1.5 Function Arithmetic

Vanessa Coffelt

1.5 Function Arithmetic

1.5.1 Function Arithmetic

In this section, we begin our study of what can be considered as the algebra of functions by defining function arithmetic.

Given two real numbers, we have four primary arithmetic operations available to us: addition, subtraction, multiplication, and division (provided we don’t divide by $0$ .) As the functions we study in this text have ranges which are sets of real numbers, it makes sense we can extend these arithmetic notions to functions.

For example, to add two functions means we add their outputs; to subtract two functions, we subtract their outputs, and so on and so forth. More formally, given two functions $f$ and $g$ , we define a new function $f+g$ whose rule is determined by adding the outputs of $f$ and $g$ . That is $(f+g)(x) = f(x) + g(x)$ . While this looks suspiciously like some kind of distributive property, it is nothing of the sort. The ` $+$ ‘ sign in the expression ` $f+g$ ‘ is part of the name of the function we are defining,^[1] whereas the plus sign ` $+$ ‘ sign in the expression $f(x) + g(x)$ represents real number addition: we are adding the output from $f$ , $f(x)$ with the output from $g$ , $g(x)$ to determine the output from the sum function, $(f+g)(x)$ .

Of course, in order to define $(f+g)(x)$ by the formula $(f+g)(x) = f(x) + g(x)$ , both $f(x)$ and $g(x)$ need to be defined in the first place; that is, $x$ must be in the domain of $f$ and the domain of $g$ . You’ll recall^[2] this means $x$ must be in the intersection of the domains of $f$ and $g$ . We define the following.

Definition 1.13

Suppose $f$ and $g$ are functions and $x$ is in both the domain of $f$ and the domain of $g$ .

The sum of $f$ and $g$ , denoted $f+g$ , is the function defined by the formula
$(f+g)(x) = f(x) + g(x)$
The difference of $f$ and $g$ , denoted $f-g$ , is the function defined by the formula
$(f-g)(x) = f(x) - g(x)$
The product of $f$ and $g$ , denoted $fg$ , is the function defined by the formula
$(fg)(x) = f(x)g(x)$
The quotient of $f$ and $g$ , denoted $\dfrac{f}{g}$ , is the function defined by the formula
$\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}, \text{ provided } g(x) \neq 0.$

We put these definitions to work for us in the next example.

Example 1.5.1

Example 1.5.1.1a

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(f+g)(1)$

Solution:

Compute and simplify $(f+g)(1)$ .

By definition, $(f+g)(1) = f(1) + g(1)$ .

We find $f(1) = 6(1)^2-2(1) = 4$ and $g(1) = 3 - \frac{1}{1} = 2$ .

So we get

$\begin{array}{rcl} (f+g)(1) &=& 4+2 \\ &=& 6. \end{array}$

Example 1.5.1.1b

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(s-f)(-1)$

Solution:

Compute and simplify $(s-f)(-1)$ .

To find $(s-f)(-1) = s(-1) - f(-1)$ , we need both $s(-1)$ and $f(-1)$ .

To get $s(-1)$ , we look to the graph of $y = s(t)$ and look for the $y$ -coordinate of the point on the graph with the $t$ -coordinate of $-1$ . While not labeled directly, we infer the point $(-1,-2)$ is on the graph which means $s(-1) = -2$ .

For $f(-1)$ , we compute: $f(-1) = 6(-1)^2-2(-1) = 8$ .

Putting it all together, we get

$\begin{array}{rcl} (s-f)(-1) &=& (-2) -(8) \\ &=& -10. \end{array}$

Example 1.5.1.1c

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(fg)(2)$

Solution:

Compute and simplify $(fg)(2)$ .

Because $(fg)(2) = f(2)g(2)$ , we first compute $f(2)$ and $g(2)$ .

We find $f(2) = 6(2)^2-2(2) = 20$ and $g(2) = 3- \frac{1}{2} = \frac{5}{2}$ .

So

$\begin{array}{rcl} (fg)(2) &=& f(2) g(2) \\ &=& (20)\left(\frac{5}{2}\right) \\ &=& 50. \end{array}$

Example 1.5.1.1d

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$\left( \frac{s}{h} \right) (0)$

Solution:

Compute and simplify $\left( \frac{s}{h} \right) (0)$ .

By definition, $\left( \frac{s}{h} \right)(0) = \frac{s(0)}{h(0)}$ .

As $(0, -2)$ is on the graph of $y=s(t)$ , we know $s(0) = -2$ .

Likewise, the ordered pair $(0, \sqrt{2}) \in h$ , so $h(0) = \sqrt{2}$ .

We get

$\begin{array}{rcl} \left( \frac{s}{h} \right)(0) &=& \frac{s(0)}{h(0)} \\ &=& \frac{-2}{\sqrt{2}} \\ &=& -\sqrt{2}. \end{array}$

Example 1.5.1.1e

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$((s+g)+h)(3)$

Solution:

Compute and simplify $((s+g)+h)(3)$ .

The expression $((s+g)+h)(3)$ involves three functions. Fortunately, they are grouped so that we can apply Definition 1.13 by first considering the sum of the two functions $(s+g)$ and $h$ , then to the sum of the two functions $s$ and $g$ : $((s+g)+h)(3) = (s+g)(3)+h(3) = (s(3)+g(3))+h(3)$ .

To get $s(3)$ , we look to the graph of $y = s(t)$ . We infer the point $(3,2)$ is on the graph of $s$ , so $s(3) = 2$ .

We compute $g(3) = 3-\frac{1}{3} = \frac{8}{3}$ .

To find $h(3)$ , we note $(3, -6) \in h$ , so $h(3) = -6$ .

Hence,

$\begin{array}{rcl} ((s+g)+h)(3) &=& (s+g)(3)+h(3) \\ &=& (s(3)+g(3))+h(3) \\ &=& \left(2+\frac{8}{3}\right) + (-6) \\ &=& -\frac{4}{3}. \end{array}$

Example 1.5.1.1f

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(s+(g+h))(3)$

Solution:

Compute and simplify $(s+(g+h))(3)$ .

The expression $(s+(g+h))(3)$ is very similar to the previous problem, $((s+g)+h)(3)$ except that the $g$ and $h$ are grouped together here instead of the $s$ and $g$ .

We proceed as above applying Definition 1.13 twice and find $(s+(g+h))(3) = s(3) + (g+h)(3) = s(3)+(g(3)+h(3))$ .

Substituting the values for $s(3)$ , $g(3)$ and $h(3)$ , we get $(s+(g+h))(3) = 2 + \left(\frac{8}{3} + (-6)\right) = -\frac{4}{3}$ , which, not surprisingly, matches our answer to the previous problem.

Example 1.5.1.1g

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$\left( \frac{f+h}{s} \right) (3)$

Solution:

Compute and simplify $\left( \frac{f+h}{s} \right) (3)$ .

Once again, we find the expression $\left(\frac{f+h}{s}\right)(3)$ has more than two functions involved. As with all fractions, we treat ` $-$ ‘ as a grouping symbol and interpret $\left(\frac{f+h}{s}\right)(3) = \frac{(f+h)(3)}{s(3)} = \frac{f(3)+h(3)}{s(3)}$ .

We compute $f(3) = 6(3)^2-2(3) = 48$ and have $h(3) = -6$ and $s(3) = 2$ from above.

Hence,

$\begin{array}{rcl}\left(\frac{f+h}{s}\right)(3) &=& \frac{f(3)+h(3)}{s(3)} \\ &=& \frac{48+(-6)}{2}\\ &=& 21. \end{array}$

Example 1.5.1.1h

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(f(g-h))(-2)$

Solution:

Compute and simplify $(f(g-h))(-2)$ .

We need to need to exercise caution in parsing $(f(g-h))(-2)$ . In this context, $f$ , $g$ , and $h$ are all functions, so we interpret $(f(g-h))$ as the function and $-2$ as the argument. We view the function $f(g-h)$ as the product of $f$ and the function $g-h$ .

Hence, $(f(g-h))(-2) = f(-2) [(g-h)(-2)] = f(-2) [g(-2) - h(-2)]$ .

We compute $f(-2) = 6(-2)^2-2(-2) = 28$ , and $g(-2)= 3 - \frac{1}{-2} = 3 + \frac{1}{2} = \frac{7}{2} = 3.5$ . Because $(-2, 0.4) \in h$ , $h(-2) = 0.4$ .

Putting this altogether, we get

$\begin{array}{rcl} (f(g-h))(-2) &=& f(-2) [(g-h)(-2)] \\ &=& f(-2) [g(-2) - h(-2)] \\ &=& 28(3.5-0.4) \\ &=& 28(3.1)\\ &=& 86.8. \end{array}$

Example 1.5.1.2a

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

State the domain of each of the following functions:

$hg$

Solution:

State the domain of $hg$ .

