4.2 Power Functions

Monomial, and, more generally, Laurent monomial functions are specific examples of a much larger class of functions called power functions, as defined below.

Definition 4.2 broadens our scope of functions to include non-integer exponents such as $f (x) = 2 x^{4 / 3}$ , $g (t) = t^{0.4}$ and $h (w) = w^{\sqrt{2}}$ . Our primary aim in this section is to ascribe meaning to these quantities.

Rational Number Exponents

The road to real number exponents starts by defining rational number exponents.

Definition

Let $r$ be a rational number where in lowest terms $r = \frac{m}{n}$ where $m$ is an integer and $n$ is a natural number.¹ If $n equals 1$ , then $x^{r} = x^{m}$ . If $n greater than 1$ , then

$x^{r} = x^{\frac{m}{n}} = {(\sqrt[n]{x})}^{m} = \sqrt[n]{x^{m}},$

whenever ${(\sqrt[n]{x})}^{m}$ is defined.²

There are quite a few items worthy of note which are consequences of Definition 4.3. First off, if $m$ is an integer, then $x^{\frac{m}{1}} = x^{m}$ so expressions like $x^{\frac{3}{1}}$ are synonymous with $x^{3}$ , as we would expect.³ Second, the definition of $x^{\frac{m}{n}}$ can be taken as just ${(\sqrt[n]{x})}^{m}$ and shown to be equal to $\sqrt[n]{x^{m}}$ (or vice-versa) courtesy of properties of radicals. We state both in Definition 4.3 to allow for the reader to choose whichever form is more convenient in a given situation. The critical point to remember is no matter which representation you choose, keep in mind the restrictions if $n$ is even, $x greater than or equal to 0$ and if $m less than 0$ , $x not equal to 0$ .

Moreover, per this definition, $x^{\frac{1}{n}} = \sqrt[n]{x^{1}} = \sqrt[n]{x}$ , so we may rewrite principal roots as exponents: $\sqrt{x} = x^{\frac{1}{2}}$ and $\sqrt[5]{x} = x^{\frac{1}{5}}$ . This makes sense from an algebraic standpoint since per Theorem 4.2, ${(\sqrt[n]{x})}^{n} = x$ . Hence if we were to assign an exponent notation to $\sqrt[n]{x}$ , say $\sqrt[n]{x} = x^{r}$ , then ${(\sqrt[n]{x})}^{n} = {(x^{r})}^{n} = x$ . If the properties of exponents are to hold, then, necessarily, ${(x^{r})}^{n} = x^{r n} = x = x^{1}$ , so $r ⁢ n equals 1$ or $r = \frac{1}{n}$ . While this argument helps motivate the notation, as we shall see shortly, great care must be exercised in applying exponent properties in these cases. The long and short of this is that root functions as defined in Section 4.1 are all members of the `power functions' family.

Another important item worthy of note in Definition 4.3 is that it is absolutely essential we express the rational number $r$ in lowest terms before applying the root-power definition. For example, consider $x to the power 0.4$ . Expressing $r$ in lowest terms, we get: $r = 0.4 = \frac{4}{10} = \frac{2}{5}$ . Hence, $x^{0.4} = x^{2 / 5} = {(\sqrt[5]{x})}^{2}$ or $the 5th root of x squared$ , either of which is defined for all real numbers $x$ . In contrast, consider the equivalence $r = 0.4 = \frac{4}{10}$ . Here, the expression ${(\sqrt[10]{x})}^{4}$ is defined only for $x greater than or equal to 0$ owing to the presence of the even indexed root, $the 10th root of x$ . Hence, ${(\sqrt[10]{x})}^{4} \neq x^{\frac{4}{10}} = x^{\frac{2}{5}}$ unless $x greater than or equal to 0$ . On the other hand, the expression $\sqrt[10]{x^{4}}$ is defined for all numbers, $x$ , since $x to the power 4 greater than or equal to 0$ for all $x$ . In fact, it can be shown that $\sqrt[10]{x^{4}} = \sqrt[5]{x^{2}}$ for all real numbers. This means $\sqrt[10]{x^{4}} = \sqrt[5]{x^{2}} = x^{\frac{2}{5}} = x^{\frac{4}{10}}$ . So, to review, in general we have: $x^{\frac{4}{10}} = \sqrt[10]{x^{4}}$ , but $x^{\frac{4}{10}} \neq {(\sqrt[10]{x})}^{4}$ unless $x greater than or equal to 0$ . Once again the easiest way to avoid confusion here is to reduce the exponent to lowest terms before converting it to root-power notation.

Likewise, we have to be careful about the properties of exponents when it comes to rational exponents. Consider, for instance, the product rule for integer exponents: $x^{m} x^{n} = x^{m + n}$ . Consider $f (x) = x^{\frac{1}{2}} x^{\frac{1}{2}}$ and $g (x) = x^{\frac{1}{2} + \frac{1}{2}}$ . In the first case, $f (x) = x^{\frac{1}{2}} x^{\frac{1}{2}} = \sqrt{x} \sqrt{x} = {(\sqrt{x})}^{2} = x$ only for $x greater than or equal to 0$ . In the second case, $g (x) = x^{\frac{1}{2} + \frac{1}{2}} = x^{\frac{2}{2}} = x^{1} = x$ for all real numbers $x$ . Even though $f (x) = g (x)$ for $x greater than or equal to 0$ , $f$ and $g$ are different functions since they have different domains.

