Power Functions

Section 2.3 Power Functions

When the power of love overcomes the love of power the world will know peace. — Jimi Hendrix, American rock guitarist, singer, and songwriter, 1942 – 1970

Question: How many different 5–digit numbers can you write by using only digits 1 and 2?

What if you use the digits 1, 2, and 3?

Solution

Thie question is equivalent to the following problem: In how many different ways can we place digits 1 and 2 in five boxes arranged in a row:

\begin{equation*} \begin{array}{|c|c|c|c|c|} \hline \hspace{0.5cm}\amp \hspace{0.5cm}\amp \hspace{0.5cm}\amp \hspace{0.5cm}\amp \hspace{0.5cm}\\ \hline \end{array} \end{equation*}

Observe that there are two choices for each box: the digit 1 or the digit 2:

\begin{equation*} \begin{array}{|c|c|c|c|c|} \hline 2 \text{ choices} \amp 2 \text{ choices} \amp 2 \text{ choices} \amp 2 \text{ choices} \amp 2 \text{ choices} \\ \hline \end{array} \end{equation*}

It follows that the number of choices is $2\cdot 2\cdot 2\cdot 2\cdot 2=2^5=32\text{.}$

If we can use three digits, 1,2, and 3, then for each box we have three choices. It follows that the number of choices is $3\cdot 3\cdot 3\cdot 3\cdot 3=3^5=243\text{.}$

Positive Integer Exponent: If $x$ is a real number and $m$ is a positive integer than

\begin{equation*} x^m=\underbrace{x\cdot x\cdot \ldots \cdot x}_{m- \text{times}} \end{equation*}

Power Function: If $m$ is a positive integer then the function

\begin{equation*} f(x)=x^m \end{equation*}

is called a power function.

For graphs of a few power functions, see Figure 2.31.

Figure 2.31. Four power functions.

Facts About Power Functions: Let a power function

\begin{equation*} f(x)=x^m, m\in \mathbb{N} \end{equation*}

be given.

What is the domain of the function $f\text{?}$
Solution
The domain of the function $f$ is the set of all real numbers $\mathbb{R}\text{.}$
What is the range of the function $f\text{?}$
Solution

Note that if $m$ is an odd positive integer then the range of the function $f$ is the set of all real numbers $\mathbb{R}\text{.}$

If $m$ is an even positive integer that the range of the function $f$ is the set of all non–negative real numbers $[0,\infty)\text{.}$
Find the zeros of the function $f\text{,}$ i.e. solve the equation $f(x)=0\text{.}$
Solution
Since $m$ is a positive real number then $f(x)=x^m=0 \Leftrightarrow x=0\text{.}$
Determine the intervals of increase and decrease for the function $f\text{.}$
Solution

We read from the graphs above:

If $m$ is an odd positive integer then $f$ is an increasing function on its domain.

If $m$ is an even positive integer than $f$ decreases on the interval $(-\infty,0]$ and increases on the interval $[0,\infty)\text{.}$
Summary:

\begin{equation*} \begin{array}{l|l|l} f(x)=x^m, m\in\mathbb{N}\amp m \text{ even}\amp m \text { odd}\\ \hline \text{Domain}\amp \mathbb{R}\amp \mathbb{R}\\ \hline \text{Range} \amp [0,\infty)\amp \mathbb{R}\\ \hline f(x)=0 \amp x=0\amp x=0\\ \hline \text{Increasing}\amp [0,\infty) \amp\mathbb{R}\\ \hline \text{Decreasing}\amp (-\infty,0]\amp \\ \hline \text{Odd or even}\amp \text{even}\amp \text{odd}\\ \end{array} \end{equation*}

Properties of Positive Integers Exponents: Let $x$ and $y$ be real numbers and let $m$ and $n$ be positive integers. Then

\begin{equation*} x^m\cdot x^n=x^{m+n} \end{equation*}

\begin{equation*} (x^m)^n=x^{mn} \end{equation*}

\begin{equation*} x^m\cdot y^m=(xy)^m. \end{equation*}

Example 2.3.1. Properties.

