Power Functions: Positive Integral Exponents

Section 4.3 Power Functions: Positive Integral Exponents

Objectives

After completing this section, you should be able to do the following.

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First objective.
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Second objective.
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Third objective.
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In the next two sections we look at the properties and the graphs of the so-called power functions.

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Definition 4.3.1. Power Function.

A function $y = f(x)$ is called a power function if $f(x)$ can be expressed in the form:

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

where $k$ and $p$ are constants and $k \neq 0\text{.}$ The constant $k$ is called the coefficient of the power function $f(x)$ and $p$ is called the exponent.

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Note: When $y$ depends on $x$ according to the formula $y = kx^p$ we say that $y$ is directly proportional (or proportional) to $x^p$ with the coefficient of proportionality $k$ If $\displaystyle y = \frac{k}{x^p}\text{,}$ we say that $y$ is inversely proportional to $x^p$ with the coefficient of proportionality $k\text{.}$ So power functions express proportionality of the dependent variable to powers of the independent variable.

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As the title suggests in this section we look at the case when the exponent $p$ is a positive integer.

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Example 4.3.2.

Which of the functions below are power functions? For those which are, rewrite in the standard form $y = kx^p\text{.}$ Identify the coefficient and the exponent.

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$\displaystyle \displaystyle f(x) = 2x^2 \cdot 4x^3$
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$\displaystyle \displaystyle g(x) = (-(x^3))^5$
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$\displaystyle \displaystyle h(x) = \frac{(x^2)^4}{7x^3}$
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$\displaystyle \displaystyle r(x) = 2x^3 + 7x^2$
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$\displaystyle \displaystyle m(x) = \frac{8x}{\pi ^2}$
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$\displaystyle \displaystyle l(x) = 3 \cdot 2^x$
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Solution.

The function $f(x)$ is a power function. Using Rules of Exponents from Section 4.1, we can rewrite:

\begin{equation*} f(x) = 2x^2 \cdot 4x^3 = (2 \cdot 4)(x^2x^3) = 8x^5. \end{equation*}

The coefficient is $k = 8$ and the exponent is $p = 5\text{.}$

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$g(x)$ is a power function:

\begin{equation*} g(x) = (-(x^3))^5 = (-1 \cdot (x^3) )^5 = (-1)^5(x^3)^5 = -1 \cdot x^{15} = -x^{15}. \end{equation*}

The coefficient is $k = -1$ and the exponent is $p = 15\text{.}$

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$h(x)$ is a power function as well:

\begin{equation*} h(x) = \frac{(x^2)^4}{7x^3} = \frac{x^8}{7x^3} = \frac{1}{7}x^{8 - 3} = \frac{1}{7}x^{5}. \end{equation*}

The coefficient is $k = \frac{1}{7}$ and the exponent is $p = 5\text{.}$

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The function $r(x)$ is not a power function. It cannot be rewritten as $r(x) = kx^p\text{.}$ The function $r(x)$ is a sum of two power functions, $2x^3$ and $7x^2\text{.}$ A function that is a sum of power functions is called a polynomial function.
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The function $m(x)$ is a power function. Remember that $\pi$ is just a constant.

\begin{equation*} m(x) = \frac{8x}{\pi ^2} = \frac{8}{\pi^2}x. \end{equation*}

The coefficient is $k = \frac{8}{\pi^2}$ and the exponent is $p = 1\text{.}$

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$l(x)$ is not a power function. Observe that in $l(x)$ the base of the power expression $2^x$ is constant and equal to 2. The exponent $x$ is a variable. In a power function, it is the other way around: the base is a variable and the exponent is constant.
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By the way...

Functions like $l(x)$ where the variable $x$ is in the exponent is the topic of discussion in Chapter 5.
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Example 4.3.3.

A ball dropped from the Empire State Building has traveled down the distance of $d(t)$ feet after $t$ seconds where:

\begin{equation*} d(t) = 16t^2. \end{equation*}

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Is the function $d(t)$ a power function? If yes, identify the exponent and the coefficient.
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The Empire State Building is $1250$ ft tall. How long will it take for the ball to hit the ground?
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Solution.

$d(t)$ is a power function with $k = 16$ and $p = 2\text{.}$
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The ball will hit the ground when it has traveled $1250$ feet. That is, for a positive $t$ such that:

\begin{equation*} d(t) = 16t^2 = 1250. \end{equation*}

To solve the equation for $t$ we divide both sides by $16\text{:}$

\begin{equation*} t^2 = \frac{1250}{16}. \end{equation*}

The two solutions to this quadratic equation are:

\begin{equation*} t = \pm\sqrt{\frac{1250}{16}}. \end{equation*}

Since $t$ has to be positive, then the solution is

\begin{equation*} t = \sqrt{\frac{1250}{16}}. \end{equation*}

Hence, the ball will hit the ground after:

\begin{equation*} t = \sqrt{\frac{1250}{16}} \approx 8.8 \hspace{3pt} \text{seconds}. \end{equation*}

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Note that in terms of proportionality, we can say that the distance $d$ is directly proportional to $t^2$ with the coefficient of proportionality $16\text{.}$

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Example 4.3.4.

Let $V(r)$ be the volume of a sphere of radius $r\text{.}$ We know from elementary geometry that:

\begin{equation*} V(r) = \frac{4}{3}\pi r^3. \end{equation*}

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Is $V(r)$ a power function? If yes, find the coefficient and the exponent.
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What radius is required for the volume to be 25 cm$^3\text{?}$
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Solution.

