Precalculus: Lecture Notes

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

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 \(3\cdot 3\cdot 3\cdot 3\cdot 3=3^5=243\text{.}\)

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

   Solution

   The domain of the function \(f\) is the set of all real numbers \(\mathbb{R}\text{.}\)

2. 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 \([0,\infty)\text{.}\)

3. 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 \(f(x)=x^m=0 \Leftrightarrow x=0\text{.}\)

4. 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 \([0,\infty)\text{.}\)

5. Summary:

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

1. $\displaystyle \displaystyle \frac{3^5}{3^2}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3}=3^3=3^{5-2}$

2. $\displaystyle \displaystyle \frac{3^2}{3^5}=\frac{3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=\frac{1}{3^3}=\frac{1}{3^{2-5}}$

3. $\displaystyle \displaystyle \frac{3^5}{3^5}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$

4. Let $x\not=0$ and $m,n\in \mathbb{N}\text{.}$ Then

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

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

   Solution

   The domain of the function $f$ is the set \(\{x\in \mathbb{R}:x\not=0\}=\mathbb{R}\backslash \{0\}\text{.}\)

2. 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 \(\{x\in\mathbb{R}:x> 0\}=(0,\infty)\text{.}\)

3. 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 \(x\in \mathbb{R}\backslash \{0\}$\text{.}\)

4. 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 \((0,\infty)\text{.}\)

5. 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.

   

6. Summary:

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

1. What is $2^3\text{?}$

2. What is $2^{-3}\text{?}$

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

4. What is $2^{\pi}\text{?}$

1. 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, \([0,\infty)\text{.}\)

2. 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, \([0,\infty)\text{.}\)

3. 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 \(x=0\text{.}\)

4. Determine the intervals of increase and decrease for the function $f\text{.}$

   Solution

   The function is increasing on its domain.

5. Summary:

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

1. 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.

2. 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.