Alg-tool Exponential Functions

Section 7.2 Exponential Functions

Subsection 1. Evaluate exponential functions

Subsubsection Examples

Powers come before products in the order of operations, so to evaluate an exponential function $f(x)=ab^x$ we evaluate $b^x$ before multiplying by $a\text{.}$

Example 7.13.

Evaluate $~f(x)=8 \cdot 4^x\text{.}$

$\displaystyle f(2)$
$\displaystyle f(-2)$
$\displaystyle f\left(\frac{1}{2}\right)$
$\displaystyle f\left(-\frac{1}{2}\right)$

Solution

Follow the order of operations: compute powers before products.

$\displaystyle f(2)=8 \cdot 4^2 = 8\cdot 16 = 128$
$\displaystyle f(-2)=8 \cdot 4^{-2} = 8\cdot \dfrac{1}{16} = \dfrac{1}{2}$
$\displaystyle f\left(\frac{1}{2}\right)=8 \cdot 4^{1/2} = 8\cdot 2 = 16$
$\displaystyle f\left(-\frac{1}{2}\right)=8 \cdot 4^{-1/2} = 8\cdot \dfrac{1}{2} = 4$

Example 7.14.

Evaluate $~g(x)=120(0.65)^x~\text{.}$ Round your answers to thousandths.

$\displaystyle g(2.3)$
$\displaystyle g(-1.8)$
$\displaystyle g(0.4)$
$\displaystyle g(-0.25)$

Solution

Follow the order of operations: use your calculator to compute powers before products. Do not round off at intermediate steps!

$\displaystyle g(2.3) = 120(0.65)^{2.3} = 120(0.37127...) = 44.554$
$\displaystyle g(-1.8) = 120(0.65)^{-1.8} = 120(2.17148...) = 260.578$
$\displaystyle g(0.4) = 120(0.65)^{0.4} = 120(0.84171...) = 101.006$
$\displaystyle g(-0.25) = 120(0.65)^{-0.25} = 120(1.11370...) = 133.645$

Subsubsection Exercises

Checkpoint 7.15.

Evaluate each function. Give your answers as common fractions.

$G(t)=15(5)^t~\text{.}$ Find $G(-3)$
$H(n)=4\left(\dfrac{1}{27}\right)^n~\text{.}$ Find $H\left(\dfrac{2}{3}\right)$
$F(x)=\dfrac{1}{2} \cdot 8^x~\text{.}$ Find $F\left(-\dfrac{1}{3}\right)$

Answer

$\displaystyle \dfrac{3}{25}$
$\displaystyle \dfrac{4}{9}$
$\displaystyle \dfrac{1}{4}$

Checkpoint 7.16.

Evaluate each function. Round your answers to hundredths.

$G(t)=15(1.5)^t~\text{.}$ Find $G(-3)$
$h(z)=1.8(0.8)^z~\text{.}$ Find $h(4)$
$F(w)=2500(1.03)^w~\text{.}$ Find $F(25)$

Answer

$\displaystyle 4.44$
$\displaystyle 0.745$
$\displaystyle 5234.44$

Subsection 2. Interpret function notation

The definitions of the variables help us interpret function notation.

Subsubsection Examples

Example 7.17.

The number of students at Salt Creek Elementary School is growing according to the formula $~f(t)=500(1.08)^t~\text{,}$ where $t$ is the number of years since the school opened in 2005.

What does the equation $~f(6)=500(1.08)^6~$ tell us about the school?
Use function notation to say that the student population was 583 in 2007.

Solution

In this equation, $t=6$ and $f(6)=793\text{.}$ In 2011 (six years after the school opened), the student population was 793.
In 2007, $t=2\text{,}$ so $f(2)=500(1.08)^2=583\text{.}$

Example 7.18.

The value of Digicorp stock has been falling according to the formula $~V(w)=48(0.96)^w~\text{,}$ where $w$ is the number of weeks since its peak value of $48 per share.

Use function notation to say that 8 weeks later the value of a share of Digicorp stock was $34.63.
What does the equation $~V(12) = 48(0.96)^{12} = 29.41~$ tell us about the stock?

Solution

We evaluate the function at $w=8$ to get $~V(8) = 48(0.96)^8 = 34.63\text{.}$
In this equation, $w=12$ and $V(12) = 29.41\text{,}$ so 12 weeks after the peak value a share of Digicorp stock was worth $29.41.

Subsubsection Exercises

Checkpoint 7.19.

The number of internet users in the United States is given by $~I(t) = 95,331,000(1.09)^t~\text{,}$ where $t=0$ in 2000. Use function notation to say that the number of internet users in 2005 was 146,679,000.

Answer

$I(t) = 146,679,000$

Checkpoint 7.20.

The percent of U.S. households that maintain a landline telephone is decreasing according to the formula $~L(t) = 95(0.96)^t~\text{,}$ where $t=0$ in 2004. What does the equation $~L(t) = 95(0.96)^{10} = 63~$ tell us about landlines?

Answer

In 2014, 63% of households maintained a landline.

Subsection 3. Solve equations graphically

We first solve equations graphically in Section 1.3, so you might want to review that section.

Subsubsection Examples

Example 7.21.

Here is a graph of $f(x)=x^2+2x-16\text{.}$ Use the graph to solve the equation $x^2+2x-16=8\text{.}$ Show your work on the graph.

Solution

To solve the equation, we want to find $x$-values that produce a function value of 8. The vertical coordinate of each point on the graph is given by the function value, $f(x)\text{.}$ So we look for points on the graph with vertical coordinate $f(x)=8\text{.}$

There are two such points, $(-6,8)$ and $(4,8)\text{.}$ Those points tell us that $f(-6)=8$ and $f(4)=8\text{.}$ Thus, the $x$-coordinates of the points, namely $-6$ and $4\text{,}$ are the solutions. To check algebraically, we can verify that $f(-6)=8$ and $f(4)=8\text{:}$

\begin{align*} f(\alert{-6}) \amp = (\alert{-6})^2+2(\alert{-6})-16 = 36-12-16 = 8\\ f(\alert{4})\amp = \alert{4}^2+2(\alert{4})-16 = 16+8-16 = 8 \end{align*}

Example 7.22.

Here is a graph of $G(x)=\dfrac{20}{(x-3)^2}\text{.}$ Use the graph to solve the equation $\dfrac{20}{(x-3)^2} = 5\text{.}$ Show your work on the graph.

Solution

We find any points on the graph with vertical coordinate $G(x)=5\text{.}$ There are two points, $(1,5)$ and $(5,5\text{.}$) The $x$-coordinates of those points, namely 1 and 5, are the solutions.