2.12 Algebraic Combination of Functions

If $f$ and $g$ are two functions, we can define new functions $f + g$ , $f - g$ , $f . g$ , and $f / g$ by the formulas:

$\begin{aligned} (f + g) (x) & = f (x) + g (x) \\ (f - g) (x) & = f (x) - g (x) \\ (f . g) (x) = & f (x) . g (x) \\ (f / g) (x) & = \frac{f (x)}{g (x)} \end{aligned}$
The domains of $f + g$ , $f - g$ , $f . g$ consist of all $x$ for which both $f (x)$ and $g (x)$ are defined:
${x | x \in D o m (f) and x \in D o m (g)}$

That is, the domains of these functions are the intersection of the domain of

f

and the domain of

g

D o m (f) \cap D o m (g)

The domain of $f / g$ is trickier. Its domain consists of all $x$ for which $f (x)$ and $g (x)$ are defined but $g (x) \neq 0$ , because division by zero is not defined. Therefore:

$D o m (f / g) = {x | x \in D o m (f) and x \in D o m (g) and g (x) \neq 0},$ which also can be written as:
$D o m (f / g) = D o m (f) \cap D o m (g) \cap {x | g (x) \neq 0} .$

Example 1

Let

f (x) = \frac{1}{x - 3}

and

g (x) = \sqrt{x - 2}

.
(a) Find the functions

f + g, f - g, f \cdot g

f / g

and

g / f

and their domains.

(b) Find $(f + g) (5), (f - g) (5), (f g) (5)$ , $(f / g) (5)$ , and $(g / f) (5)$ .

Solution

(a) The domain of

f

{x | x \neq 3}

R - {3}

because division by zero is not defined. The square root of negative numbers is not defined [square root of a negative number becomes imaginary], thus the domain of

g

{x | x \geq 2}

. The intersection of the domains of

f

and

g

{x | x \geq 2 and x \neq 3} = [2, 3) \cup (3, \infty) .

Thus we have

\begin{aligned} (f + g) (x) & = f (x) + g (x) = \frac{1}{x - 3} + \sqrt{x - 2} \\ (f - g) (x) & = f (x) - g (x) = \frac{1}{x - 3} - \sqrt{x - 2} \\ (f g) (x) & = f (x) \cdot g (x) = \frac{\sqrt{x - 2}}{x - 3}, \end{aligned}

and the domains of all of these functions are

{x | x \geq 2 and x \neq 3} = [2, 3) \cup (3, \infty) . (try them!)

To determine the domain of

(f / g) (x)

, from this set of numbers we need to exclude 2 because

g (2) = 0.

That is, the domain of

(f / g) (x)

{x | x > 2 and x \neq 3} = (2, 3) \cup (3, \infty),

and

(\frac{f}{g}) (x) = \frac{1}{(x - 3) \sqrt{x - 2}} .

Because

f

does not vanish (

f (x) \neq 0

) in its domain,* the domain of

(g / f) (x)

is the same as

D o m (f) \cap D o m (g) = [2, 3) \cup (3, \infty)

,and

(\frac{g}{f}) (x) = \frac{\sqrt{x - 2}}{\frac{1}{x - 3}} = (x - 3) \sqrt{x - 2} .

(b) Because $x = 5$ is in the domain of each function, each of these values exists:
$\begin{aligned} (f + g) (5) & = f (5) + g (5) = \frac{1}{5 - 3} + \sqrt{5 - 2} = \frac{1}{2} + \sqrt{3} \\ (f - g) (5) & = f (5) - g (5) = \frac{1}{5 - 3} - \sqrt{5 - 2} = \frac{1}{2} - \sqrt{3} \\ (f g) (5) & = f (5) \cdot g (5) = \frac{1}{5 - 3} \cdot \sqrt{5 - 2} = \frac{\sqrt{3}}{2} \\ (\frac{f}{g}) (5) & = \frac{f (5)}{g (5)} = \frac{\frac{1}{5 - 3}}{\sqrt{5 - 2}} = \frac{1}{2 \sqrt{3}} \\ (\frac{g}{f}) (5) & = \frac{g (5)}{f (5)} = \frac{\sqrt{5 - 2}}{\frac{1}{5 - 3}} = 2 \sqrt{3} \end{aligned}$

* You may sketch the graph of $y = 1 / (x - 3)$ to see that. Note that the graph of $y = 1 / (x - 3)$ is the graph of $y = 1 / x$ shifted horizontally 3 units to the right. See Section 2.11 for more details.

Example 2

What is the natural domain of

H (x) = \sqrt{x - 1} + \sqrt{2 - x}

Solution

We have already determined the domain of this function in Example 1 in the Section on Natural Domain and Range of a Function. But now, let’s use what we learned in this section. Let

f (x) = \sqrt{x - 1}

g (x) = \sqrt{2 - x}

, and

H = f + g

. The function

f

is defined for all

x

such that

x - 1 \geq 0

. That is,

D o m (f) = {x | x \geq 1} = [1, \infty)

. The domain of

g

consists of all

x

for which

2 - x \geq 0

, i.e.