Similarly, the power rule for integer exponents: ${(x^{n})}^{m} = x^{n m}$ does not hold in general for rational exponents. To see this, consider the three functions: $f (x) = {(x^{\frac{1}{2}})}^{2}$ , $g (x) = x^{\frac{2}{2}}$ , and $h (x) = {(x^{2})}^{\frac{1}{2}}$ . In the first case, $f (x) = {(x^{\frac{1}{2}})}^{2} = {(\sqrt{x})}^{2} = x$ for $x greater than or equal to 0$ only (this is the same function $f$ above.) In the second case, the rational number $r = \frac{2}{2} = 1$ , so $g (x) = x^{\frac{2}{2}} = x^{\frac{1}{1}} = x^{1} = x$ for all real numbers, $x$ (this is the same function $g$ from above.) In the last case, $h (x) = {(x^{2})}^{\frac{1}{2}} = \sqrt{x^{2}} = | x |$ for all real numbers, $x$ . Once again, despite $f (x) = g (x) = h (x)$ for all $x greater than or equal to 0$ , $f$ , $g$ and $h$ and are three different functions. We graph $f$ , $g$ , and $h$ below.

Figure: , — **Figure 4.57** $f (x) = x^{\frac{1}{2}} x^{\frac{1}{2}} = {(x^{\frac{1}{2}})}^{2} = x$ , $x greater than or equal to 0$

Coordinate-plane figure. — **Figure 4.58** $g (x) = x^{\frac{1}{2} + \frac{1}{2}} = x^{\frac{2}{2}} = x$

In general, the properties of integer exponents do not extend to rational exponents unless the bases involved represent non-negative real numbers or the roots involved are odd. We have the following:

Theorem

Let $r$ and $s$ are rational numbers. The following properties hold provided none of the computations results in division by $0$ and either $r$ and $s$ have odd denominators or $x greater than or equal to 0$ and $y greater than or equal to 0$ :

Product Rules: $x^{r} x^{s} = x^{r + s}$ and ${(x y)}^{r} = x^{r} y^{r}$ .
Quotient Rules: $\frac{x^{r}}{x^{s}} = x^{r - s}$ and ${(\frac{x}{y})}^{r} = \frac{x^{r}}{y^{r}}$
Power Rule: ${(x^{r})}^{s} = x^{r s}$

Next, we turn our attention to the graphs of $f (x) = x^{r} = x^{\frac{m}{n}}$ for varying values of $m$ and $n$ . When $n$ is even, the domain is restricted owing to the presence of the even indexed root to $open bracket 0, ∞ close parenthesis$ . The range is likewise $open bracket 0, ∞ close parenthesis$ , a fact leave to the reader. All of the functions below are increasing on their domains, and it turns out this is always the case provided $r greater than 0$ . There is, however, is a difference in how the functions are increasing - and this is the concept of concavity. As with many concepts we've encountered so far in the text, concavity is most precisely defined using Calculus terminology, but we can nevertheless get a sense of concavity geometrically. For us, a curve is concave up over an interval if it resembles a portion of a ` $⌣$ ' shape. Similarly, a curve is called concave down over an interval if resembles part of a ` $⌢$ ' shape. When $0 less than r less than 1$ , the graphs of $f (x) = x^{r}$ resemble the left half of $⌢$ and so are concave down; when $r greater than 1$ , the graphs resemble the right half of a ` $⌣$ ' and are hence described as `concave up.'

Below we graph several examples of $f (x) = x^{r} = x^{\frac{m}{n}}$ where $n$ is odd. Here, the domain is $open parenthesis minus ∞, ∞ close parenthesis$ since the index on the root here is odd. Note that when $m$ is even, the graphs appear to be symmetric about the $y$ -axis and the range looks to be $open bracket 0, ∞ close parenthesis$ . When $m$ is odd, the graphs appear to be symmetric about the origin with range $open parenthesis minus ∞, ∞ close parenthesis$ . We leave verification of these facts to the reader. Note here also that for $x greater than or equal to 0$ , the graphs are down for $0 less than r less than 1$ and concave up for $r greater than 1$ .

When $r less than 0$ , we have variables appear in the denominator which open the opportunities for vertical and horizontal asymptotes. Below are graphed two examples

Unsurprisingly, Theorem 4.1, which, as stated, applied to root functions, generalizes to all rational powers.

Theorem

For real numbers $a$ , $b$ , $h$ , and $k$ and rational number $r$ with $a, b, r not equal to 0$ , the graph of $F (x) = a {(b x - h)}^{r} + k$ can be obtained from the graph of $f (x) = x^{r}$ by performing the following operations, in sequence:

add $h$ to each of the $x$ -coordinates of the points on the graph of $f$ . This results in a horizontal shift to the right if $h greater than 0$ or left if $h less than 0$ .
NOTE: This transforms the graph of $y = x^{r}$ to $y = {(x - h)}^{r}$ .
divide the $x$ -coordinates of the points on the graph obtained in Step 1 by $b$ . This results in a horizontal scaling, but may also include a reflection about the $y$ -axis if $b less than 0$ .
NOTE: This transforms the graph of $y = {(x - h)}^{r}$ to $y = {(b x - h)}^{r}$ .
multiply the $y$ -coordinates of the points on the graph obtained in Step 2 by $a$ . This results in a vertical scaling, but may also include a reflection about the $x$ -axis if $a less than 0$ .
NOTE: This transforms the graph of $y = {(b x - h)}^{r}$ to $y = a {(b x - h)}^{r}$ .
add $k$ to each of the $y$ -coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if $k greater than 0$ or down if $k less than 0$ .
NOTE: This transforms the graph of $y = a {(b x - h)}^{r}$ to $y = a {(b x - h)}^{r} + k$ .

The proof of Theorem 4.4 is identical to that of Theorem 4.1, and we suggest the reader work through the details. We give Theorem 4.4 a test run in the following example.

Example 1

Use the given graphs of $f$ and $g$ below long with Theorem 4.4 to graph $F$ and $G$ . State the domain and range of $F$ and $G$ using interval notation.

Graph $F (x) = {(2 x - 1)}^{2.6}$ .
Graph $G (t) = 1 - 2 {(t + 3)}^{\frac{3}{8}}$ .