Evaluate:

$\displaystyle 3^4\cdot 3^3-(3^3)^2$
$\displaystyle (3^3+6^3)^2- 2\cdot 18^3$
$(x^n+y^n)^2- 2\cdot (xy)^n\text{,}$ where $n\in \mathbb{N}\text{.}$

Solution

We use the fact that $(a+b)^2=a^2+2ab+b^2$ and the above listed properties of positive integers exponents:

$3^4\cdot 3^3-(3^3)^2= 3^{4+3}-3^{3\cdot 2}=3^7-3^6=3\cdot3^6-3^6=(3-1)\cdot 3^6=2\cdot 3^6.$
$(3^3+6^3)^2- 2\cdot 18^3=(3^3)^2+2\cdot 3^3\cdot 6^3+(6^3)^2-2\cdot 18^3$

$=3^6+2\cdot (3\cdot 6)^3+6^6-2\cdot 18^3=3^6+2\cdot 18^3+6^6-2\cdot 18^3=3^6+6^6.$
$\displaystyle (x^n+y^n)^2- 2\cdot (xy)^n=(x^n)^2+2\cdot x^n\cdot y^n+(y^n)^2-2\cdot x^n\cdot y^n=x^{2n}+y^{2n}.$

Detective J: Evaluate:

$\displaystyle \displaystyle \frac{3^5}{3^2}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3}=3^3=3^{5-2}$
$\displaystyle \displaystyle \frac{3^2}{3^5}=\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=\frac{1}{3^3}=\frac{1}{3^{2-5}}$
$\displaystyle \displaystyle \frac{3^5}{3^5}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$
Let $x\not=0$ and $m,n\in \mathbb{N}\text{.}$ Then

\begin{equation*} \frac{x^m}{x^n}=\left\{ \begin{array}{lll} x^{m-n}\amp \text{ if }\amp m\gt n\\ {\displaystyle \frac{1}{x^{n-m}}}\amp \text{ if }\amp m\lt n\\ 1\amp \text{ if }\amp m=n\\ \end{array} \right. \end{equation*}

Two Definitions: Let $x\not=0$ and $m\in \mathbb{N}\text{.}$ Then, by definition

\begin{equation*} x^{-m}=\frac{1}{x^m} \end{equation*}

and

\begin{equation*} x^0=1. \end{equation*}

More power functions: See Figure 2.32.

Figure 2.32. Four power functions.

More Facts About Power Functions: Let a power function

\begin{equation*} f(x)=x^{-m}=\frac{1}{x^m}, m\in \mathbb{N} \end{equation*}

be given.

What is the domain of the function $f\text{?}$
Solution
The domain of the function $f$ is the set $\{x\in \mathbb{R}:x\not=0\}=\mathbb{R}\backslash \{0\}\text{.}$
What is the range of the function $f\text{?}$
Solution

If $m$ is an odd integer then the range of the function $f$ is the set $\{x\in\mathbb{R}:x\not= 0\}=(-\infty,0)\cup (0,\infty)\text{.}$

If $m$ is an even integer then the range of the function $f$ is the set $\{x\in\mathbb{R}:x> 0\}=(0,\infty)\text{.}$
Find the zeros of the function $f\text{,}$ i.e. solve the equation $f(x)=0\text{.}$
Solution
Note that the numerator of the function $f$ equals to 1, so $f(x)\not=0$ for all $x\in \mathbb{R}\backslash \{0\}$\text{.}$
Determine the intervals of increase and decrease for the function $f\text{.}$
Solution

If $m$ is an odd positive integer then $f$ is a decreasing function on $(-\infty,0)$ and $(0,\infty)\text{.}$

If $m$ is an even positive integer then $f$ is a decreasing function on $(-\infty,0)$ and increasing on $(0,\infty)\text{.}$
Determine positive integers $m$ for which the function $f(x)=\frac{1}{x^m}$ can be the inverse function of another function.
Solution

Observe that if $m$ is an even number then $f$ fails the horizontal line test and hence is not a one–to–one function. This implies that in this case it does not have an inverse and thus cannot be the inverse function of another function.

