Lesson 4 of 7Rates of Change & Combining Functions

Unit 6 · Lesson 4 of 7

Combining Functions

Add, subtract, multiply and divide two functions to build a new one — and work out the domain of the result from the domains of the pieces.

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Key things to remember

1 $(f \pm g) (x) = f (x) \pm g (x)$ and $(f g) (x) = f (x) \cdot g (x)$ ; the domain of each is $D_{f} \cap D_{g}$ .
2 $(\frac{f}{g}) (x) = \frac{f ( x )}{g ( x )}$ , with domain $D_{f} \cap D_{g}$ minus every $x$ where $g (x) = 0$ .
3Domain restrictions are read off the original functions, never off the simplified result.
4A function is even if $f (- x) = f (x)$ (symmetric about the $y$ -axis) and odd if $f (- x) = - f (x)$ (symmetric about the origin).
5Even $\times$ odd $=$ odd, odd $\times$ odd $=$ even, even $\times$ even $=$ even — the signs multiply.
6On graphs, combine outputs vertically: add, subtract, multiply or divide the two heights at each input.

A small business tracks its revenue with one function and its costs with another, both depending on how many units it sells. Profit is neither of those functions — it is revenue minus cost, a brand new function built from the two. This happens constantly: total surface area is one area plus another, average cost is total cost divided by quantity, power is voltage times current. Rather than inventing a formula from scratch each time, we define arithmetic directly on functions.

The four operations

Name	Notation	Definition	Domain
Sum	$(f + g) (x)$	$f (x) + g (x)$	$D_{f} \cap D_{g}$
Difference	$(f - g) (x)$	$f (x) - g (x)$	$D_{f} \cap D_{g}$
Product	$(f g) (x)$	$f (x) \cdot g (x)$	$D_{f} \cap D_{g}$
Quotient	$(\frac{f}{g}) (x)$	$\frac{f ( x )}{g ( x )}$	$D_{f} \cap D_{g}$ , and $g (x) \neq = 0$

The four function operations. The symbol for intersection means values that belong to both domains.

Fig 1Two functions feeding one operation box to produce a single new function.

VideoAn Ontario Grade 12 lesson on combining two functions by adding and subtracting them.Watch on YouTube

The domain of a combination

To evaluate $(f + g) (x)$ you need both $f (x)$ and $g (x)$ to exist. If either one is undefined at some input, the sum is undefined there too. So the domain of any combination starts as the intersection of the two domains — the overlap, the values allowed by both. For a quotient there is one extra hazard: even where both functions are defined, division by zero is still illegal, so every $x$ with $g (x) = 0$ must be thrown out as well.

Find the restrictions before you simplify

Work out the domain from the original

f

and

g

, not from the tidied-up answer. If

f (x) = x^{2} - 4

and

g (x) = x + 2

, then

\frac{f}{g}

simplifies to

x - 2

, which looks perfectly happy at

x = - 2

. But

g (- 2) = 0

, so

x = - 2

was excluded before you cancelled and it stays excluded afterward. Cancelling never restores a lost point.

Fig 2Reading a domain off a graph as the shadow cast on the horizontal axis.

Reading combinations off graphs and tables

You do not need equations. Given two graphs on the same axes, $(f + g) (3)$ is just the height of one curve at $x = 3$ added to the height of the other at $x = 3$ . Do that for several inputs and you can sketch the sum curve point by point. Where one graph crosses the $x$ -axis, the sum simply equals the other function there; where both are positive the sum sits above both; and where $g$ crosses zero, the quotient $\frac{f}{g}$ has a break in it.

x	-2	-1	0	1	2	3
f(x)	5	4	1	0	3	5
g(x)	1	2	2	-1	4	0

Values of two functions at six inputs. Every combination can be built column by column.

$(f + g) (2) = f (2) + g (2) = 3 + 4 = 7$ .
$(f - g) (- 1) = f (- 1) - g (- 1) = 4 - 2 = 2 .$
$(f g) (0) = f (0) \cdot g (0) = 1 \times 2 = 2$ .
$(\frac{f}{g}) (1) = \frac{0}{- 1} = 0$ — a zero numerator is fine; only a zero denominator is fatal.
$(\frac{f}{g}) (3)$ is undefined, because $g (3) = 0$ .