Figure 4.70

f (x) = x^{2.6}

Figure 4.71

g (t) = t^{\frac{3}{8}}

The expression $F (x) = {(2 x - 1)}^{2.6}$ is given to us in the form prescribed by Theorem 4.4, and we identify $r equals 2.6$ , $a equals 1$ , $b equals 2$ , $h equals 1$ , and $k equals 0$ . Even though the graph of $f (x) = x^{2.6}$ is given to us, it's worth taking a moment to reinforce some concepts. Since, in lowest terms, $2.6 = \frac{26}{10} = \frac{13}{5}$ , it makes sense the domain and range of $f (x) = x^{2.6}$ are both all real numbers and the graph is symmetric about the origin.⁴ Moreover, since $2.6 greater than 1$ , the concavity matches what we would expect, too. We proceed as we have several times in the past, beginning with the horizontal shift.
Step 1: add $1$ to each of the $x$ -coordinates of each of the points on the graph of $y = x^{2.6}$ :
Figure 4.72 $y = x^{2.6}$
$→ add 1 to each x -coordinate$
Figure 4.73 $y = {(x - 1)}^{2.6}$
$open parenthesis minus 1, 1 close parenthesis, open parenthesis 0, 0 close parenthesis, open parenthesis 1, 1 close parenthesis$
$open parenthesis 0, minus 1 close parenthesis, open parenthesis 1, 0 close parenthesis, open parenthesis 2, 1 close parenthesis$
Step 2: divide each of the $x$ -coordinates of each of the points on the graph of $y = {(x - 1)}^{2.6}$ by $2$ :
Figure 4.74 $y = {(x - 1)}^{2.6}$
$→ divide to each x -coordinate by 2$
Figure 4.75 $F (x) = {(2 x - 1)}^{2.6}$
$open parenthesis 0, minus 1 close parenthesis, open parenthesis 1, 0 close parenthesis, open parenthesis 2, 1 close parenthesis$
$(0, - 1), (\frac{1}{2}, 0), (1, 1)$
We get the domain and range here are both $open parenthesis minus ∞, ∞ close parenthesis$ .
We first need to rewrite $G (t) = 1 - 2 {(t + 3)}^{\frac{3}{8}}$ in the form required by Theorem 4.4: $G (t) = - 2 {(t + 3)}^{\frac{3}{8}} + 1$ . We identify $r = \frac{3}{8}$ , $a = - 2$ , $b equals 1$ , $h = - 3$ , and $k equals 1$ . Since $\frac{3}{8}$ is in lowest terms and has an even denominator, it makes sense the domain and range of $g (t) = t^{\frac{3}{8}}$ is $open bracket 0, ∞ close parenthesis$ , since the root here, $8$ is even. Also, since $0 < \frac{3}{8} < 1$ , the graph of $y = t^{\frac{3}{8}}$ is concave down, as we would expect. As usual, we start with the horizontal shift.
Step 1: add $minus 3$ to each of the $t$ -coordinates of each of the points on the graph of $y = t^{\frac{3}{8}}$ :
Figure 4.76 $y = t^{\frac{3}{8}}$
$→ add minus 3 to each t -coordinate$
Figure 4.77 $y = {(t + 3)}^{\frac{3}{8}}$
$open parenthesis 0, 0 close parenthesis, open parenthesis 1, 1 close parenthesis$
$(- 3, 0), (- 2, 1)$
Step 2: Since $b equals 1$ , we proceed directly to Step 3.
Step 3: multiply each of the $y$ -coordinates of each of the points on the graph of $y = {(t + 3)}^{\frac{3}{8}}$ by $minus 2$ :
Figure 4.78 $y = {(t + 3)}^{\frac{3}{8}}$
$\overset{multiply each y -coordinate by - 2}{\overset{}{\to}}$
Figure 4.79 $y = - 2 {(t + 3)}^{\frac{3}{8}}$
$(- 3, 0), (- 2, 1)$
$(- 3, 0), (- 2, - 2)$
Step 4: add $1$ to each of the $y$ -coordinates of each of the points on the graph of $y = - 2 {(t + 3)}^{\frac{3}{8}}$ :
Figure 4.80 $y = - 2 {(t + 3)}^{\frac{3}{8}}$
$→ add 1 to each y -coordinate$
Figure 4.81 $y = - 2 {(t + 3)}^{\frac{3}{8}} + 1$
$(- 3, 0), (- 2, - 2)$
$(- 3, 1), (- 2, - 1)$
From the graph, we get the domain is $open bracket minus 3, ∞ close parenthesis$ and the range is $open parenthesis minus ∞, 1 close bracket$ .

We now turn our attention to more complicated functions involving rational exponents.

Example 2

For the following functions:

Analytically:
- find the domain.
- find the axis intercepts.
- analyze the end behavior.
Graph the function with help from a graphing utility and determine:
- the range.
- the local extrema, if they exist.
- intervals of increase.
- intervals of decrease.
Construct a sign diagram for each function using the intercepts and graph.

$f (x) = 3 x^{2} {(x^{3} - 8)}^{- \frac{2}{3}}$
$g (t) = \frac{{(t^{2} - 4)}^{\frac{3}{2}}}{t^{2} - 36}$