To find the domain of $hg$ , we need to find the real numbers in both the domain of $h$ and the domain of $g$ .

The domain of $h$ is $\{ -3, -2, 0, 3 \}$ and the domain of $g$ is $\{ t \in \mathbb{R} \, | \, t > 0 \}$ so the only real number in common here is $3$ .

Hence, the domain of $hg$ is $\{ 3\}$ , which may be small, but it’s better than nothing.^[3]

Example 1.5.1.2b

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

State the domain of each of the following functions:

$\frac{f}{s}$

Solution:

State the domain of $\frac{f}{s}$ .

To find the domain of $\frac{f}{s}$ , we first note the domain of $f$ is all real numbers, but that the domain of $s$ , based on the graph, is just $[-2, \infty)$ .

Moreover, $s(t) = 0$ when $t=1$ , so we must exclude this value from the domain of $\frac{f}{s}$ .

Hence, we are left with $[-2, 1) \cup (1, \infty)$ .

Example 1.5.1.3a

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Determine expressions for the functions below. State the domain for each.

$(fg)(x)$

Solution:

Determine an expression for $(fg)(x)$ . Then state the domain of the function.

By definition, $(fg)(x) = f(x)g(x)$ .

We are given $f(x) = 6x^2-2x$ and $g(t) = 3 - \frac{1}{t}$ so $g(x) = 3 - \frac{1}{x}$ . Hence,

$\begin{array}{rclr} (fg)(x) & = & f(x)g(x) & \\ & = & \left( 6x^2-2x \right) \left( 3 - \dfrac{1}{x} \right) & \\ & = & 6x^2(3) - 6x^2 \left(\dfrac{1}{x}\right) - 2x(3) + 2x \left(\dfrac{1}{x}\right) & \text{distribute} \\ & = & 18x^2 - 6x - 6x + 2 & \\ & = & 18x^2 - 12x + 2 & \\ \end{array}$

To find the domain of $fg$ , we note the domain of $f$ is all real numbers, $(-\infty, \infty)$ whereas the domain of $g$ is restricted to $\{ t \in \mathbb{R} \, | \, t > 0 \} = (0, \infty)$ .

Hence, the domain of $fg$ is likewise restricted to $(0, \infty)$ .

Note if we relied solely on the simplified formula for $(fg)(x) = 18x^2 - 12x + 2$ , we would have obtained the incorrect answer for the domains of $fg$ .

Example 1.5.1.3b

Consider the following functions:

$f(x) = 6x^2 - 2x$
$g(t) = 3-\dfrac{1}{t}$ , $t > 0$
$h = \{ (-3,2), (-2,0.4), (0,\sqrt{2}), (3, -6) \}$
$s$ whose graph is given below:

Determine expressions for the functions below. State the domain for each.

$\left(\dfrac{g}{f}\right)(t)$

Solution:

Determine an expression for $\left( \frac{g}{f} \right) (t)$ . Then state the domain of the function.

To find an expression for $\left(\frac{g}{f}\right)(t) = \frac{g(t)}{f(t)}$ we first note $f(t) = 6t^2-2t$ and $g(t) = 3 - \frac{1}{t}$ .

Hence:

$\begin{array}{rclr} \left( \dfrac{g}{f}\right)(t) & = & \dfrac{g(t)}{f(t)} & \\ & = & \dfrac{3-\dfrac{1}{t}}{6t^2 - 2t} & \\[12pt] &=& \dfrac{3-\dfrac{1}{t}}{6t^2 - 2t} \cdot \dfrac{t}{t} & \text{simplify compound fractions} \\[12pt] & = & \dfrac{\left(3-\dfrac{1}{t}\right) t}{\left(6t^2 - 2t\right)t} & \\[12pt] &=& \dfrac{3t-1}{\left(6t^2 - 2t\right)t} & \\[12pt] & = & \dfrac{3t-1}{2t^2(3t-1)} & \\[12pt] &=& \dfrac{\cancelto{1}{(3t-1)}}{2t^2\cancel{(3t-1)}} & \text{factor and divide out} \\[12pt] & = & \dfrac{1}{2t^2} & \text{provided } t\neq \dfrac{1}{3} \\ \end{array}$

A few remarks are in order. First, in number 1 parts 1e through 1h, we first encountered combinations of three functions despite Definition 1.13 only addressing combinations of two functions at a time. It turns out that function arithmetic inherits many of the same properties of real number arithmetic. For example, we showed above that $((s+g)+h)(3) = (s+(g+h))(3)$ . In general, given any three functions $f$ , $g$ , and $h$ , $(f+g)+h = f +(g+h)$ that is, function addition is assocative. To see this, choose an element $x$ common to the domains of $f$ , $g$ , and $h$ . Then

$\begin{array}{rclr} ((f+g)+h)(x) & = & (f+g)(x)+h(x) & \text{definiton of $((f+g)+h)(x)$} \\ & = & (f(x)+g(x))+h(x) & \text{definition of $(f+g)(x)$} \\ & = & f(x) + (g(x)+h(x)) & \text{associative property of real number addition} \\ & = & f(x) + (g+h)(x) & \text{definition of $(g+h)(x)$} \\ & = & (f+(g+h))(x) & \text{definition of $(f+(g+h))(x)$} \\ \end{array}$

The key step to the argument is that $(f(x)+g(x))+h(x) = f(x) + (g(x)+h(x))$ which is true courtesy of the associative property of real number addition. And just like with real number addition, because function addition is associative, we may write $f+g+h$ instead of $(f+g)+h$ or $f+(g+h)$ even though, when it comes down to computations, we can only add two things together at a time.^[4]

For completeness, we summarize the properties of function arithmetic in the theorem below. The proofs of the properties all follow along the same lines as the proof of the associative property and are left to the reader. We investigate some additional properties in the exercises.

Theorem 1.5

Suppose $f$ , $g$ and $h$ are functions.

Commutative Law of Addition: $f+g =g+f$
Associative Law of Addition: $(f+g) + h = f+(g+h)$
Additive Identity: The function $Z(x) = 0$ satisfies: $f+Z = Z+f = f$ for all functions $f$ .
Additive Inverse: The function $F(x) = -f(x)$ for all $x$ in the domain of $f$ satisfies:
$f+F=F+f=Z.$
Commutative Law of Multiplication: $fg = gf$
Associative Law of Multiplication: $(fg)h =f(gh)$
Multiplicative Identity: The function $I(x) = 1$ satisfies: $fI = If = f$ for all functions $f$ .
Multiplicative Inverse: If $f(x) \neq 0$ for all $x$ in the domain of $f$ , then $F(x) = \dfrac{1}{f(x)}$ satisfies:
$fF=Ff = I$
Distributive Law of Multiplication over Addition: $f(g+h) = fg+fh$

In the next example, we decompose given functions into sums, differences, products and/or quotients of other functions. Note that there are infinitely many different ways to do this, including some trivial ones. For example, suppose we were instructed to decompose $f(x) = x+2$ into a sum or difference of functions. We could write $f = g+h$ where $g(x) = x$ and $h(x) = 2$ or we could choose $g(x) = 2x+3$ and $h(x) = -x-1$ . More simply, we could write $f = g+h$ where $g(x) = x+2$ and $h(x) = 0$ . We’ll call this last decomposition a `trivial’ decomposition. Likewise, if we ask for a decomposition of $f(x) = 2x$ as a product, a nontrivial solution would be $f = gh$ where $g(x) = 2$ and $h(x) = x$ whereas a trivial solution would be $g(x) = 2x$ and $h(x) = 1$ . In general, non-trivial solutions to decomposition problems avoid using the additive identity, $0$ , for sums and differences and the multiplicative identity, $1$ , for products and quotients.

Example 1.5.2

Example 1.5.2.1a

For $f(x) = x^2 - 2x$ , find functions $g$ , $h$ and $k$ to decompose $f$ nontrivially as:

$f=g-h$

Solution:

Decompose $f=g-h$ .

To decompose $f = g-h$ , we need functions $g$ and $h$ so $f(x) = (g-h)(x) = g(x) - h(x)$ .

Given $f(x) = x^2 - 2x$ , one option is to let $g(x) = x^2$ and $h(x) = 2x$ .

To check, we find $(g-h)(x) = g(x) - h(x) = x^2-2x = f(x)$ as required. In addition to checking the formulas match up, we also need to check domains.

There isn’t much work here as the domains of $g$ and $h$ are all real numbers which combine to give the domain of $f$ which is all real numbers.

Example 1.5.2.1b

For $f(x) = x^2 - 2x$ , find functions $g$ , $h$ and $k$ to decompose $f$ nontrivially as:

$f=g+h$

Solution:

Decompose $f=g+h$ .

In order to write $f = g+h$ , we need $f(x) = (g+h)(x) = g(x) + h(x)$ .

One way to accomplish this is to write $f(x) = x^2 - 2x = x^2+(-2x)$ and identify $g(x) = x^2$ and $h(x) = -2x$ .