In the case that $m$ is odd then, by the horizontal line test, $f$ is one–to–one and hence there is $f^{-1}(x)\text{.}$ Recall that $(f^{-1})^{-1}=f\text{.}$

Can we find the formula for $f^{-1}(x)$ where $f(x)=\frac{1}{x^m}\text{,}$ $m$ and odd positive integer and $x\not= 0\text{?}$

We start with

\begin{equation*} y=\frac{1}{x^m} \end{equation*}

and then interchange the roles of “$x$” and “$y$” to obtain

\begin{equation*} x=\frac{1}{y^m} \Leftrightarrow y^m=\frac{1}{x}, y\not=0,\ x\not= 0. \end{equation*}

Since $m$ is odd it follows that

\begin{equation*} y=\left(\frac{1}{x}\right)^{\frac{1}{m}}=\sqrt[m]{\frac{1}{x}}=\frac{1}{\sqrt[m]{x}}, \ x\not= 0. \end{equation*}

Therefore $\displaystyle f^{-1}(x)=\frac{1}{\sqrt[m]{x}}\text{,}$ $x\not=0\text{.}$

Note that if $m=1$ then $f^{-1}(x)=f(x)=\frac{1}{x}\text{.}$ Otherwise $f^{-1}\not= f\text{.}$ See Figure 2.33.

Figure 2.33. Functions $f(x)=\frac{1}{x^3}$ and $f^{-1}(x)=\frac{1}{\sqrt[3]{x}}\text{.}$
Summary:

\begin{equation*} \begin{array}{l|l|l} f(x)=x^{-m}, m\in\mathbb{N}\amp m \text{ even}\amp m \text { odd}\\ \hline \text{Domain}\amp \mathbb{R}\backslash \{0\}\amp \mathbb{R}\backslash \{0\}\\ \hline \text{Range} \amp (0,\infty)\amp (-\infty,0)\cup (0,\infty)\\ \hline f(x)=0 \amp \text{never}\amp \text{never}\\ \hline \text{Increasing}\amp (0,\infty)\amp \text{never}\\ \hline \text{Decreasing}\amp (-\infty,0)\amp (0,\infty)\text{ and } (0,\infty)\\ \hline \text{Odd or even}\amp \text{even}\amp \text{odd}\\ \end{array} \end{equation*}

More detective work:

What is $2^3\text{?}$
What is $2^{-3}\text{?}$
What is $\displaystyle 2^{\frac{22}{7}}\text{?}$
What is $2^{\pi}\text{?}$

Definition: For $m\in \mathbb{N}\backslash\{1\}$ and $x\in \mathbb{R}$ we define $\displaystyle x^{\frac{1}{m}}$ as the real number for which

\begin{equation*} \left( x^{\frac{1}{m}}\right)^m=x \end{equation*}

if such number exists. If there are two numbers with this property (this may happened if $m$ is even) we choose one that is non-negative.

The number $\displaystyle x^{\frac{1}{m}}$ is called the $m^{\text{th}}$ root of $x\text{.}$

Notation:

\begin{equation*} x^{\frac{1}{m}}=\sqrt[m]{x}, m\geq 3 \end{equation*}

\begin{equation*} x^{\frac{1}{2}}=\sqrt{x} \end{equation*}

Example 2.3.2. Rational powers.

Evaluate:

$\displaystyle \displaystyle 16^{\frac{1}{4}}$
$\displaystyle \displaystyle \sqrt[4]{16}$
$\displaystyle \displaystyle (-8)^{\frac{1}{3}}=\sqrt[3]{-8}.$
$\displaystyle \displaystyle 4^{\frac{1}{2}}=\sqrt{4}$
$\displaystyle \displaystyle (-4)^{\frac{1}{2}}=\sqrt{-4}$

Solution

Say $\displaystyle 16^{\frac{1}{4}}=t\text{.}$ Then, by definition $\displaystyle t^4=\left( 16^{\frac{1}{4}}\right)^4=16$ if such a number $t$ exists. We note that there are two numbers with this property, $t=-2$ and $t=2\text{,}$ i.e. $(-2)^4=16$ and $2^4=16\text{.}$ By definition, we choose the non-negative value $t=2$ and we conclude that $\displaystyle 16^{\frac{1}{4}}=2\text{.}$
By definition

\begin{equation*} \sqrt[4]{16}=16^{\frac{1}{4}}=2. \end{equation*}
Since $(-2)^3=-8\text{,}$ by definition $\displaystyle (-8)^{\frac{1}{3}}=\sqrt[3]{-8}=-2\text{.}$
By definition $\displaystyle 4^{\frac{1}{2}}=\sqrt{4}=2\text{.}$
Say $\displaystyle (-4)^{\frac{1}{2}}=t\text{.}$ Then, by definition $\displaystyle t^2=\left( (-4)^{\frac{1}{2}}\right)^2=-4$ if such a number $t$ exists. But $t^2\geq 0$ for all real numbers $t$ and $-4\lt 0\text{.}$

Hence there is no real number $t$ such that $t^2=-4$ and, by definition, $\displaystyle (-4)^{\frac{1}{2}}$ does not exist.