We first note that, owing to the negative exponent, the quantity ${(x^{3} - 8)}^{\frac{2}{3}}$ is in the denominator, alerting us to a potential domain issue. Rewriting ${(x^{3} - 8)}^{\frac{2}{3}}$ we set about solving $\sqrt[3]{{(x^{3} - 8)}^{2}} = 0$ . Cubing both sides and extracting square roots gives $x^{3} - 8 = 0$ or $x equals 2$ . Hence, $x equals 2$ is excluded from the domain.⁵ Since the root involved here is odd ( $3$ ), the only issue we have is with the denominator, hence our domain is $open brace x ∈ ℝ vertical bar x not equal to 2 close brace$ or $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ .
While not required to do so, we analyze the behavior of $f$ near $x equals 2$ . As $x \to 2^{-}$ , $3 ⁢ x squared approximately equals 12$ and $x^{3} - 8 \approx small (-)$ . Hence, ${(x^{3} - 8)}^{\frac{2}{3}} = \sqrt[3]{{(x^{3} - 8)}^{2}} \approx \sqrt[3]{{(small (-))}^{2}} \approx \sqrt[3]{small (+)} \approx small (+)$ . As such, $f (x) \approx \frac{12}{small (+)} \approx big (+)$ . We conclude $lim_{x \to 2^{-}} f (x) = \infty$ . As $x \to 2^{+}$ , $3 ⁢ x squared approximately equals 12$ and $x^{3} - 8 \approx small (+)$ , and we likewise get $lim_{x \to 2^{+}} f (x) = \infty$ . Since the unbounded behavior agrees from both directions as $x → 2$ , we write $lim x → 2 f ⁢ open parenthesis x close parenthesis equals ∞$ . Our analysis suggests $x equals 2$ is a vertical asymptote to the graph.
To find the $x$ -intercepts, we set $f (x) = 3 x^{2} {(x^{3} - 8)}^{- \frac{2}{3}} = 0$ , so that $3 ⁢ x squared equals 0$ or $x equals 0$ . We get $open parenthesis 0, 0 close parenthesis$ is our only $x$ - (and $y$ -)intercept.
For end behavior, note that in the denominator the $x^{3}$ term dominates so as $x → minus ∞$ and $x → ∞$ ,
$f (x) = 3 x^{2} {(x^{3} - 8)}^{- \frac{2}{3}} = \frac{3 x^{2}}{{(x^{3} - 8)}^{\frac{2}{3}}} \approx \frac{3 x^{2}}{{(x^{3})}^{\frac{2}{3}}} = \frac{3 x^{2}}{\sqrt[3]{{(x^{3})}^{2}}} = \frac{3 x^{2}}{\sqrt[3]{x^{6}}} = \frac{3 x^{2}}{x^{2}} = 3 .$
This suggests $lim_{x \to - \infty} f (x) = 3$ and $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 3$ so $y equals 3$ is a horizontal asymptote to the graph.
Graphing $y = f (x)$ below on the right bears out our analysis regarding zeros and asymptotes. The range appears to be $open bracket 0, ∞ close parenthesis$ , with the graph of $y = f (x)$ crossing its horizontal asymptote between $x equals 1$ and $x equals 2$ . We see we have a single local minimum at $open parenthesis 0, 0 close parenthesis$ with $f$ is decreasing on $open parenthesis minus ∞, 0 close bracket$ and $open parenthesis 2, ∞ close parenthesis$ and increasing on $open bracket 0, 2 close parenthesis$ .
For the sign diagram, we note that $f$ has only one zero, $x equals 0$ and is undefined at $x equals 2$ . For all $x$ values between these two numbers, $f ⁢ open parenthesis x close parenthesis greater than 0$ or $(+)$ . Our sign diagram for $f (x)$ is below on the right.
Figure 4.82
Figure 4.83
The graph of $y = f (x)$
Sign Diagram for $f (x)$
To find the domain of $g (t) = \frac{{(t^{2} - 4)}^{\frac{3}{2}}}{t^{2} - 36}$ , we have two issues to address: the denominator and an even (square) root. Solving $t^{2} - 36 = 0$ gives two excluded values, $t = \pm 6$ . For the numerator, we may rewrite ${(t^{2} - 4)}^{\frac{3}{2}} = {(\sqrt{t^{2} - 4})}^{3}$ , so we require $t squared minus 4 greater than or equal to 0$ , or $t squared greater than or equal to 4$ . Extracting square roots, we have $\sqrt{t^{2}} \geq \sqrt{4}$ or $vertical bar t vertical bar greater than or equal to 2$ which means $t less than or equal to minus 2$ or $t greater than or equal to 2$ . Taking into account our excluded values $t = \pm 6$ , we get the domain of $g$ is $(- \infty, - 6) \cup (- 6, - 2] \cup [2, 6) \cup (6, \infty)$ .
Looking near $t = - 6$ , we note that as $t → minus 6$ , ${(t^{2} - 4)}^{\frac{3}{2}} \approx 32^{\frac{3}{2}} = 32^{1.5}$ , a positive number. As $t \to - 6^{-}$ , $t^{2} - 36 \approx small (+)$ , so $g (t) \approx \frac{32^{1.5}}{small (+)} \approx big (+)$ . This suggests $lim_{t \to - 6^{-}} g (t) = \infty$ . On the other hand, as $t \to - 6^{+}$ , $t^{2} - 36 \approx small (-)$ , so $g (t) \approx \frac{32^{1.5}}{small (-)} \approx big (-)$ , suggesting $lim_{t \to - 6^{+}} g (t) = - \infty$ . Similarly, we find as $lim_{t \to 6^{-}} g (t) = - \infty$ and as $lim_{t \to 6^{+}} g (t) = \infty$ . This suggests we have two vertical asymptotes to the graph of $y = g (t)$ : $t = - 6$ and $t equals 6$ .
To find the $t$ -intercepts, we set $g ⁢ open parenthesis t close parenthesis equals 0$ and solve ${(t^{2} - 4)}^{\frac{3}{2}} = 0$ . This reduces to $t^{2} - 4 = 0$ or $t = \pm 2$ . As these are (just barely!) in the domain of $g$ , we have two $t$ -intercepts, $open parenthesis minus 2, 0 close parenthesis$ and $open parenthesis 2, 0 close parenthesis$ . The graph of $g$ has no $y$ -intercepts, since $0$ is not in the domain of $g$ , so $g ⁢ open parenthesis 0 close parenthesis$ is undefined.
Regarding end behavior, as $t → minus ∞$ and $t → ∞$ , the $t^{2}$ in both numerator and denominator dominate the constant terms, so we have
$g (t) = \frac{{(t^{2} - 4)}^{\frac{3}{2}}}{t^{2} - 36} \approx \frac{{(t^{2})}^{\frac{3}{2}}}{t^{2}} = \frac{{(\sqrt{t^{2}})}^{3}}{t^{2}} = \frac{{| t |}^{3}}{t^{2}} = \frac{| t | {| t |}^{2}}{t^{2}} = \frac{| t | t^{2}}{t^{2}} = | t | .$
This suggests that as $t → minus ∞$ and $t → ∞$ , the graph of $y = g (t)$ resembles $y = | t |$ . Hence, $lim_{t \to - \infty} g (t) = \infty$ and $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ . Using the piecewise definition of $| t |$ , we have that as $t → minus ∞$ , $g (t) \approx - t$ and as $t → ∞$ , $g (t) \approx t$ . In other words, the graph of $y = g (t)$ has two slant asymptotes with slopes $\pm 1$ .
Graphing $y = g (t)$ below on the left verifies our analysis. From the graph, the range appears to be $open parenthesis minus ∞, 0 close bracket union open bracket 14.697, ∞ close parenthesis$ . The points $open parenthesis minus 10, 14.697 close parenthesis$ and $open parenthesis 10, 14.697 close parenthesis$ are local minimums. $g$ appears to be decreasing on $(- \infty, - 10]$ , $open bracket 2, 6 close parenthesis$ , and $open parenthesis 6, 10 close bracket$ . Likewise, $g$ is increasing on $[- 10, - 6)$ , $(- 6, - 2]$ and $open bracket 10, ∞ close parenthesis$ . The graph of $y = g (t)$ certainly appears to be symmetric about the $y$ -axis. We leave it to the reader to show $g$ is, indeed, an even function.
For the sign diagram for $g (t)$ , we note that $g$ has zeros $t = \pm 2$ and is undefined at $t = \pm 6$ . Moreover, there is a gap in the domain for all values in the interval $open parenthesis minus 2, 2 close parenthesis$ , so we excise that portion of the real number line for our discussion. We find $g ⁢ open parenthesis t close parenthesis greater than 0$ or $(+)$ on the intervals $(- \infty - 6)$ and $open parenthesis 6, ∞ close parenthesis$ while $g ⁢ open parenthesis t close parenthesis less than 0$ or $(-)$ on $(- 6, - 2)$ and $open parenthesis 2, 6 close parenthesis$ . Our sign diagram for $g (t)$ is below on the right.
Figure 4.84
Figure 4.85
The graph of $y = g (t)$
Sign Diagram for $g (t)$

Real Number Exponents

We wish now to extend the concept of `exponent' from rational to all real numbers which means we need to discuss how to interpret an irrational exponent. Once again, the notions presented here are best discussed using the language of Calculus or Analysis, but we nevertheless do what we can with the notions we have.

Consider the wildly famous irrational number $π$ . The number $π$ is defined geometrically as the ratio of the circumference of a circle to that circle's diameter.⁶ The reason we use the symbol ` $π$ ' instead of any numerical expression is that $π$ is an irrational number, and, as such, its decimal representation neither terminates nor repeats. Hence we approximate $π$ as $π approximately equals 3.14$ or $π approximately equals 3.14159265$ . No matter how many digits we write, however, what we have is a rational number approximation of $π$ .

The good news is we can approximate $π$ to any desired accuracy using rational numbers by taking enough digits, so while we'll never `reach' the exact value of $π$ with rational numbers, we can get as close as we like to $π$ using rational numbers. That being said, we assume $π$ exists on the real number line, despite the fact the list of digits to pinpoint its location is, in some sense, infinite.

We take this tack when defining the value of a number raised to an irrational exponent. Consider, for instance, $2 to the power π$ . We can compute $2 cubed equals 8$ , $2^{3.1} = 2^{\frac{31}{10}} = \sqrt[10]{2^{31}} \approx 8.574$ , $2^{3.14} = 2^{\frac{314}{100}} = 2^{\frac{157}{50}} = \sqrt[50]{2^{157}} \approx 8.8512$ , and so on, so one way to define $2 to the power π$ as the unique real number we obtain as the exponents `approach' $π$ .

It is with this understanding that we present the notion of a `power function,' as described in Definition 4.2: $f (x) = a x^{p}$ where $a$ and $p$ are nonzero real number parameters. Here the exponent $p$ is open to any (nonzero) real number. Because of how we define real number exponents, if $p$ is irrational, then $x greater than or equal to 0$ to avoid having negatives under even-indexed roots as we go through the approximation process.⁷

In general, real number exponents inherit their properties from rational number exponents. For instance, Theorem 4.3 also holds for all real number exponents and the graphs of power functions inherit their behavior from graphs of rational exponent functions. More specifically, the graphs of functions of the form $f (x) = x^{p}$ where $p greater than 0$ all contain the points $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ . Moreover, these functions are increasing and their graphs are concave down if $0 less than p less than 1$ and concave up if $p greater than 1$ .

Figure: for varying values of . — **Figure 4.86** $f (x) = x^{p}$ for varying values of $p$ .

Theorem 4.4 generalizes to real number power functions, so, for instance to graph $F (x) = {(x - 2)}^{π}$ , one need only start with $y = x^{π}$ and shift horizontally two units to the right. (See the Exercises.)

We close this section with an application to economics. According to the US Census , Table 2, the share of money income (2014-2015) is given in the table below on the left. From these data, we can create a cumulative distribution, $y = L (x)$ called the Lorenz Curve.