To check, $(g+h)(x) = g(x) + h(x) = x^2 - 2x = f(x)$ .

Again, the domains for both $g$ and $h$ are all real numbers which combine to give $f$ its domain of all real numbers.

Example 1.5.2.1c

For $f(x) = x^2 - 2x$ , find functions $g$ , $h$ and $k$ to decompose $f$ nontrivially as:

$f=gh$

Solution:

Decompose $f=gh$ .

To write $f = gh$ , we require $f(x) = (gh)(x) = g(x) h(x)$ .

In other words, we need to factor $f(x)$ . We find $f(x) = x^2-2x = x(x-2)$ , so one choice is to select $g(x) = x$ and $h(x) = x-2$ . Then $(gh)(x) = g(x)h(x) = x(x-2) = x^2-2x = f(x),$ as required.

As above, the domains of $g$ and $h$ are all real numbers which combine to give $f$ the correct domain of $(-\infty, \infty)$ .

Example 1.5.2.1d

For $f(x) = x^2 - 2x$ , find functions $g$ , $h$ and $k$ to decompose $f$ nontrivially as:

$f = g(h-k)$

Solution:

Decompose $f = g(h-k)$ .

We need to be careful here interpreting the equation $f = g(h-k)$ . What we have is an equality of functions so the parentheses here do not represent function notation here, but, rather function multiplication. The way to parse $g(h-k)$ , then, is the function $g$ times the function $h-k$ . Hence, we seek functions $g$ , $h$ , and $k$ so that $f(x) = [g(h-k)](x) = g(x) [(h-k)(x)] = g(x) (h(x) - k(x))$ .

From the previous example, we know we can rewrite $f(x) = x(x-2)$ , so one option is to set $g(x) = h(x) = x$ and $k(x) = 2$ so that

$[g(h-k)](x) = g(x) [(h-k)(x)] = g(x) (h(x) - k(x)) = x(x-2) = x^2-2x = f(x),$

as required.

As above, the domain of all constituent functions is $(-\infty, \infty)$ which matches the domain of $f$ .

Example 1.5.2.2a

For $F(t) = \dfrac{2t+1}{\sqrt{t^2-1}}$ , find functions $G$ , $H$ and $K$ to decompose $F$ nontrivially as:

$F= \frac{G}{H}$

Solution:

Decompose $F= \frac{G}{H}$ .

To write $F = \frac{G}{H}$ , we need $G(t)$ and $H(t)$ so $F(t)= \left(\frac{G}{H}\right)(t) = \frac{G(t)}{H(t)}$ .

We choose $G(t) = 2t+1$ and $H(t) = \sqrt{t^2-1}$ . Sure enough, $\left(\frac{G}{H}\right)(t) = \frac{G(t)}{H(t)} = \frac{2t+1}{ \sqrt{t^2-1}} = F(t)$ as required.

When it comes to the domain of $F$ , owing to the square root, we require $t^2-1 \geq 0$ . We have a denominator as well, therefore we require $\sqrt{t^2-1} \neq 0$ . The former requirement is the same restriction on $H$ , and the latter requirement comes from Definition 1.13. Starting with the domain of $G$ , all real numbers, and working through the details, we arrive at the correct domain of $F$ , $(-\infty, -1) \cup (1, \infty)$ .

Example 1.5.2.2b

For $F(t) = \dfrac{2t+1}{\sqrt{t^2-1}}$ , find functions $G$ , $H$ and $K$ to decompose $F$ nontrivially as:

$F=GH$

Solution:

Decompose $F=GH$ .

Next, we are asked to find functions $G$ and $H$ so $F (t) = (GH)(t) = G(t) H(t)$ . This means we need to rewrite the expression for $F(t)$ as a product. One way to do this is to convert radical notation to exponent notation:

$F(t) = \dfrac{2t+1}{\sqrt{t^2-1}} = \dfrac{2t+1}{\left(t^2-1 \right)^{\frac{1}{2}}} = (2t+1) \left(t^2-1\right)^{-\frac{1}{2}}.$

Choosing $G(t) = 2t+1$ and $H(t) = \left(t^2-1\right)^{-\frac{1}{2}}$ , we see $(GH)(t) = G(t) H(t) = (2t+1) \left(t^2-1\right)^{-\frac{1}{2}}$ as required.

The domain restrictions on $F$ stem from the presence of the square root in the denominator – both are addressed when finding the domain of $H$ . Hence, we obtain the correct domain of $F$ as $(- \infty, -1) \cup (1, \infty)$ .

Example 1.5.2.2c

For $F(t) = \dfrac{2t+1}{\sqrt{t^2-1}}$ , find functions $G$ , $H$ and $K$ to decompose $F$ nontrivially as:

$F = G+ H$

Solution:

Decompose $F = G+ H$ .

To express $F$ as a sum of functions $G$ and $H$ , we could rewrite

$F(t) = \dfrac{2t+1}{\sqrt{t^2-1}} = \dfrac{2t}{\sqrt{t^2-1}} + \dfrac{1}{\sqrt{t^2-1}},$

so that $G(t) = \frac{2t}{\sqrt{t^2-1}}$ and $H(t) = \frac{1}{\sqrt{t^2-1}}$ .

Indeed, $(G+H)(t) = G(t)+H(t) = \frac{2t}{\sqrt{t^2-1}} + \frac{1}{\sqrt{t^2-1}} = \frac{2t+1}{\sqrt{t^2-1}} = F(t)$ , as required.

Moreover, the domain restrictions for $F$ are the same for both $G$ and $H$ , so we get agreement on the domain being $(- \infty, -1) \cup (1, \infty)$ .

Example 1.5.2.2d

For $F(t) = \dfrac{2t+1}{\sqrt{t^2-1}}$ , find functions $G$ , $H$ and $K$ to decompose $F$ nontrivially as:

$F = \frac{G+H}{K}$

Solution:

Decompose $F = \frac{G+H}{K}$ .

Last, but not least, to write $F = \frac{G+H}{K}$ , we require $F(t) =\left(\frac{G+H}{K}\right)(t) = \frac{(G+H)(t)}{K(t)} = \frac{G(t)+H(t)}{K(t)}$ .

Identifying $G(t) = 2t$ , $H(t) = 1$ , and $K(t) = \sqrt{t^2-1}$ , we get

$\left(\dfrac{G+H}{K}\right)(t) = \dfrac{(G+H)(t)}{K(t)} = \dfrac{G(t) + H(t)}{K(t)} = \dfrac{2t+1}{\sqrt{t^2-1}} = F(t).$

Concerning domains, the domain of both $G$ and $H$ are all real numbers, but the domain of $K$ is restricted to $t^2-1 \geq 0$ . Coupled with the restriction stated in Definition 1.13 that $K(t) \neq 0$ , we recover the domain of $F$ , $(-\infty, -1) \cup (1, \infty)$ .

1.5.2 Function Composition

We just saw how the arithmetic of real numbers carried over into an arithmetic of functions. In this section, we discuss another way to combine functions which is unique to functions and isn’t shared with real numbers – function composition.

Definition 1.14

Let $f$ and $g$ be functions where the real number $x$ is in the domain of $f$ and the real number $f(x)$ is in the domain of $g$ . The composite of $g$ with $f$ , denoted $g \circ f$ , and read ` $g$ composed with $f$ ‘ is defined by the formula: $(g \circ f) (x) = g(f(x))$ .

To compute $(g \circ f)(x)$ , we use the formula given in Defintion 1.14: $(g \circ f) (x) = g(f(x))$ . However, from a procedural viewpoint, Defintion 1.14 tells us the output from $g \circ f$ is found by taking the output from $f$ , $f(x)$ , and then making that the input to $g$ . From this perspective, we see $g \circ f$ as a two step process taking an input $x$ and first applying the procedure $f$ then applying the procedure $g$ . Abstractly, we have

A function mapping with three function bubbles. The first bubble contains x and there is an arrow representing the function f to the f(x) in the middle bubble. Then there was another arrow from f(x) to g(f(x)) in the third bubble representing the function g. Below the bubbles is an arrow from x in the first bubble to g(f(x)) in the third bubble representing the composition of the g with f. — Composition of Functions

In the expression $g(f(x))$ , the function $f$ is often called the `inside’ function while $g$ is often called the `outside’ function. When evaluating composite function values we present two methods in the example below: the `inside out’ and `outside in’ methods.

Example 1.5.3

Example 1.5.3.1

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Compute the indicated function value:

$(g \circ f)(1)$

Solution:

Compute $(f \circ g)(1)$ .

Using Definition 1.14, $(g \circ f)(1) = g(f(1))$ .

To start $f(1) = (1)^2 - 4(1) = -3$ .