Four more graphs: See Figure 2.34.

Figure 2.34. $\displaystyle y=x^{\frac{1}{m}}=\sqrt[m]{x}, m=2,3,4,5$

Even More Facts About Power Functions: Let a power function

\begin{equation*} f(x)=x^{\frac{1}{m}}=\sqrt[m]{x}, m\in \mathbb{N}\backslash\{1\} \end{equation*}

be given.

What is the domain of the function $f\text{?}$
Solution

If $m$ is an odd positive integer then the domain of $f$ is the set of all real numbers.

If $m$ is an even positive integer then then the domain of $f$ is the set of all non-negative real numbers, $[0,\infty)\text{.}$
What is the range of the function $f\text{?}$
Solution

If $m$ is an odd positive integer then the range of $f$ is the set of all real numbers.

If $m$ is an even positive integer then then the range of $f$ is the set of all non-negative real numbers, $[0,\infty)\text{.}$
Find the zeros of the function $f\text{,}$ i.e., solve the equation $f(x)=0\text{.}$
Solution
Note that $f(x)=0$ if and only if $x=0\text{.}$
Determine the intervals of increase and decrease for the function $f\text{.}$
Solution
The function is increasing on its domain.
Summary:

\begin{equation*} \begin{array}{l|l|l} f(x)=\sqrt[m]{x}, m\in\mathbb{N}\backslash\{1\}\amp m \text{ even}\amp m \text { odd}\\ \hline \text{Domain}\amp [0,\infty)\amp \mathbb{R}\\ \hline \text{Range} \amp [0,\infty)\amp \mathbb{R}\\ \hline f(x)=0 \amp x=0\amp x=0\\ \hline \text{Increasing}\amp \text{on its domain}\amp \text{on its domain}\\ \hline \text{Decreasing}\amp \text{never} \amp \text{never} \\ \hline \text{Odd or even}\amp \text{neither}\amp \text{odd}\\ \end{array} \end{equation*}

Functions and their inverses: See Figure 2.35.

Figure 2.35. Functions and their inverses

Two facts:

If $m$ is an even integer then $\displaystyle f(x)=x^m\text{,}$ $x\geq 0\text{,}$ is an one-to-one function and its inverse function is $\displaystyle f^{-1}(x)=x^{\frac{1}{m}}=\sqrt[m]{x}\text{.}$ The domain of $f^{-1}$ is the set of nonnegative real numbers.
If $m$ is an odd integer then $\displaystyle f(x)=x^m\text{,}$ $x\in \mathbb{R}\text{,}$ is an one–to–one function and its inverse function is $\displaystyle f^{-1}(x)=x^{\frac{1}{m}}=\sqrt[m]{x}\text{.}$ The domain of $f^{-1}$ is the set of all real numbers.

Definition: If $m\in \mathbb{N}\text{,}$ $n\in \mathbb{Z}\text{,}$ and $x\in \mathbb{R}$ then

\begin{equation*} x^{\frac{n}{m}}=\left( x^{\frac{1}{m}}\right)^n \end{equation*}

whenever this is defined.

Must Know: Let $x$ and $y$ be real numbers and let $m$ and $n$ be rational numbers. Then

\begin{equation*} x^m\cdot x^n=x^{m+n} \end{equation*}

\begin{equation*} (x^m)^n=x^{mn} \end{equation*}

\begin{equation*} x^0=1 \end{equation*}

\begin{equation*} \frac{x^m}{x^n}=x^{m-n}, \ x\not= 0 \end{equation*}

\begin{equation*} x^m\cdot y^m=(xy)^m \end{equation*}

\begin{equation*} \left( \frac{x}{y}\right)^m=\frac{x^m}{y^m}, \ y\not= 0 \end{equation*}

Example 2.3.3. Power rules.

Simplify $\displaystyle \left( \frac{x^3y^2z}{x^2y^2z^3}\right)^3=$

Solution

By the rules listed above

\begin{equation*} \left( \frac{x^3y^2z}{x^2y^2z^3}\right)^3=\left(x^{3-2}y^{2-2}z^{1-3}\right)^2=(xy^0z^{-2})^3=(x\cdot 1\cdot z^{-2})^3=x^3z^{-6}=\frac{x}{z^6}. \end{equation*}