The number $L (x)$ gives the percentage of the total national income earned by the bottom $x$ percent of wage earners, ranked from lowest income to highest income. Since the population here is separated into `quintiles,' each data point corresponds to $20 %$ of the population. So, for example, $L ⁢ open parenthesis 20 close parenthesis$ is the percentage of money income earned by the lowest $20 %$ of wage earners. In this case, we see $L ⁢ open parenthesis 20 close parenthesis equals 3.1$ . The number $L ⁢ open parenthesis 40 close parenthesis$ is the percentage of the money income earned by the bottom $40 %$ of wage earners - so this includes not only the money from the Second Quintile, but also the Lowest Quintile: $L (40) = 8.2 + L (20) = 8.2 + 3.1 = 11.3$ . Likewise, $L ⁢ open parenthesis 60 close parenthesis$ is the total income share of the bottom $60 %$ of wage earners which includes the income from the Middle, Second, and Lowest Quintiles: $L (60) = 14.3 + L (40) = 14.3 + (8.2 + 3.1) = 25.6$ .

Continuing in this manner, we get $L ⁢ open parenthesis 80 close parenthesis equals 48.8$ and $L ⁢ open parenthesis 100 close parenthesis equals 100$ , which is what we would expect: $100 %$ of the income is earned by $100 %$ of the population. We summarize these findings below on the right.

Table 4.1

Portion of Population, Percent of Money Income; Lowest Quintile, 3.1; Second Quintile, 8.2; Middle Quintile, 14.3; Fourth Quintile, 23.2; Highest Quintile, 51.2

percent wage earners, x, percent income, L ⁢ open parenthesis x close parenthesis; 20, 3.1; 40, 11.3; 60, 25.6; 80, 48.8; 100, 100

Example 3

Find power function to model the Lorenz Curve: $L (x) = a x^{p}$ . Comment on the goodness of fit.
Find and interpret $L ⁢ open parenthesis 90 close parenthesis$ .

Using desmos , we get $L (x) = 0.00027901 x^{2.7738}$ with $R squared equals 0.993$ , indicating a pretty good fit.
Figure 4.87
We compute $L (90) = 0.00027901 {(90)}^{2.7738} \approx 73.5$ meaning that the bottom $90 %$ of the wage earners brought home $73.5 %$ of the total income. Said differently, the top $10 %$ of wage earners made over $25 %$ of the total national income.

Exercises

In Exercises -, use the given graphs along with Theorem 4.4 to graph the given function. Track at least two points and state the domain and range using interval notation.

$F (x) = {(x - 2)}^{\frac{2}{3}} - 1$
$G (t) = {(t + 3)}^{π} + 1$
$F (x) = 3 - x^{\frac{2}{3}}$
$G (t) = {(1 - t)}^{π} - 2$
$F (x) = {(2 x + 5)}^{\frac{2}{3}} + 1$
$G (t) = {(\frac{t + 3}{2})}^{π} - 1$

In Exercises -, find a formula for each function below in the form $F (x) = a {(b x - h)}^{\frac{2}{3}} + k$ .

NOTE: There may be more than one solution!

$y = F (x)$
Figure 4.90
$y = F (x)$
Figure 4.91

For each function in Exercises - below

Analytically:
- find the domain.
- find the axis intercepts.
- analyze the end behavior.
Graph the function with help from a graphing utility and determine:
- the range.
- the local extrema, if they exist.
- intervals of increase/decrease.
- any `unusual steepness' or `local' verticality.
- vertical asymptotes.
- horizontal / slant asymptotes.
Construct a sign diagram for each function using the intercepts and graph.
Comment on any observed symmetry.
$f (x) = x^{\frac{2}{3}} {(x - 7)}^{\frac{1}{3}}$
$f (x) = x^{\frac{3}{2}} {(x - 7)}^{\frac{1}{3}}$
$g (t) = 2 t {(t + 3)}^{- \frac{1}{3}}$
$g (t) = t^{\frac{3}{2}} {(t - 2)}^{- \frac{1}{2}}$
$f (x) = x^{0.4} {(3 - x)}^{0.6}$
$f (x) = x^{0.5} {(3 - x)}^{0.5}$
$g (t) = 4 t {(9 - t^{2})}^{- \sqrt{2}}$
$g (t) = 3 {(t^{2} + 1)}^{- π}$
For each function $f (x)$ listed below, compute the average rate of change over the indicated interval.⁸ What trends do you observe? How do your answers manifest themselves graphically? Compare the results of this exercise with those of Exercise in Section 2.1 and Exercise in Section 3.1
$\begin{matrix} f (x) & [0.9, 1.1] & [0.99, 1.01] & [0.999, 1.001] & [0.9999, 1.0001] \\ x^{\frac{1}{2}} \\ x^{\frac{2}{3}} \\ x^{- 0.23} \\ x^{π} \end{matrix}$
The National Weather Service uses the following formula to calculate the wind chill:
$W = 35.74 + 0.6215 T_{a} - 35.75 V^{0.16} + 0.4275 T_{a} V^{0.16}$
where $W$ is the wind chill temperature in $∘$ F, $T_{a}$ is the air temperature in $∘$ F, and $V$ is the wind speed in miles per hour. Note that $W$ is defined only for air temperatures at or lower than $50 degrees$ F and wind speeds above $3$ miles per hour.
1. Suppose the air temperature is $42 degrees$ and the wind speed is $7$ miles per hour. Find the wind chill temperature. Round your answer to two decimal places.
2. Suppose the air temperature is $37 degrees$ F and the wind chill temperature is $30 degrees$ F. Find the wind speed. Round your answer to two decimal places.
As a follow-up to Exercise, suppose the air temperature is $28 degrees$ F.
1. Use the formula from Exercise to find an expression for the wind chill temperature as a function of the wind speed, $W (V)$ .
2. Solve $W ⁢ open parenthesis V close parenthesis equals 0$ , round your answer to two decimal places, and interpret.
3. Graph the function $W$ using a graphing utility and check your answer to part.
Suppose Fritzy the Fox, positioned at a point $(x, y)$ in the first quadrant, spots Chewbacca the Bunny at $open parenthesis 0, 0 close parenthesis$ . Chewbacca begins to run along a fence (the positive $y$ -axis) towards his warren. Fritzy, of course, takes chase and constantly adjusts his direction so that he is always running directly at Chewbacca. If Chewbacca's speed is $v sub 1$ and Fritzy's speed is $v sub 2$ , the path Fritzy will take to intercept Chewbacca, provided $v sub 2$ is directly proportional to, but not equal to, $v sub 1$ is modeled by
$y = \frac{1}{2} (\frac{x^{1 + v_{1} / v_{2}}}{1 + v_{1} / v_{2}} - \frac{x^{1 - v_{1} / v_{2}}}{1 - v_{1} / v_{2}}) + \frac{v_{1} v_{2}}{v_{2}^{2} - v_{1}^{2}}$
1. Determine the path that Fritzy will take if he runs exactly twice as fast as Chewbacca; that is, $v sub 2 equals 2 ⁢ v sub 1$ . Use your calculator to graph this path for $x greater than or equal to 0$ . What is the significance of the $y$ -intercept of the graph?
2. Determine the path Fritzy will take if Chewbacca runs exactly twice as fast as he does; that is, $v sub 1 equals 2 ⁢ v sub 2$ . Use a graphing utility to graph this path for $x greater than 0$ . Describe the behavior of $y$ as $x \to 0^{+}$ and interpret this physically.
3. With the help of your classmates, generalize parts (a) and (b) to two cases: $v sub 2 greater than v sub 1$ and $v sub 2 less than v sub 1$ . We will discuss the case of $v sub 1 equals v sub 2$ in Exercise in Section 7.6.