Then $g(-3) = 2 - \sqrt{(-3)+3} = 2$ , so we have

$\begin{array}{rcl} (g \circ f)(1) &=& g(f(1)) \\ &=& g(-3) \\ &=& 2. \end{array}$

Example 1.5.3.2

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Compute the indicated function value:

$(f \circ g)(1)$

Solution:

Compute $(f \circ g)(1)$ .

By definition, $(f \circ g)(1) = f(g(1))$ .

We find $g(1) = 2 - \sqrt{1+3} = 0$ , and

$f(0) = (0)^2-4(0) = 0$ , so

$\begin{array}{rcl} (f \circ g)(1) &=& f(g(1)) \\ &=& f(0) \\ &=& 0. \end{array}$

Comparing this with our answer to the last problem, we see that $(g \circ f)(1) \neq (f \circ g)(1)$ which tells us function composition is not commutative,^[5]

Example 1.5.3.3

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Compute the indicated function value:

$(g \circ g)(6)$

Solution:

Compute $(g \circ g)(6)$ .

We note $(g \circ g)(6) = g(g(6))$ , and thus we `iterate’ the process $g$ : that is, we apply the process $g$ to $6$ , then apply the process $g$ again.

We find $g(6) = 2 - \sqrt{6+3} = -1$ , and

$g(-1) = 2 - \sqrt{(-1)+3} = 2 - \sqrt{2}$ , so

$\begin{array}{rcl} (g \circ g)(6) &=& g(g(6)) \\ &=& g(-1) \\ &=& 2-\sqrt{2}. \end{array}$

Example 1.5.3.4

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(g \circ f)(x)$

Solution:

Determine and simplify $(g \circ f)(x)$ .

By definition, $(g \circ f)(x) = g(f(x))$ . We now illustrate two ways to approach this problem.

inside out: We substitute $f(x) = x^2-4x$ in for $t$ in the expression $g(t)$ and get
$\begin{array}{rcl} (g \circ f)(x) &=& g(f(x)) \\ &=& g\left(x^2-4x\right) \\ &=& 2 - \sqrt{\left(x^2-4x\right)+3} \\ &=& 2 - \sqrt{x^2-4x+3} \end{array}$

Hence, $(g \circ f)(x) = 2 - \sqrt{x^2-4x+3}.$
outside in: We use the formula for $g$ first to get
$\begin{array}{rcl} (g \circ f)(x) &=& g(f(x)) \\ &=& 2 - \sqrt{f(x)+3} \\ &=& 2 - \sqrt{\left(x^2-4x\right)+3} \\ &=& 2 - \sqrt{x^2-4x+3} \end{array}$

We get the same answer as before, $(g \circ f)(x) = 2 - \sqrt{x^2-4x+3}.$

To find the domain of $g \circ f$ , we need to find the elements in the domain of $f$ whose outputs $f(x)$ are in the domain of $g$ .

The domain of $f$ is all real numbers, allowing us to focus on finding the range elements compatible with $g$ . Owing to the presence of the square root in the formula $g(t) = 2 - \sqrt{t+3}$ we require $t \geq -3$ .

Hence, we need $f(x) \geq -3$ or $x^2-4x \geq -3$ . To solve this inequality we rewrite as $x^2-4x + 3 \geq 0$ and use a sign diagram. Letting $r(x) = x^2-4x+3$ , we find the zeros of $r$ to be $x = 1$ and $x = 3$ and obtain

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Our solution to $x^2-4x+3 \geq 0$ , and hence the domain of $g \circ f$ , is $(-\infty, 1] \cup [3,\infty)$ .

Example 1.5.3.5

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(f \circ g)(t)$

Solution:

Determine and simplify $(f \circ g)(t)$ .

To find $(f \circ g)(t)$ , we find $f(g(t))$ .

inside out: We substitute the expression $g(t) = 2 - \sqrt{t+3}$ in for $x$ in the formula $f(x)$ and get
$\begin{array}{rcl} (f \circ g)(t) & = & f(g(t)) \\ &=& f\left(2-\sqrt{t+3}\right) \\ [2pt] & = & \left(2-\sqrt{t+3}\right)^2 - 4\left(2-\sqrt{t+3}\right) \\[2pt] & = & 4 - 4\sqrt{t+3} + \left(\sqrt{t+3}\right)^2 - 8 + 4 \sqrt{t+3} \\ [2pt] & = & 4 + t+3 - 8 \\ & = & t-1 \\ \end{array}$
outside in: We use the formula for $f(x)$ first to get
$\begin{array}{rclr} (f \circ g)(t) & = & f(g(t)) & \\ &=& \left(g(t)\right)^2 - 4\left(g(t)\right) & \\ [2pt] & = & \left(2-\sqrt{t+3}\right)^2 - 4\left(2-\sqrt{t+3}\right) & \\[2pt] & = & t-1 & \text{same algebra as before} \\ \end{array}$

Thus we get $(f \circ g)(t) = t-1$ .

To find the domain of $f \circ g$ , we look for the elements $t$ in the domain of $g$ whose outputs, $g(t)$ are in the domain of $f$ . As mentioned previously, the domain of $g$ is limited by the presence of the square root to $\{ t \in \mathbb{R} \, | \, t \geq -3\}$ , while the domain of $f$ is all real numbers.

Hence, the domain of $f \circ g$ is restricted only by the domain of $g$ and is $\{ t \in \mathbb{R} \, | \, t \geq -3\}$ or, using interval notation, $[-3, \infty)$ .

Note that as with Example 1.5.1 of this section, had we used the simplified formula for $(f \circ g)(t) = t-1$ to determine domain, we would have arrived at the incorrect answer.

Example 1.5.3.6

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(g \circ h)(s)$

Solution:

Determine and simplify $(g \circ h)(s)$ .

To find $(g \circ h)(s)$ , we compute $g(h(s))$ .

inside out: We substitute $h(s)$ in for $t$ in the expression $g(t)$ to get
$\begin{array}{rclr} (g \circ h)(s) & = & g(h(s)) & \\ &=& g\left(\dfrac{2s}{s+1}\right) & \\ [12pt] & = & 2 - \sqrt{\left(\dfrac{2s}{s+1}\right)+3} & \\[12pt] & = & 2 - \sqrt{\dfrac{2s}{s+1} + \dfrac{3(s+1)}{s+1}} & \text{get common denominators}\\ [12pt] & = & 2 - \sqrt{\dfrac{5s+3}{s+1}} & \\ \end{array}$
outside in: We use the formula for $g(t)$ first to get
$\begin{array}{rclr} (g \circ h)(s) & = & g(h(s)) & \\ &=& 2 - \sqrt{h(s)+3}& \\ [2pt] & = & 2 - \sqrt{\left(\dfrac{2s}{s+1}\right)+3} & \\[12pt] & = & 2 - \sqrt{\dfrac{5s+3}{s+1}} & \text{get common denominators as before}\\ \end{array}$

To find the domain of $g \circ h$ , we need the elements in the domain of $h$ so that $h(s)$ is in the domain of $g$ .

Owing to the $s+1$ in the denominator of the expression $h(s)$ , we require $s \neq -1$ . Once again, because of the square root in $g(t) = 2 - \sqrt{t+3}$ , we need $t \geq -3$ or, in this case $h(s) \geq -3$ . We rearrange this inequality:

$\begin{array}{rclr} \dfrac{2s}{s+1} & \geq & -3 & \\ [10pt] \dfrac{2s}{s+1} +3 & \geq & 0 & \\ [10pt] \dfrac{5s+3}{s+1} & \geq & 0 & \text{get common denominators as before} \\ \end{array}$

Defining $r(s) = \frac{5s+3}{s+1}$ , we see $r$ is undefined at $s=-1$ (a carry over from the domain restriction of $h$ ) and $r(s) = 0$ at $s = -\frac{3}{5}$ . Our sign diagram is

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hence our domain is $(-\infty, -1) \cup \left[-\frac{3}{5}, \infty\right)$ .

Example 1.5.3.7

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(h \circ g)(t)$

Solution:

Determine and simplify $(h \circ g)(t)$ .

We find $(h \circ g)(t)$ by finding $h(g(t))$ .

inside out: We substitute the expression $g(t)$ for $s$ in the formula $h(s)$
$\begin{array}{rcl} (h \circ g)(t) & = & h(g(t)) \\ &=& h\left(2-\sqrt{t+3}\right)\\[2pt] & = & \dfrac{2 \left(2-\sqrt{t+3} \right)}{\left(2-\sqrt{t+3}\right)+1} \\[12pt] & = & \dfrac{4-2\sqrt{t+3}}{3-\sqrt{t+3}}\\ \end{array}$
outside in: We use the formula for $h(s)$ first to get
$\begin{array}{rcl} (h \circ g)(t) & = & h(g(t)) \\ &=& \dfrac{2 \left(g(t)\right)}{\left( g(t)\right) + 1}\\ [12pt] & = & \dfrac{2 \left(2-\sqrt{t+3} \right)}{\left(2-\sqrt{t+3}\right)+1}\\[12pt] & = & \dfrac{4-2\sqrt{t+3}}{3-\sqrt{t+3}} \\ \end{array}$

To find the domain of $h \circ g$ , we need the elements of the domain of $g$ so that $g(t)$ is in the domain of $h$ . As we’ve seen already, for $t$ to be in the domain of $g$ , $t \geq -3$ . For $s$ to be in the domain of $h$ , $s \neq -1$ , so we require $g(t) \neq -1$ .

