0.4em] x \in \operatorname{dom} f \cap \operatorname{dom} g. \end{gathered}

In plain words At each allowed input, the height of $f + g$ is the height of $f$ plus the height of $g$ .

Example Let $f(x) = \sqrt{x}$ and $g(x) = \sqrt{4 - x}$ . The first needs $x \ge 0$ and the second needs $x \le 4$ , so $f + g$ has domain $[0, 4]$ . At $x = 1$ it gives $1 + \sqrt{3}$ , and at $x = 2$ it gives $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$ .

Difference of functions

Definition The difference of $f$ and $g$ is the function $f - g$ defined by

\begin{gathered} (f - g)(x) = f(x) - g(x), \\[0.4em] x \in \operatorname{dom} f \cap \operatorname{dom} g. \end{gathered}

Order matters, since $(g - f)(x) = -(f - g)(x)$ .

In plain words Subtract the second output from the first at the same input. Swapping the two functions flips the sign of every answer.

Example For $f(x) = x^2 + 1$ and $g(x) = 2x$ , both defined everywhere,

\begin{aligned} (f - g)(x) &= x^2 - 2x + 1 \\[0.4em] &= (x - 1)^2. \end{aligned}

At $x = 3$ this gives $10 - 6 = 4$ , while $(g - f)(3) = -4$ .

Product of functions

Definition The product of $f$ and $g$ is the function $fg$ defined by

\begin{gathered} (fg)(x) = f(x)\,g(x), \\[0.4em] x \in \operatorname{dom} f \cap \operatorname{dom} g. \end{gathered}

Here $fg$ means multiplying outputs. It is not the composition $f \circ g$ , given by $(f \circ g)(x) = f(g(x))$ , which feeds the output of $g$ into $f$ .

In plain words Multiply the two heights at the same input. Plugging one function into the other is a different operation with a different notation.

Example Let $f(x) = \sqrt{x}$ and $g(x) = x - 4$ . The product

(fg)(x) = \sqrt{x}\,(x - 4)

has domain $[0, \infty)$ , and $(fg)(9) = 3 \cdot 5 = 15$ . The composition gives $f(g(9)) = \sqrt{5}$ instead, and $f(g(x)) = \sqrt{x - 4}$ has the different domain $[4, \infty)$ .

Quotient of functions

Definition The quotient of $f$ by $g$ is the function $\frac{f}{g}$ defined by

\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}

for every $x$ in $\operatorname{dom} f \cap \operatorname{dom} g$ with $g(x) \ne 0$ . The function $g$ is called the denominator function. Every input where the denominator function equals zero is excluded, in addition to the inputs outside the common domain.

In plain words Divide the outputs at the same input, but only where both functions work and the bottom output is not zero.

Example Let $f(x) = \sqrt{x + 1}$ and $g(x) = x^2 - 4$ . The common domain is $[-1, \infty)$ , and $g(x) = 0$ at $x = 2$ and $x = -2$ . Only $2$ lies in $[-1, \infty)$ , so the domain of $\frac{f}{g}$ is $[-1, 2) \cup (2, \infty)$ . At $x = 3$ the quotient is $\frac{2}{5}$ .

Domain restriction

Definition The domain of a combined function is fixed by the original functions, as described above. If the formula is then simplified, the simplified formula agrees with the combined function only on that domain. Every restriction inherited from $f$ , from $g$ and from $g(x) \ne 0$ stays in force, even when the simplified formula would make sense at more inputs.

In plain words Find the domain first, from the functions as given. Simplifying may tidy the formula, but it never adds inputs back.

Example Let $f(x) = x - 1$ and $g(x) = \sqrt{x} - 1$ . The function $g$ needs $x \ge 0$ , and $g(x) = 0$ at $x = 1$ , so $\frac{f}{g}$ has domain $[0, 1) \cup (1, \infty)$ . On that domain

\begin{aligned} \frac{x - 1}{\sqrt{x} - 1} &= \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} - 1} \\[0.4em] &= \sqrt{x} + 1. \end{aligned}

The formula $\sqrt{x} + 1$ is fine at $x = 1$ , but the quotient is not defined there, so its graph has a hole at $(1, 2)$ . In the same way $\sqrt{x} \cdot \sqrt{x} = x$ holds only for $x \ge 0$ , so that product is the line $y = x$ kept only on $[0, \infty)$ .

Check yourself

Let $f(x) = \sqrt{x + 3}$ and $g(x) = \sqrt{2 - x}$ . Find the domain of $f + g$ and the value $(f + g)(1)$ .

Show the answer

The function $f$ needs $x \ge -3$ and $g$ needs $x \le 2$ , so the domain is $[-3, 2]$ . The value is $\sqrt{4} + \sqrt{1} = 3$ .

Let $f(x) = x^2 - 4$ and $g(x) = x^2 + x - 2$ . Find the domain of $\frac{f}{g}$ , simplify it and evaluate it at $x = 0$ .

Show the answer

The denominator factors as $(x - 1)(x + 2)$ , so the domain is all real numbers except $1$ and $-2$ . For those inputs

\frac{(x - 2)(x + 2)}{(x - 1)(x + 2)} = \frac{x - 2}{x - 1},

and $x = -2$ stays excluded. At $x = 0$ the value is $\frac{-2}{-1} = 2$ .

For $f(x) = \sqrt{x}$ and $g(x) = x + 5$ , compute $(fg)(4)$ and $f(g(4))$ .

Show the answer

The product multiplies outputs, so $(fg)(4) = 2 \cdot 9 = 18$ . The composition feeds $g(4) = 9$ into $f$ , so $f(g(4)) = 3$ .

Adding, Multiplying, and Dividing Functions

Key terms

Pointwise operation

Common domain

Sum of functions

Difference of functions

Product of functions

Quotient of functions

Domain restriction

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus