7.1 Exponential Functions

Of all of the functions we study in this text, exponential functions are possibly the ones which impact everyday life the most. This section introduces us to these functions while the rest of the chapter will more thoroughly explore their properties.

Up to this point, we have dealt with functions which involve terms like $x^{3}$ , $x^{\frac{3}{2}}$ , or $x^{π}$ - in other words, terms of the form $x^{p}$ where the base of the term, $x$ , varies but the exponent of each term, $p$ , remains constant.

In this chapter, we study functions of the form $f (x) = b^{x}$ where the base $b$ is a constant and the exponent $x$ is the variable. We start our exploration of these functions with the time-honored classic, $f (x) = 2^{x}$ .

We make a table of function values, plot enough points until we are more or less confident with the shape of the curve, and connect the dots in a pleasing fashion.

$\begin{matrix} x & f (x) & (x, f (x)) \\ - 3 & 2^{- 3} = \frac{1}{8} & (- 3, \frac{1}{8}) \\ - 2 & 2^{- 2} = \frac{1}{4} & (- 2, \frac{1}{4}) \\ - 1 & 2^{- 1} = \frac{1}{2} & (- 1, \frac{1}{2}) \\ 0 & 2^{0} = 1 & (0, 1) \\ 1 & 2^{1} = 2 & (1, 2) \\ 2 & 2^{2} = 4 & (2, 4) \\ 3 & 2^{3} = 8 & (3, 8) \end{matrix}$

Coordinate-plane figure. — **Figure 7.1** $y = f (x) = 2^{x}$

A few remarks about the graph of $f (x) = 2^{x}$ are in order. As $x → minus ∞$ and takes on values like $x = - 100$ or $x = - 1000$ , the function $f (x) = 2^{x}$ takes on values like $f (- 100) = 2^{- 100} = \frac{1}{2^{100}}$ or $f (- 1000) = 2^{- 1000} = \frac{1}{2^{1000}}$ .

In other words, as $x → minus ∞$ , $2^{x} \approx \frac{1}{very big (+)} \approx very small (+)$ That is, as $x → minus ∞$ , $2^{x} \to 0^{+}$ , so $lim_{x \to - \infty} 2^{x} = 0$ . This produces the $x$ -axis, $y equals 0$ as a horizontal asymptote to the graph as $x → minus ∞$ .

On the flip side, as $x → ∞$ , we find $f (100) = 2^{100}$ , $f (1000) = 2^{1000}$ , and so on, thus $lim_{x \to \infty} 2^{x} = \infty$ .

We note that by `connecting the dots in a pleasing fashion,' we are implicitly using the fact that $f (x) = 2^{x}$ is not only defined for all real numbers,¹ but is also continuous. Moreover, we are assuming $f (x) = 2^{x}$ is increasing: that is, if $a < b$ , then $2^{a} < 2^{b}$ . While these facts are true, the proofs of these properties are best left to Calculus. For us, we assume these properties in order to state the domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ , the range of $f$ is $open parenthesis 0, ∞ close parenthesis$ and, since $f$ is increasing, $f$ is one-to-one, hence invertible.

Suppose we wish to study the family of functions $f (x) = b^{x}$ . Which bases $b$ make sense to study? We find that we run into difficulty if $b less than 0$ . For example, if $b = - 2$ , then the function $f (x) = {(- 2)}^{x}$ has trouble, for instance, at $x = \frac{1}{2}$ since ${(- 2)}^{1 / 2} = \sqrt{- 2}$ is not a real number. In general, if $x$ is any rational number with an even denominator,² then ${(- 2)}^{x}$ is not defined, so we must restrict our attention to bases $b greater than or equal to 0$ .

What about $b equals 0$ ? The function $f (x) = 0^{x}$ is undefined for $x less than or equal to 0$ because we cannot divide by $0$ and $0 to the power 0$ is an indeterminate form. For $x greater than 0$ , $0^{x} = 0$ so the function $f (x) = 0^{x}$ is the same as the function $f ⁢ open parenthesis x close parenthesis equals 0$ , $x greater than 0$ . Since we know everything about this function, we ignore this case.

The only other base we exclude is $b equals 1$ , since the function $f (x) = 1^{x} = 1$ for all real numbers $x$ , since, once again, a function we have already studied. We are now ready for our definition of exponential functions.

Definition

An exponential function is the function of the form

$f (x) = b^{x}$

where $b$ is a real number, $b greater than 0$ , $b not equal to 1$ . The domain of an exponential function $open parenthesis minus ∞, ∞ close parenthesis$ .

NOTE: More specifically, $f (x) = b^{x}$ is called the `base $b$ exponential function.'

We leave it to the reader to verify³ that if $b greater than 1$ , then the exponential function $f (x) = b^{x}$ will share the same basic shape and characteristics as $f (x) = 2^{x}$ .