Hence, we solve $g(t) = 2-\sqrt{t+3} = -1$ with the intent of excluding these solutions. Isolating the radical expression gives $\sqrt{t+3} = 3$ or $t = 6$ . Sure enough, we check $g(6)=-1$ , so we exclude $t=6$ from the domain of $h \circ g$ .

Our final answer is $[-3, 6) \cup (6, \infty)$ .

Example 1.5.3.8

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(h \circ h)(x)$

Solution:

Determine and simplify $(h \circ h)(s)$ .

To find $(h \circ h)(s)$ we find $h(h(s))$ :

inside out: We substitute the expression $h(s)$ for $s$ in the expression $h(s)$ into $h$ to get
$\begin{array}{rclr} (h \circ h)(s) & = & h(h(s)) & \\[8pt] &=& h\left(\dfrac{2s}{s+1}\right) & \\ [15pt] &=& \dfrac{2\left(\dfrac{2s}{s+1}\right)}{\left(\dfrac{2s}{s+1}\right)+1} & \\ [25pt] & = & \dfrac{\dfrac{4s}{s+1}}{\dfrac{2s}{s+1}+1} \cdot \dfrac{(s+1)}{(s+1)} & \\ [25pt] & = & \dfrac{\dfrac{4s}{s+1} \cdot (s+1)}{\left(\dfrac{2s}{s+1}\right)\cdot(s+1)+1\cdot(s+1)} & \\ [25pt] & = & \dfrac{\dfrac{4s}{\cancelto{1}{(s+1})} \cdot \cancel{(s+1)}}{\dfrac{2s}{\cancelto{1}{(s+1)}}\cdot\cancel{(s+1)}+s+1} & \\ [25pt] & = & \dfrac{4s}{3s+1} & \\ \end{array}$
outside in: This approach yields
$\begin{array}{rclr} (h \circ h)(s) & = & h(h(s)) & \\[8pt] &=& \dfrac{2 (h(s))}{h(s) + 1} & \\ [.25in] & = & \dfrac{2\left(\dfrac{2s}{s+1}\right)}{\left(\dfrac{2s}{s+1}\right)+1} & \\[.35in] & = & \dfrac{4s}{3s+1} & \text{same algebra as before}\\ \end{array}$

To find the domain of $h \circ h$ , we need to find the elements in the domain of $h$ so that the outputs, $h(s)$ are also in the domain of $h$ .

The only domain restriction for $h$ comes from the denominator: $s \neq -1$ , so in addition to this, we also need $h(s) \neq -1$ . To this end, we solve $h(s) = -1$ and exclude the answers. Solving $\frac{2s}{s+1} = -1$ gives $s = -\frac{1}{3}$ .

The domain of $h \circ h$ is $(-\infty, -1) \cup \left(-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, \infty\right)$ .

Example 1.5.3.9

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$(h \circ (g \circ f))(x)$

Solution:

Determine and simplify $(h \circ (g \circ f))(x)$ ..

The expression $(h \circ (g \circ f))(x)$ indicates that we first find the composite, $g \circ f$ and then compose the function $h$ with the result.

We know from number 4 that $(g \circ f)(x) = 2 - \sqrt{x^2-4x+3}$ with domain $(-\infty, 1] \cup [3,\infty)$ . We now proceed as usual.

inside out: We substitute the expression $(g \circ f)(x)$ for $s$ in the expression $h(s)$ first to get
$\begin{array}{rclr} (h \circ (g \circ f))(x) & = & h((g \circ f)(x)) & \\ &=& h\left(2 - \sqrt{x^2-4x+3}\right) & \\ [5pt] & = & \dfrac{2 \left(2 - \sqrt{x^2-4x+3}\right)}{\left(2 - \sqrt{x^2-4x+3}\right)+1} & \\ [20pt] & = & \dfrac{4 - 2\sqrt{x^2-4x+3}}{3 - \sqrt{x^2-4x+3}} & \\ \end{array}$
outside in: We use the formula for $h(s)$ first to get
$\begin{array}{rclr} (h \circ (g \circ f))(x) & = & h((g \circ f)(x)) & \\ &=& \dfrac{2 \left( (g \circ f)(x)\right)}{ \left( (g \circ f)(x)\right) + 1} & \\ [15pt] & = & \dfrac{2 \left(2 - \sqrt{x^2-4x+3}\right)}{\left(2 - \sqrt{x^2-4x+3}\right)+1} & \\ [20pt] & = & \dfrac{4 - 2\sqrt{x^2-4x+3}}{3 - \sqrt{x^2-4x+3}} & \\ \end{array}$

To find the domain of $h \circ (g \circ f)$ , we need the domain elements of $g \circ f$ , $(-\infty, 1] \cup [3,\infty)$ , so that $(g \circ f)(x)$ is in the domain of $h$ .

As we’ve seen several times already, the only domain restriction for $h$ is $s \neq -1$ , so we set $(g \circ f)(x) = 2 - \sqrt{x^2-4x+3} = -1$ and exclude the solutions. We get $\sqrt{x^2-4x+3} = 3$ , and, after squaring both sides, we have $x^2-4x+3 = 9$ . We solve $x^2-4x-6 = 0$ using the quadratic formula and obtain $x = 2 \pm \sqrt{10}$ . The reader is encouraged to check that both of these numbers satisfy the original equation, $2 - \sqrt{x^2-4x+3} = -1$ and also belong to the domain of $g \circ f$ , $(-\infty, 1] \cup [3,\infty)$ , and so must be excluded from our final answer.^[6]

Our final domain for $h \circ (f \circ g)$ is $(-\infty, 2 -\sqrt{10}) \cup (2 - \sqrt{10}, 1] \cup \left[3, 2 + \sqrt{10}\right) \cup \left(2+\sqrt{10}, \infty\right)$ .

Example 1.5.3.10

Let $f(x) = x^2-4x$ , $g(t) = 2-\sqrt{t+3}$ , and $h(s) = \dfrac{2s}{s+1}$ .

Determine and simplify the indicated composite functions. State the domain of each.

$((h \circ g) \circ f)(x)$

Solution:

Determine and simplify $((h \circ g) \circ f)(x)$ .

The expression $((h \circ g) \circ f)(x)$ indicates that we first find the composite $h \circ g$ and then compose that with $f$ . From number 7, we have

$(h \circ g)(t) = \frac{4-2\sqrt{t+3}}{3-\sqrt{t+3}}$

with domain $[-3, 6) \cup (6, \infty)$ .

We know from number 4 that $(g \circ f)(x) = 2 - \sqrt{x^2-4x+3}$ with domain $(-\infty, 1] \cup [3,\infty)$ . We now proceed as usual.

inside out: We substitute the expression $(g \circ f)(x)$ for $s$ in the expression $h(s)$ first to get
$\begin{array}{rcl} ((h \circ g) \circ f)(x) & = & (h \circ g)(f(x)) \\ &=& (h \circ g)\left(x^2-4x\right) \\ [2pt] & = & \dfrac{4-2\sqrt{\left(x^2-4x\right)+3}}{3-\sqrt{\left(x^2-4x\right)+3}} \\ [20pt] & = & \dfrac{4 - 2\sqrt{x^2-4x+3}}{3 - \sqrt{x^2-4x+3}} \\ \end{array}$
outside in: We use the formula for $(h \circ g)(t)$ first to get
$\begin{array}{rcl} ((h \circ g) \circ f)(x) & = & (h \circ g)(f(x)) \\ &=& \dfrac{4-2\sqrt{(f(x))+3}}{3-\sqrt{f(x))+3}} \\ [20pt] & = & \dfrac{4 - 2\sqrt{\left(x^2-4x\right)+3}}{3 - \sqrt{\left(x^2-4x\right)+3}} \\[20pt] & = & \dfrac{4 - 2\sqrt{x^2-4x+3}}{3 - \sqrt{x^2-4x+3}} \\ \end{array}$

The domain of $f$ is all real numbers, thus the challenge here in computing the domain of $(h \circ g) \circ f$ is to determine the values $f(x)$ which are in the domain of $h \circ g$ , $[-3, 6) \cup (6, \infty)$ .