Even and odd functions

Some functions have a symmetry that survives these operations. A function is even when replacing $x$ with $- x$ changes nothing, that is $f (- x) = f (x)$ , which makes its graph symmetric about the $y$ -axis. It is odd when replacing $x$ with $- x$ flips the sign of every output, that is $f (- x) = - f (x)$ , which makes its graph symmetric about the origin — rotate it $180°$ and it lands on itself. Most functions are neither.

AEven functions

f(−x) = f(x) for every x in the domain
Symmetric about the y-axis
Polynomials with only even powers: x², x⁴, x² + 7
cos x is even

BOdd functions

f(−x) = −f(x) for every x in the domain
Symmetric about the origin (180° rotation)
Polynomials with only odd powers: x, x³, x³ − 4x
sin x and tan x are odd

How to test, and what survives

Always substitute

- x

and simplify all the way before deciding — guessing from the shape of the formula fails on things like

x^{2} + x

, which is neither. Combinations follow rules too: even plus even is even, odd plus odd is odd, and even times odd is odd, exactly like multiplying signs.

VideoAn MHF4U lesson on deciding whether a function is even, odd or neither from its symmetry.Watch on YouTube

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Given

f (x) = x^{2} - 4

and

g (x) = x + 2

, find

(f + g) (x)

(f g) (x)

and

(\frac{f}{g}) (x)

, each with its domain.

Example 2

Let

f (x) = x - 1

and

g (x) = 5 - x

. State the domain of

(f + g)

and the domain of

\frac{f}{g}

Example 3

Classify each function as even, odd, or neither:

f (x) = x^{3} - 4 x

g (x) = x^{4} - 3 x^{2} + 1

p (x) = x^{2} + x

Example 4

Using the table of values above, evaluate

(f - g) (- 2)

(f g) (- 1)

and

(\frac{f}{g}) (3)

Where marks get lost

Reading the domain off the simplified expression, so the restriction $x \neq = - 2$ disappears after the factor $x + 2$ cancels.
Taking the union of the two domains instead of the intersection — a combination needs values allowed by both functions, not by either one.
Forgetting the extra $g (x) \neq = 0$ condition on a quotient, especially when $g$ is a radical or a trig function that is zero at several inputs.
Dropping the bracket when subtracting, so with $f (x) = x^{2}$ and $g (x) = x - 5$ they write $(f - g) (x) = x^{2} - x - 5$ instead of the correct $x^{2} - (x - 5) = x^{2} - x + 5$ .
Declaring a function odd because it has a negative term, or even because it looks symmetric — the only valid test is substituting $- x$ and simplifying.

Flashcards 8

(f + g) (x)

Domain of

f + g

Domain of

\frac{f}{g}

Even function

+ 4 more

Practice

From the table in this lesson,

f (2) = 3

and

g (2) = 4

. Find

(f + g) (2)

From the table,

f (- 1) = 4

and

g (- 1) = 2

. Find

(f g) (- 1)

f (x) = x + 3

and

g (x) = x^{2} - 9

, what is the domain of

\frac{f}{g}

f (x) = x^{4} - 2 x^{2}

even, odd, or neither?

Classify

f (x) = x^{3} + x

For

f (x) = x - 2

and

g (x) = 10 - x

, what is the smallest value in the domain of

(f + g)

For the same

f (x) = x - 2

and

g (x) = 10 - x

, how many integer values lie in the domain of

(f + g)

Order the steps for finding the domain of the quotient f/g:

1Find the domain of g on its own
2Remove every x where g(x) = 0
3Take the intersection of the two domains
4Find the domain of f on its own

Match each notation to what it means.

A. $f (x) - g (x)$
B. $f (x)$ divided by $g (x)$ , with $g (x) \neq = 0$
C. $f (x)$ multiplied by $g (x)$
D.The inputs allowed by both functions

(f - g) (x)

(f g) (x)

(\frac{f}{g}) (x)

D_{f} \cap D_{g}

f

is even and

g

is odd, what is

f g

Let

f (x) = 2 x + 1

and

g (x) = x^{2} - 3

. Find

(f - g) (4)

f (x) = \frac{1}{x}

and

g (x) = x - 5

. State the domain of

(f + g)

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