D o m (g) = {x | x \leq 2} = (- \infty, 2]

. The intersection of the domains of

f

and

g

\begin{aligned} D o m (H) & = {x | x \leq 2 and x \geq 1} \\ = {x | 1 \leq x \leq 2} \\ = [1, 2], \end{aligned}

D o m (H) = D o m (f) \cap D o m (g) = [1, \infty) \cap (- \infty, 2] = [1, 2],

which is exactly the same as we saw in Example 1 in Section 2.3.

Example 3

Are the following functions equal?

(a) $h_{1} (x) = \sqrt{x - 1} \sqrt{2 - x}$ and $h_{2} (x) = \sqrt{(x - 1) (2 - x)}$

(b) $u_{1} (x) = \sqrt{x - 1} \sqrt{x - 2}$ and $u_{2} (x) = \sqrt{(x - 1) (x - 2)}$

Solution

The functions

h_{1} (x)

and

h_{2} (x)

, and

u_{1} (x)

and

u_{2} (x)

can be equal functions if they have the same (natural) domains. [Note that if

r

and

s

are both nonnegative, then

r s \geq 0

. Therefore if

A = \sqrt{r} \sqrt{s}

, then we can also write

A

A = \sqrt{r s} .

However, if

B = \sqrt{p q}

, without knowing that

p

and

q

are both nonnegative, we cannot write

B

\sqrt{p} \sqrt{q} .

For example, if

p = - 2

, and

q = - 3

, then we can calculate

\sqrt{(- 2) \cdot (- 3)}

but

\sqrt{- 2}

\sqrt{- 3}

are not defined.] To determine the domains of these functions, we note that:

\sqrt{} is defined if the sign of is not negative

and

sign (\times ◊) = sign () \times sign (◊)

(a) First, we should determine the signs of all expressions under

\sqrt{.}

From what we learned in this section:

\begin{aligned} D o m (h_{1}) & = D o m (\sqrt{x - 1}) \cap D o m (\sqrt{2 - x}) \\ = [1, \infty) \cap (- \infty, 2] = [1, 2], \end{aligned}

and from the above table:

D o m (h_{2}) = D o m (\sqrt{(x - 1) (2 - x)} = [1, 2] .

Therefore

h_{1} (x) = h_{2} (x)

everywhere they are defined. In other words, they are equal functions.

(b) Similarly, using the following table:

$\begin{aligned} D o m (u_{1}) & = D o m (\sqrt{x - 1}) \cap D o m (\sqrt{x - 2}) \\ = [1, \infty) \cap [2, \infty) = [2, \infty) \end{aligned}$
but
$D o m (u_{2}) = (- \infty, 1] \cup [2, \infty) .$ Therefore $u_{1} (x)$ and $u_{2} (x)$ are equal only when $x \in [2, \infty)$ . Because the domains of $u_{1} (x)$ and $u_{2} (x)$ are not equal, they are not equal functions.

Example 4

Determine the natural domain of the following functions:
(a)

f (x) = \frac{\sqrt{1 - x}}{\log x}

(b) $g (u) = \frac{e^{u}}{\sqrt{| 1 + u |}}$

Solution

Recall that

\log x

is defined when

x > 0

and

\log x = 0 \Rightarrow x = 1

(a)

\begin{aligned} D o m (f) & = D o m (\sqrt{1 - x}) \cap D o m (\log x) \cap {x | \log x \neq 0} \\ = (- \infty, 1] \cap (0, \infty) \cap {x | x \neq 1} \\ = (0, 1) . \end{aligned}

(b)

\begin{aligned} D o m (g) & = D o m (e^{u}) \cap D o m (\sqrt{| 1 + u |}) \cap {u | \sqrt{| 1 + u |} \neq 0} \\ = R \cap R \cap {u | | 1 + u | \neq 0} \\ = R \cap R \cap {u | u \neq \pm 1} \\ = {u | u \neq \pm 1} \\ = R - {- 1, 1}, \end{aligned}

which can also be written as

D o m (g) = (- \infty, - 1) \cup (- 1, 1) \cup (1, \infty) .

Graphs of Combined Functions

Suppose the graphs of $f$ and $g$ are known:

to obtain the graph of the function $f + g$ , we just add the corresponding $y$ -coordinates $f (x)$ and $g (x)$ at each $x$ that belongs both to the domain of $f$ and the domain of $g$
to obtain the graph of the function $f + g$ , we just subtract the $y$ -coordinate $g (x)$ from the corresponding $y$ -coordinate $f (x)$ at each $x$ that belongs both to the domain of $f$ and the domain of $g$
to obtain the graph of the function $f g$ , we multiply the corresponding $y$ -coordinates $f (x)$ and $g (x)$ at each $x$ that belongs both to the domain of $f$ and the domain of $g$ .