At first glance, it appears as if we have two (or three!) inequalities to solve: $-3 \leq f(x) < 6$ and $f(x) > 6$ . Alternatively, we could solve $f(x) = x^2-4x \geq -3$ and exclude the solutions to $f(x) = x^2-4x = 6$ which is not only easier from a procedural point of view, but also easier because we’ve already done both calculations. In number 4, we solved $x^2-4x \geq -3$ and obtained the solution $(-\infty, 1] \cup [3, \infty)$ and in number 9, we solved $x^2-4x-6 = 0$ and obtained $x = 2 \pm \sqrt{10}$ .

Hence, the domain of $(h \circ g) \circ f$ is $(-\infty, 2 -\sqrt{10}) \cup (2 - \sqrt{10}, 1] \cup \left[3, 2 + \sqrt{10}\right) \cup \left(2+\sqrt{10}, \infty\right)$ .

As previously mentioned, it should be clear from Example 1.5.3 that, in general, $g \circ f \neq f \circ g$ , in other words, function composition is not commutative. However, numbers 9 and 10 demonstrate the associative property of function composition. That is, when composing three (or more) functions, as long as we keep the order the same, it doesn’t matter which two functions we compose first. We summarize the important properties of function composition in the theorem below.

Theorem 1.6: Properties of Function Composition

Suppose $f$ , $g$ , and $h$ are functions

Associative Law of Composition: $h \circ (g \circ f) = (h \circ g) \circ f$ , provided the composite functions are defined.
Composition Identity: The function $I(x) = x$ satisfies: $I \circ f = f \circ I =f$ for all functions, $f$ .

By repeated applications of Definition 1.14, we find $(h \circ (g \circ f))(x) = h((g \circ f)(x)) = h(g(f(x)))$ . Similarly, $((h \circ g) \circ f)(x) = (h \circ g)(f(x)) = h(g(f(x)))$ . This establishes that the formulas for the two functions are the same. We leave it to the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality $h \circ (g \circ f) = (h \circ g) \circ f$ . A consequence of the associativity of function composition is that there is no need for parentheses when we write $h \circ g \circ f$ . The second property can also be verified using Definition 1.14. Recall that the function $I(x) = x$ is called the identity function and was introduced in Exercise 76 in Section 1.3. If we compose the function $I$ with a function $f$ , then we have $(I \circ f)(x) = I(f(x)) = f(x)$ , and a similar computation shows $(f\circ I)(x) = f(I(x)) = f(x)$ . This establishes that we have an identity for function composition much in the same way the function $I(x) = 1$ is an identity for function multiplication.

As we know, not all functions are described by formulas, and, moreover, not all functions are described by just one formula. The next example applies the concept of function composition to functions represented in various and sundry ways.

Example 1.5.4

Example 1.5.4.1a

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(g \circ f)(2)$

Solution:

Compute and simplify $(g \circ f)(2)$ .

To find $(g \circ f)(2) = g(f(2))$ , we first find $f(2) = 6(2)-(2)^2 = 8$ .

Considering $8 \geq 3$ , we use the rule $g(t) = t^2$ so $g(8) = (8)^2 = 64$ .

Hence, $(g \circ f)(3) = g(f(3)) = g(8) = 64$ .

Example 1.5.4.1b

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(h \circ g)(-1)$

Solution:

Compute and simplify $(h \circ g)(-1)$ .

As $(h\circ g)(-1) = h(g(-1))$ , we first need $g(-1)$ .

Given $-1 \leq -1 < 3$ , we use the rule $g(t) = 2t-1$ and find $g(-1) = 2(-1)-1 = -3$ .

Next, we need $h(-3)$ . As $(-3, 1) \in h$ , we have that $h(-3) =1$ .

Putting this all together, we find $(h\circ g)(-1) = h(g(-1)) = h(-3) = 1$ .

Example 1.5.4.1c

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(h \circ s)(-2)$

Solution:

Compute and simplify $(h \circ s)(-2)$ .

To find $(h \circ s)(-2) = h(s(-2))$ , we first need $s(-2)$ .

We see the point $(-2,3)$ is on the graph of $s$ , so $s(-2) = 3$ .

Next, we see $(3,-1) \in h$ , so $h(3) = -1$ .

Hence, $(h \circ s)(-2) = h(s(-2)) = h(3) = -1$ .

Example 1.5.4.1d

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Compute and simplify the following function values:

$(f \circ s)(0)$

Solution:

Compute and simplify $(f \circ s)(0)$ .

To find $(f \circ s)(0) = f(s(0))$ , we infer from the graph of $s$ that it contains the point $(0,3)$ , so $s(0) = 3$ .

Then $f(3) = 6(3) - (3)^2 = 9$ .

Thus we have $(f \circ s)(0) = f(s(0)) = f(3) = 9$ .

Example 1.5.4.2

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Determine and simplify a formula for $(g \circ f)(x)$ .

Solution:

Determine and simplify a formula for $(g \circ f)(x)$ .

To determine a formula for $(g \circ f)(x) = g(f(x))$ , we substitute $f(x) = 6x-x^2$ in for $t$ in the formula for $g(t)$ :

$\begin{array}{rcl} (g \circ f)(x) &=& g(f(x)) \\[3pt] &=& g(6x-x^2) \\[12pt] &=& \left\{ \begin{array}{rc} 2(6x-x^2) -1 & \text{if } -1 \leq 6x-x^2 < 3, \\[8pt] (6x-x^2)^2 & \text{if }6x-x^2 \geq 3. \\ \end{array} \right. \end{array}$

Simplifying each expression, we get $2(6x-x^2) -1 = -2x^2+12x-1$ for the first piece and $(6x-x^2)^2 = x^4 - 12x^3 +36x^2$ for the second piece.

The real challenge comes in solving the inequalities $-1 \leq 6x-x^2 < 3$ and $6x-x^2 \geq 3$ . While we could solve each individually using a sign diagram, a graphical approach works best here. We graph the parabola $y = 6x-x^2$ , identifying the vertex is $(3, 9)$ with intercepts $(0,0)$ and $(6,0)$ along with the horizontal lines $y = -1$ and $y=3$ below.

We determine the intersection points by solving $6x-x^2=-1$ and $6x-x^2=3$ .

Using the quadratic formula, we find the solutions to each equation are $x = 3 \pm \sqrt{10}$ and $x = 3 \pm \sqrt{6}$ , respectively.

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From the graph, we see the parabola $y = 6x - x^2$ is between the lines $y = -1$ and $y=3$ from $x = 3- \sqrt{10}$ to $x= 3 - \sqrt{6}$ and again from $x = 3+\sqrt{6}$ to $x=3+\sqrt{10}$ . Hence the solution to $-1 \leq 6x-x^2 < 3$ is $[3- \sqrt{10}, 3 - \sqrt{6}) \cup (3+\sqrt{6}, 3+\sqrt{10}]$ .

We also note $y = 6x-x^2$ is above the line $y=3$ for all $x$ between $x=3-\sqrt{6}$ and $3+\sqrt{6}$ . Hence, the solution to $6x-x^2 \geq 3$ is $[3-\sqrt{6}, 3+\sqrt{6}]$ .

Hence,

$(g \circ f)(x) = \left\{ \begin{array}{rc} -2x^2+12x-1 & \text{if } x \in [3- \sqrt{10}, 3 - \sqrt{6}) \cup (3+\sqrt{6}, 3+\sqrt{10}], \\[8pt] x^4 - 12x^3 +36x^2 & \text{if } x \in [3-\sqrt{6}, 3+\sqrt{6}]. \\ \end{array} \right.}$

Example 1.5.4.3

Consider the following functions:

$f(x) = 6x - x^2$
$g(t) = \left\{ \begin{array}{rc} 2t-1 & \text{if } -1 \leq t < 3, \\ t^2 & \text{if } t \geq 3. \\ \end{array}.$
$h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$
$s$ whose graph is given below:

Write $s \circ h$ as a set of ordered pairs.

Solution:

Write $s \circ h$ as a set of ordered pairs.

Last but not least, we are tasked with representing $s \circ h$ as a set of ordered pairs.

$h$ is described by the discrete set of points, $h = \{ (-3,1), (-2,6), (0,-2), (1,5), (3,-1) \}$ , so we will find $s \circ h$ point by point. We keep the graph of $s$ handy and construct the table below to help us organize our work.

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Neither $6$ nor $5$ are in the domain of $s$ , therefore $-2$ and $1$ are not in the domain of $s \circ h$ .

Hence, we get $s \circ h = \{ (-3,3), (0,3), (3,3) \}$ .

A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions which our last example illustrates. As with Example 1.52, we want to avoid trivial decompositions, which, when it comes to function composition, are those involving the identity function $I(x) = x$ as described in Theorem 1.6.

Example 1.5.5

Example 1.5.5.1a

Write each of the following functions as a composition of two or more (non-identity) functions. Check your answer by performing the function composition.

$F(x) = |3x-1|$

Solution:

There are many approaches to this kind of problem, and we showcase a different methodology in each of the solutions below.

Write $F(x) = |3x-1|$ as a composition of two or more functions.