Answers

$F (x) = {(x - 2)}^{\frac{2}{3}} - 1$
Figure 4.92 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open bracket minus 1, ∞ close parenthesis$
$G (t) = {(t + 3)}^{π} + 1$
Figure 4.93 Domain: $open bracket minus 3, ∞ close parenthesis$ , Range: $open bracket 1, ∞ close parenthesis$
$F (x) = 3 - x^{\frac{2}{3}} = (- 1) x^{\frac{2}{3}} + 3$
Figure 4.94 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open parenthesis minus ∞, 3 close bracket$
$G (t) = {(1 - t)}^{π} - 2 = {((- 1) t + 1)}^{π} - 2$
Figure 4.95 Domain: $open parenthesis minus ∞, 1 close bracket$ , Range: $open bracket minus 2, ∞ close parenthesis$
$F (x) = {(2 x + 5)}^{\frac{2}{3}} + 1$
Figure 4.96 Domain: $open parenthesis minus ∞, ∞ close parenthesis$ , Range: $open bracket 1, ∞ close parenthesis$
$G (t) = {(\frac{t + 3}{2})}^{π} - 1 = {(\frac{1}{2} t + \frac{3}{2})}^{π} - 1$
Figure 4.97 Domain: $open bracket minus 3, ∞ close parenthesis$ , Range: $open bracket minus 1, ∞ close parenthesis$
One solution is: $F (x) = 2 {(x - 1)}^{\frac{2}{3}} - 2$
One solution is: $F (x) = - {(x + 1)}^{\frac{2}{3}} + 4$
$f (x) = x^{\frac{2}{3}} {(x - 7)}^{\frac{1}{3}}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ Intercepts: $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 7, 0 close parenthesis$ Graph:
Figure 4.98
$lim_{x \to - \infty} f (x) = - \infty$ , $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$ ⁹ Range: $open parenthesis minus ∞, ∞ close parenthesis$ Local minimum: $\approx (4.667, - 3.704)$ Local maximum: $open parenthesis 0, 0 close parenthesis$ (this is a cusp) Increasing: $open parenthesis minus ∞, 0 close bracket$ , $approximately equals open bracket 4.667, ∞ close parenthesis$ Decreasing: $open bracket 0, 4.667 close bracket$ Unusual steepness at $x equals 7$
Sign Diagram:
Figure 4.99
$f (x) = x^{\frac{3}{2}} {(x - 7)}^{\frac{1}{3}}$ Graph:
Figure 4.100
Domain: $open bracket 0, ∞ close parenthesis$ Intercepts: $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 7, 0 close parenthesis$ $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$ Range: $\approx [- 14.854, \infty)$ Local minimum: $\approx (5.727, - 14.854)$ Increasing: $approximately equals open bracket 5.727, ∞ close parenthesis$ Decreasing: $approximately equals open bracket 0, 5.727 close bracket$ Unusual steepness at $x equals 7$ Sign Diagram:
Figure 4.101
$g (t) = 2 t {(t + 3)}^{- \frac{1}{3}}$ Graph:
Figure 4.102
Domain: $(- \infty, - 3) \cup (- 3, \infty)$ Intercept: $open parenthesis 0, 0 close parenthesis$ $lim_{t \to - \infty} g (t) = \infty$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ Range: $open parenthesis minus ∞, ∞ close parenthesis$ Local minimum: $\approx (- 4.5, 7.862)$ Increasing: $\approx [- 4.5, - 3)$ , $open parenthesis minus 3, ∞ close parenthesis$ Decreasing: $\approx (- \infty, - 4.5]$ Vertical Asymptote: $t = - 3$ Sign Diagram:
Figure 4.103
$g (t) = t^{\frac{3}{2}} {(t - 2)}^{- \frac{1}{2}}$ Domain: $open parenthesis 2, ∞ close parenthesis$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ Graph:
Figure 4.104
$lim t → ∞ g ⁢ open parenthesis t close parenthesis equals ∞$ ¹⁰ Range: $approximately equals open bracket 5.196, ∞ close parenthesis$ Local minimum: $approximately equals open parenthesis 3, 5.196 close parenthesis$ Increasing: $approximately equals open bracket 3, ∞ close parenthesis$ Decreasing: $approximately equals open parenthesis 2, 3 close bracket$ Vertical asymptote: $t equals 2$ Sign Diagram:
Figure 4.105
$f (x) = x^{0.4} {(3 - x)}^{0.6}$ Graph:
Figure 4.106
Domain: $open bracket 0, 3 close bracket$ Intercepts: $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 3, 0 close parenthesis$ Range: $approximately equals open bracket 0, 1.5 close bracket$ Increasing: $approximately equals open bracket 0, 1.2 close bracket$ Decreasing: $approximately equals open bracket 1.2, 3 close bracket$ Unusual Steepness:¹¹ $x equals 0$ , $x equals 3$ Sign Diagram:
Figure 4.107
$f (x) = x^{0.5} {(3 - x)}^{0.5}$ Graph:
Figure 4.108