Yes, $V(r)$ is a power function. The coefficient $k$ is $\frac{4\pi}{3}$ and the exponent $p$ is $3\text{.}$
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We are looking for the radius $r\text{,}$ in centimeters, such that:

\begin{equation*} \frac{4}{3}\pi r^3 = 25 \end{equation*}

Dividing both sides by $4\pi$ and multiplying by 3, then

\begin{equation*} r^3 = \frac{3}{4\pi}25 \end{equation*}

Now we take the power $\frac{1}{3}$ of both sides (in other words, the cube root) and obtain:

\begin{equation*} (r^3)^{\frac{1}{3}} = \left( \frac{3}{4\pi}25 \right)^{\frac{1}{3}} \end{equation*}

Since $(r^3)^{\frac{1}{3}} = r\text{,}$ then

\begin{equation*} r = \left( \frac{3}{4\pi}25 \right)^{\frac{1}{3}} \approx 1.814 \text{ cm} \end{equation*}

The volume is $25 \text{ cm}^3$ when the radius is $1.814$ cm.

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Subsection Graphs of Power Functions: Positive Integral Exponents

Graphs of power functions $y = kx^p$ with exponents $p$ that are positive integers are different for $p$ even and for $p$ odd.

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Even Positive Exponents

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If $p$ is even, then $x^p$ is positive for all $x$ except for $x = 0$ where $x^p = 0\text{.}$ For example, if $p=6$ then $1^6 = 1\text{,}$ $(-1)^6 = 1\text{,}$ $2^6 = 64\text{,}$ $(-2)^6 = 64\text{,}$ and so on. Hence, the graph of the power function $f(x) = kx^p$ for any even $p$ is symmetric about the $y$-axis as $f(-x) = f(x)\text{.}$

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Here are the graphs of the functions $y = kx^p$ for $p = 2, 4, 6$ and $k = 1\text{.}$ All graphs are U-shaped and reminiscent of the quadratic parabola $y = x^2\text{.}$

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$A short description.$

The coefficient $k$ in $y = kx^p\text{,}$ stretches or shrinks the graph of $y = x^p$ vertically. Additionally, if $k \lt 0\text{,}$ the graph is reflected over the $x$-axis.

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$A short description.$

Odd Positive Exponents

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If $p$ is odd, then $x^p$ is positive for $x \gt 0\text{,}$ it is $0$ for $x = 0\text{,}$ and negative for $x \lt 0\text{.}$ For example, let $p=3\text{.}$ Then $1^3 = 1\text{,}$ $(-1)^3 = -1\text{,}$ $2^3 = 8\text{,}$ $(-2)^3 = -8\text{,}$ and so on. Hence, the graph of the power function $f(x) = kx^p$ for any odd $p$ is symmetric about the origin $(0, 0)$ as $f(-x) = -f(x)\text{.}$

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Here are the graphs of the functions $y = kx^p$ for $p = 3, 5$ and $k = 1\text{.}$ This time, the graphs are S-shaped.

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$A short description.$

The coefficient $k$ in $y = kx^p\text{,}$ stretches or shrinks the graph of $y = x^p$ vertically. Additionally, if $k \lt 0\text{,}$ the graph is reflected over the $x$-axis.

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$A short description.$

Here is a summary of how graphs of power functions look for positive integer exponents $p\text{,}$ even and odd, and for coefficients $k$ positive and negative.

$f(x)=kx^p$

$p$ even

$p$ odd

Positive coefficient, $k \gt 0$

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$Graph of a representative power function with even exponent and positive coefficient.$

$Graph of a representative power function with odd exponent and positive coefficient.$

Negative coefficient, $k \lt 0$

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$Graph of a representative power function with even exponent and negative coefficient.$

$Graph of a representative power function with odd exponent and negative coefficient.$

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Exercises Exercises

. The braking distance is proportional to the square of the car’s speed and it depends on the coefficient of friction, $\mu\text{,}$ between the tires and the road surface. Let $D$ denote distance, in feet, and $S$ speed in mph. The formula for the braking distance is:

\begin{equation*} D=\frac{0.034}{\mu}S^2 \end{equation*}

Note that the braking distance does not include a driver’s reaction time

https://en.wikipedia.org/wiki/Braking_distance, accessed: 7/5/20

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7.

Is the braking distance a power function of speed? If yes, give the coefficient $k$ and the exponent $p\text{.}$ Assume that $\mu$ is a given constant.

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Solution.

Yes; $k=\frac{0.034}{\mu}\text{;}$ $p=2$

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8.

The coefficient of friction under normal conditions when the road is dry is $\mu=0.7\text{.}$ What is the braking distance of a car that travels on a dry road at $35$ mph? What is the braking distance at $70$ mph?

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Solution.

$59.5$ feet; $238$ feet

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9.

By what factor does the braking distance increase when speed $S$ doubles?

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Solution.

Factor of $4\text{.}$

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10.

The coefficient of friction on a wet road is $\mu=0.4\text{.}$ Calculate the braking distance of a car traveling at $70$ mph on a wet road.

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Solution.

$416.5$ feet

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11.

Below are the graphs of four power functions $y=kx^p$ where $p$ is a positive integer. In each of the graphs, is the exponent $p$ even or odd? Is the coefficient $k$ positive or negative?

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$A short description.$