Our goal is to express the function $F$ as $F = g \circ f$ for functions $g$ and $f$ .

From Definition 1.14, we know $F(x) = g(f(x))$ , and we can think of $f(x)$ as being the `inside’ function and $g$ as being the `outside’ function.

Looking at $F(x) = |3x-1|$ from an `inside versus outside’ perspective, we can think of $3x-1$ being inside the absolute value symbols. Taking this cue, we define $f(x) = 3x-1$ .

At this point, we have $F(x) = |f(x)|$ . What is the outside function? The function which takes the absolute value of its input, $g(x) = |x|$ .

Sure enough, this checks:

$\begin{array}{rcl} (g \circ f)(x) &=& g(f(x)) \\ &=& |f(x)| \\ &=& |3x-1| \\ &=& F(x). \end{array}$

Example 1.5.5.1b

Write each of the following functions as a composition of two or more (non-identity) functions. Check your answer by performing the function composition.

$G(t) = \dfrac{2}{t^2+1}$

Solution:

There are many approaches to this kind of problem, and we showcase a different methodology in each of the solutions below.

Write $G(t) = \dfrac{2}{t^2+1}$ as a composition of two or more functions.

We attack deconstructing $G$ from an operational approach.

Given an input $t$ , the first step is to square $t$ , then add $1$ , then divide the result into $2$ . We will assign each of these steps a function so as to write $G$ as a composite of three functions: $f$ , $g$ and $h$ .

Our first function, $f$ , is the function that squares its input, $f(t) = t^2$ .

The next function is the function that adds $1$ to its input, $g(t) = t+1$ .

Our last function takes its input and divides it into $2$ , $h(t) = \frac{2}{t}$ .

The claim is that $G = h \circ g \circ f$ which checks:

$\begin{array}{rcl} (h \circ g \circ f)(t) &=& h(g(f(t))) \\[4pt] &=& h(g\left(t^2\right)) \\[4pt] &=& h\left(t^2+1\right) \\[4pt] &=& \frac{2}{t^2+1} \\[4pt] &=& G(x). \end{array}$

Example 1.5.5.1c

Write each of the following functions as a composition of two or more (non-identity) functions. Check your answer by performing the function composition.

$H(s) = \dfrac{\sqrt{s}+1}{\sqrt{s}-1}$

Solution:

There are many approaches to this kind of problem, and we showcase a different methodology in each of the solutions below.

Write $H(s) = \dfrac{\sqrt{s}+1}{\sqrt{s}-1}$ as a composition of two or more functions.

If we look $H(s) = \frac{\sqrt{s}+1}{\sqrt{s}-1}$ with an eye towards building a complicated function from simpler functions, we see the expression $\sqrt{s}$ is a simple piece of the larger function.

If we define $f(s) = \sqrt{s}$ , we have $H(s) = \frac{f(s)+1}{f(s)-1}$ .

If we want to decompose $H = g \circ f$ , then we can glean the formula for $g(s)$ by looking at what is being done to $f(s)$ .

We take $g(s) = \frac{s+1}{s-1}$ , and check below:

$\begin{array}{rcl} (g \circ f)(s) &=& g(f(s)) \\[4pt] &=& \frac{f(s)+1}{f(s)-1} \\[4pt] &=& \frac{\sqrt{s}+1}{\sqrt{s}-1} \\[4pt] &=& H(s). \end{array}$

Example 1.5.5.2

For $F(x) = \sqrt{\dfrac{2x-1}{x^2+4}}$ , find functions $f$ , $g$ , and $h$ to decompose $F$ nontrivially as $F = f \circ \left(\dfrac{g}{h} \right)$ .

Solution:

For $F(x) = \sqrt{\dfrac{2x-1}{x^2+4}}$ , find functions $f$ , $g$ , and $h$ to decompose $F$ nontrivially as $F = f \circ \left(\dfrac{g}{h} \right)$ .

To write $F = f \circ \left(\frac{g}{h} \right)$ means

$\begin{array}{rcl} F(x) &=& \sqrt{\dfrac{2x-1}{x^2+4}} \\[8pt] &=& \left( f \circ \left(\dfrac{g}{h} \right) \right)(x) \\[8pt] &=& f \left( \left(\dfrac{g}{h} \right)(x) \right) \\[8pt] &=& f \left( \dfrac{g(x)}{h(x)} \right). \end{array}$

Working from the inside out, we have a rational expression with numerator $g(x)$ and denominator $h(x)$ .

Looking at the formula for $F(x)$ , one choice is $g(x) = 2x-1$ and $h(x) = x^2+4$ . Making these identifications, we have

$F(x) = \sqrt{\dfrac{2x-1}{x^2+4}} = \sqrt{\dfrac{g(x)}{h(x)}}.$

$F$ takes the square root of $\frac{g(x)}{h(x)}$ , which tells us that our last function, $f$ , is the function that takes the square root of its input, i.e., $f(x) = \sqrt{x}$ .

We leave it to the reader to check that, indeed, $F = f \circ \left(\frac{g}{h} \right)$ .

We close this section of a real-world application of function composition.

Example 1.5.6

The surface area of a sphere is a function of its radius $r$ and is given by the formula $S(r) = 4 \pi r^2$ . Suppose the sphere is being inflated so that the radius of the sphere is increasing according to the formula $r(t) = 3t^2$ , where $t$ is measured in seconds, $t \geq 0$ , and $r$ is measured in inches. Find and interpret $(S \circ r)(t)$ .

Solution:

If we look at the functions $S(r)$ and $r(t)$ individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time.

So, given a specific time, $t$ , we could find the radius at that time, $r(t)$ and feed that into $S(r)$ to compute the surface area at that time.

From this we see that the surface area $S$ is ultimately a function of time $t$ and we conclude

$\begin{array}{rcl} (S \circ r)(t) &=& S(r(t)) \\ &=& 4 \pi (r(t))^2 \\ &=& 4 \pi \left(3t^2\right)^2 \\ &=& 36 \pi t^{4}. \end{array}$

This formula allows us to compute the surface area directly given the time without going through the `intermediary variable’ $r$ .

1.5.3 Section Exercises

In Exercises 1 – 10, use the pair of functions $f$ and $g$ to find the following values if they exist.

$(f+g)(2)$
$(f-g)(-1)$
$(g-f)(1)$
$(fg)\left(\frac{1}{2}\right)$
$\left(\frac{f}{g}\right)(0)$
$\left(\frac{g}{f}\right)\left(-2\right)$

$f(x) = 3x+1$ and $g(t) = 4-t$
$f(x) = x^2$ and $g(t) = -2t+1$
$f(x) = x^2 - x$ and $g(t) = 12-t^2$
$f(x) = 2x^3$ and $g(t) = -t^2-2t-3$
$f(x) = \sqrt{x+3}$ and $g(t) = 2t-1$
$f(x) = \sqrt{4-x}$ and $g(t) = \sqrt{t+2}$
$f(x) = 2x$ and $g(t) = \dfrac{1}{2t+1}$
$f(x) = x^2$ and $g(t) = \dfrac{3}{2t-3}$
$f(x) = x^2$ and $g(t) = \dfrac{1}{t^2}$
$f(x) = x^2+1$ and $g(t) = \dfrac{1}{t^2+1}$

Exercises 11 – 20 refer to the functions $f$ and $g$ whose graphs are below.

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$(f + g)(-4)$
$(f + g)(0)$
$(f- g)(4)$
$(fg)(-4)$
$(fg)(-2)$
$(fg)(4)$
$\left(\dfrac{f}{g}\right)(0)$
$\left(\dfrac{f}{g}\right)(2)$
$\left(\dfrac{g}{f}\right)(-1)$
Compute the domains of $f+g$ , $f-g$ , $fg$ , $\dfrac{f}{g}$ and $\dfrac{g}{f}$ .

In Exercises 21 – 32, let $f$ be the function defined by

$f = \{(-3, 4), (-2, 2), (-1, 0), (0, 1), (1, 3), (2, 4), (3, -1)\}$

and let $g$ be the function defined by

$g = \{(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)\}$

Compute the indicated value if it exists.

$(f + g)(-3)$
$(f - g)(2)$
$(fg)(-1)$
$(g + f)(1)$
$(g - f)(3)$
$(gf)(-3)$
$\left(\frac{f}{g}\right)(-2)$
$\left(\frac{f}{g}\right)(-1)$
$\left(\frac{f}{g}\right)(2)$
$\left(\frac{g}{f}\right)(-1)$
$\left(\frac{g}{f}\right)(3)$
$\left(\frac{g}{f}\right)(-3)$

In Exercises 33 – 42, use the pair of functions $f$ and $g$ to find the domain of the indicated function then find and simplify an expression for it.