Domain: $open bracket 0, 3 close bracket$ Intercepts: $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 3, 0 close parenthesis$ Range: $approximately equals open bracket 0, 1.5 close bracket$ Increasing: $approximately equals open bracket 0, 1.5 close bracket$ Decreasing: $approximately equals open bracket 1.5, 3 close bracket$ Unusual Steepness:¹² $x equals 0$ , $x equals 3$ Sign Diagram:
Figure 4.109
$g (t) = 4 t {(9 - t^{2})}^{- \sqrt{2}}$ Graph:
Figure 4.110
Domain: $open parenthesis minus 3, 3 close parenthesis$ Intercepts: $open parenthesis 0, 0 close parenthesis$ Range: $open parenthesis minus ∞, ∞ close parenthesis$ $lim_{t \to - 3^{+}} g (t) = - \infty$ , $lim_{t \to 3^{-}} g (t) = \infty$ Vertical asymptotes: $t = - 3$ and $t equals 3$ Increasing: $open parenthesis minus 3, 3 close parenthesis$ Sign Diagram:
Figure 4.111
Note: $g$ is odd
$g (t) = 3 {(t^{2} + 1)}^{- π}$ Domain: $open parenthesis minus ∞, ∞ close parenthesis$ Graph:
Figure 4.112
Intercept: $open parenthesis 0, 3 close parenthesis$ $lim_{t \to - \infty} g (t) = 0$ $lim t → ∞ g ⁢ open parenthesis t close parenthesis equals 0$ Range: $open parenthesis 0, 3 close bracket$ Increasing: $open parenthesis minus ∞, 0 close bracket$ Decreasing: $open bracket 0, ∞ close parenthesis$ Horizontal asymptote: $y equals 0$ Sign Diagram:
Figure 4.113
Note: $g$ is even
As in Exercise in Section 2.1 and Exercise in Section 3.1, the slopes of these curves near $x equals 1$ approach the value of the exponent on $x$ .
$\begin{matrix} f (x) & [0.9, 1.1] & [0.99, 1.01] & [0.999, 1.001] & [0.9999, 1.0001] \\ x^{\frac{1}{2}} & 0.5006 & \approx \frac{1}{2} & \approx \frac{1}{2} & \approx \frac{1}{2} \\ x^{\frac{2}{3}} & 0.6672 & 0.6667 & \approx \frac{2}{3} & \approx \frac{2}{3} \\ x^{- 0.23} & - 0.2310 & \approx - 0.23 & \approx - 0.23 & \approx - 0.23 \\ x^{π} & 3.1544 & 3.1417 & \approx π & \approx π \end{matrix}$
1. $W approximately equals 37.55 degrees$ F.
2. $V approximately equals 9.84$ miles per hour.
1. $W (V) = 53.142 - 23.78 V^{0.16}$ . Since we are told in Exercise that wind chill is only effect for wind speeds of more than 3 miles per hour, we restrict the domain to $V greater than 3$ .
2. $W ⁢ open parenthesis V close parenthesis equals 0$ when $V approximately equals 152.29$ . This means, according to the model, for the wind chill temperature to be $0 degrees$ F, the wind speed needs to be $152.29$ miles per hour.
3. The graph of $y = W (V)$ is below.
  Figure 4.114
1. $y = \frac{1}{3} x^{\frac{3}{2}} - x^{\frac{1}{2}} + \frac{2}{3}$ . The point $(0, \frac{2}{3})$ is when Fritzy's path crosses Chewbacca's path - in other words, where Fritzy catches Chewbacca.
  Figure 4.115
  $y = \frac{1}{3} x^{\frac{3}{2}} - x^{\frac{1}{2}} + \frac{2}{3}$
2. $y = \frac{1}{6} x^{3} + \frac{1}{2} x^{- 1} - \frac{2}{3}$ . We find as $x \to 0^{+}$ , $y → ∞$ which means, in this case, Fritzy's pursuit never ends; he never catches Chewbacca. This makes sense since Chewbacca has a head start and is running faster than Fritzy.
  Figure 4.116
  $y = \frac{1}{6} x^{3} + \frac{1}{2} x^{- 1} - \frac{2}{3}$

Recall `lowest terms' means $m$ and $n$ have no common factors other than $1$ . ↩
That is, if $n$ is even, $x greater than or equal to 0$ and if $m less than 0$ , $x not equal to 0$ . ↩
Either $n equals 1$ is a special case in Definition 4.3 or we need to define what is meant by $the 1th root of x$ . The authors chose the former. ↩
The domain is all real numbers since the denominator (root) $5$ is odd; the range is all real numbers since the numerator (power) $13$ is odd. Since both power and root are odd, the function itself is an odd function, hence the symmetry about the origin. ↩
In general if $u^{p} = 0$ where $p greater than 0$ , then $u equals 0$ . ↩
This works for each and every circle, by the way, regardless of how large or small the circle is! ↩
or $x greater than 0$ if $p$ is negative. ↩
See Definition 1.8 in Section 1.2.4 for a review of this concept, as needed. ↩
Using Calculus it can be shown that $y = x - \frac{7}{3}$ is a slant asymptote of this graph. ↩
Using Calculus it can be shown that $y = t + 1$ is a slant asymptote of this graph. ↩
Note you may need to zoom in to see this. ↩
Note you may need to zoom in to see this. ↩

Adapted from Precalculus, Preliminary 4th Edition (integrated calculus), by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.