$(f+g)(x)$
$(f-g)(x)$
$(fg)(x)$
$\left(\frac{f}{g}\right)(x)$

$f(x) = 2x+1$ and $g(x) = x-2$
$f(x) = 1-4x$ and $g(x) = 2x-1$
$f(x) = x^2$ and $g(x) = 3x-1$
$f(x) = x^2-x$ and $g(x) = 7x$
$f(x) = x^2-4$ and $g(x) = 3x+6$
$f(x) = -x^2+x+6$ and $g(x) = x^2-9$
$f(x) = \dfrac{x}{2}$ and $g(x) = \dfrac{2}{x}$
$f(x) =x-1$ and $g(x) = \dfrac{1}{x-1}$
$f(x) = x$ and $g(x) = \sqrt{x+1}$
$f(x) =\sqrt{x-5}$ and $g(x) = \sqrt{x-5}$

In Exercises 43 – 47, write the given function as a nontrivial decomposition of functions as directed.

For $p(z) = 4z-z^3$ , find functions $f$ and $g$ so that $p=f-g$ .
For $p(z) = 4z-z^3$ , find functions $f$ and $g$ so that $p=f+g$ .
For $g(t) = 3t|2t-1|$ , find functions $f$ and $h$ so that $g = fh$ .
For $r(x) = \dfrac{3-x}{x+1}$ , find functions $f$ and $g$ so $r = \dfrac{f}{g}$ .
For $r(x) = \dfrac{3-x}{x+1}$ , find functions $f$ and $g$ so $r = fg$ .
Can $f(x) = x$ be decomposed as $f = g-h$ where $g(x) = x+\dfrac{1}{x}$ and $h(x) = \dfrac{1}{x}$ ?

In Exercises 49 – 60, use the given pair of functions to find the following values if they exist.

$(g\circ f)(0)$
$(f\circ g)(-1)$
$(f \circ f)(2)$
$(g\circ f)(-3)$
$(f\circ g)\left(\frac{1}{2}\right)$
$(f \circ f)(-2)$

$f(x) = x^2$ and $g(t) = 2t+1$
$f(x) = 4-x$ and $g(t) = 1-t^2$
$f(x) = 4-3x$ and $g(t) = |t|$
$f(x) = |x-1|$ and $g(t) = t^2-5$
$f(x) = 4x+5$ and $g(t) = \sqrt{t}$
$f(x) = \sqrt{3-x}$ and $g(t) = t^2+1$
$f(x) = 6-x-x^2$ and $g(t) = t\sqrt{t+10}$
$f(x) = \sqrt[3]{x+1}$ and $g(t) = 4t^2-t$
$f(x) = \dfrac{3}{1-x}$ and $g(t) = \dfrac{4t}{t^2+1}$
$f(x) = \dfrac{x}{x+5}$ and $g(t) = \dfrac{2}{7-t^2}$
$f(x) = \dfrac{2x}{5-x^2}$ and $g(t) = \sqrt{4t+1}$
$f(x) =\sqrt{2x+5}$ and $g(t) = \dfrac{10t}{t^2+1}$

In Exercises 61 – 72, use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using interval notation.

$(g \circ f)(x)$
$(f \circ g)(t)$
$(f \circ f)(x)$

$f(x) = 2x+3$ and $g(t) = t^2-9$
$f(x) = x^2 -x+1$ and $g(t) = 3t-5$
$f(x) = x^2-4$ and $g(t) = |t|$
$f(x) = 3x-5$ and $g(t) = \sqrt{t}$
$f(x) = |x+1|$ and $g(t) = \sqrt{t}$
$f(x) = 3-x^2$ and $g(t) = \sqrt{t+1}$
$f(x) = |x|$ and $g(t) = \sqrt{4-t}$
$f(x) = x^2-x-1$ and $g(t) = \sqrt{t-5}$
$f(x) = 3x-1$ and $g(t) = \dfrac{1}{t+3}$
$f(x) = \dfrac{3x}{x-1}$ and $g(t) =\dfrac{t}{t-3}$
$f(x) = \dfrac{x}{2x+1}$ and $g(t) = \dfrac{2t+1}{t}$
$f(x) = \dfrac{2x}{x^2-4}$ and $g(t) =\sqrt{1-t}$

In Exercises 73 – 78, use $f(x) = -2x$ , $g(t) = \sqrt{t}$ and $h(s) = |s|$ to find and simplify expressions for the following functions and state the domain of each using interval notation.

$(h\circ g \circ f)(x)$
$(h\circ f \circ g)(t)$
$(g\circ f \circ h)(s)$
$(g\circ h \circ f)(x)$
$(f\circ h \circ g)(t)$
$(f\circ g \circ h)(s)$

In Exercises 79 – 91, let $f$ be the function defined by

$f = \{(-3, 4), (-2, 2), (-1, 0), (0, 1), (1, 3), (2, 4), (3, -1)\}$

and let $g$ be the function defined by

$g = \{(-3, -2), (-2, 0), (-1, -4), (0, 0), (1, -3), (2, 1), (3, 2)\}.$

Find the following, if it exists.

$(f \circ g)(3)$
$f(g(-1))$
$(f \circ f)(0)$
$(f \circ g)(-3)$
$(g \circ f)(3)$
$g(f(-3))$
$(g \circ g)(-2)$
$(g \circ f)(-2)$
$g(f(g(0)))$
$f(f(f(-1)))$
$f(f(f(f(f(1)))))$
$\underbrace{(g \circ g \circ \cdots \circ g)}_{n \mbox{ times}}(0)$
Find the domain and range of $f \circ g$ and $g \circ f$ .

In Exercises 92 – 98, use the graphs of $y=f(x)$ and $y=g(x)$ below to find the following if it exists.

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$(g\circ f)(1)$
$(f \circ g)(3)$
$(g\circ f)(2)$
$(f\circ g)(0)$
$(f\circ f)(4)$
$(g \circ g)(1)$
Find the domain and range of $f \circ g$ and $g \circ f$ .

In Exercises 99 – 108, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.)

$p(x) = (2x+3)^3$
$P(x) = \left(x^2-x+1\right)^5$
$h(t) = \sqrt{2t-1}$
$H(t) = |7-3t|$
$r(s) = \dfrac{2}{5s+1}$
$R(s) = \dfrac{7}{s^2-1}$
$q(z) = \dfrac{|z|+1}{|z|-1}$
$Q(z) = \dfrac{2z^3+1}{z^3-1}$
$v(x) = \dfrac{2x+1}{3-4x}$
$w(x) = \dfrac{x^2}{x^4+1}$
Write the function $F(x) = \sqrt{\dfrac{x^{3} + 6}{x^{3} - 9}}$ as a composition of three or more non-identity functions.
Let $g(x) = -x, \, h(x) = x + 2, \, j(x) = 3x$ and $k(x) = x - 4$ . In what order must these functions be composed with $f(x) = \sqrt{x}$ to create $F(x) = 3\sqrt{-x + 2} - 4$ ?
What linear functions could be used to transform $f(x) = x^{3}$ into $F(x) = -\frac{1}{2}(2x - 7)^{3} + 1$ ? What is the proper order of composition?
Let $f(x) = 3x+1$ and let $g(x) = \left\{ \begin{array}{rc} 2x-1 & \text{if } x \leq 3 \\ 4-x & \text{if } x > 3 \\ \end{array} \right.}$ . Find expressions for $(f \circ g)(x)$ and $(g \circ f)(x)$ .
The volume $V$ of a cube is a function of its side length $x$ . Let’s assume that $x = t + 1$ is also a function of time $t$ , where $x$ is measured in inches and $t$ is measured in minutes. Find a formula for $V$ as a function of $t$ .
Suppose a local vendor charges per hot dog and that the number of hot dogs sold per hour is given by , where is the number of hours since AM, .
1. Find an expression for the revenue per hour $R$ as a function of $x$ .
2. Find and simplify $\left(R \circ x\right)(t)$ . What does this represent?
3. What is the revenue per hour at noon?

Section 1.5 Exercise Answers can be found in the Appendix … Coming soon

We could have just as easily called this new function $S(x)$ for `sum' of $f$ and $g$ and defined $S$ by $S(x) = f(x) + g(x)$ . ↵
see Section 0.4. ↵
Due to the fact that $(hg)(3) = h(3)g(3) = (-6)\left(\frac{8}{3}\right) = -16$ , we can write $hg = \{ (3,-16) \}$ . ↵
Addition is a `binary' operation - meaning it is defined only on two objects at once. Even though we write $1+2+3 = 6$ , mentally, we add just two of numbers together at any given time to get our answer: for example, $1+2+3 = (1+2)+3 = 3+3 = 6$ . ↵
That is, in general, $g \circ f \neq f \circ g$ . This shouldn't be too surprising, because, in general, the order of processes matters: adding eggs to a cake batter then baking the cake batter has a much different outcome than baking the cake batter then adding eggs. ↵
We can approximate $\sqrt{10} \approx 3$ so $2-\sqrt{10} \approx -1$ and $2+\sqrt{10} \approx 5$ . ↵