What if $0 less than b less than 1$ ? Consider $g (x) = {(\frac{1}{2})}^{x}$ . We could certainly build a table of values and connect the points, or we could take a step back and note that $g (x) = {(\frac{1}{2})}^{x} = {(2^{- 1})}^{x} = 2^{- x} = f (- x)$ , where $f (x) = 2^{x}$ . Per Section 5.4, the graph of $f (- x)$ is obtained from the graph of $f (x)$ by reflecting it across the $y$ -axis.

$\overset{\overset{reflect across y -axis}{\overset{}{\to}}}{multiply each x -coordinate by - 1}$

We see that the domain and range of $g$ match that of $f$ , namely $open parenthesis minus ∞, ∞ close parenthesis$ and $open parenthesis 0, ∞ close parenthesis$ , respectively. Like $f$ , $g$ is also one-to-one. Whereas $f$ is always increasing, $g$ is always decreasing. As a result, as $lim_{x \to - \infty} g (x) = \infty$ , and on the flip side, $lim x → ∞ g ⁢ open parenthesis x close parenthesis equals 0$ . (More specifically, $x → ∞$ , $g (x) \to 0^{+}$ .) It shouldn't be too surprising that for all choices of the base $0 less than b less than 1$ , the graph of $y = b^{x}$ behaves similarly to the graph of $g$ .

We summarize the basic properties of exponential functions in the following theorem.

Theorem

Properties of Exponential Functions: Suppose $f (x) = b^{x}$ .

The domain of $f$ is $open parenthesis minus ∞, ∞ close parenthesis$ and the range of $f$ is $open parenthesis 0, ∞ close parenthesis$ .
$f ⁢ open parenthesis x close parenthesis greater than 0$ for all $x$ .
$open parenthesis 0, 1 close parenthesis$ is on the graph of $f$ and $y equals 0$ is a horizontal asymptote to the graph of $f$ .
$f$ is one-to-one, continuous and smooth⁴

If $b greater than 1$ :
- $f$ is always increasing
- $lim_{x \to - \infty} f (x) = 0$
- $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals ∞$
- The graph of $f$ resembles:
  Figure 7.4 $y = b^{x}$ , $b greater than 1$

If $0 less than b less than 1$ :
- $f$ is always decreasing
- $lim_{x \to - \infty} f (x) = \infty$
- $lim x → ∞ f ⁢ open parenthesis x close parenthesis equals 0$
- The graph of $f$ resembles:
  Figure 7.5 $y = b^{x}$ , $0 less than b less than 1$

Exponential functions also inherit the basic properties of exponents from Theorem 4.3. We formalize these below and use them as needed in the coming examples.

Theorem

(Algebraic Properties of Exponential Functions) Let $f (x) = b^{x}$ be an exponential function ( $b greater than 0$ , $b not equal to 1$ ) and let $u$ and $w$ be real numbers.

Product Rule: $f (u + w) = f (u) f (w)$ . In other words, $b^{u + w} = b^{u} b^{w}$
Quotient Rule: $f (u - w) = \frac{f (u)}{f (w)}$ . In other words, $b^{u - w} = \frac{b^{u}}{b^{w}}$
Power Rule: ${(f (u))}^{w} = f (u w)$ . In other words, ${(b^{u})}^{w} = b^{u w}$

In addition to base $2$ which is important to computer scientists,⁵ two other bases are used more often than not in scientific and economic circles. The first is base $10$ . Base $10$ is called the `common base' and is important in the study of intensity (sound intensity, earthquake intensity, acidity, etc.)

The second base is an irrational number, $e$ . Like $the square root of 2$ or $π$ , the decimal expansion of $e$ neither terminates nor repeats, so we represent this number by the letter ` $e$ .' A decimal approximation of $e$ is $e approximately equals 2.718$ , so the function $f (x) = e^{x}$ is an increasing exponential function.

The number $e$ is called the `natural base' for lots of reasons, one of which is that it `naturally' arises in the study of growth functions in Calculus. We will more formally discuss the origins of $e$ in Section 7.6.

It is time for an example.

Example 1

Graph the following functions by starting with a basic exponential function and using transformations, Theorem 5.11. Track at least three points and the horizontal asymptote through the transformations.
1. $F (x) = 2 {(\frac{1}{3})}^{x - 1}$
2. $G (t) = 2 - e^{- t}$
Find a formula for the graph of the function below. Assume the base of the exponential is $2$ .
Figure 7.6 $y = F (x)$

1. Since the base of the exponent in $F (x) = 2 {(\frac{1}{3})}^{x - 1}$ is $\frac{1}{3}$ , we start with the graph of $f (x) = {(\frac{1}{3})}^{x}$ .
  To use Theorem 5.11, we first need to choose some `control points' on the graph of $f (x) = {(\frac{1}{3})}^{x}$ . Since we are instructed to track three points (and the horizontal asymptote, $y equals 0$ ) through the transformations, we choose the points corresponding to $x = - 1$ , $x equals 0$ , and $x equals 1$ : $open parenthesis minus 1, 3 close parenthesis$ , $open parenthesis 0, 1 close parenthesis$ , and $(1, \frac{1}{3})$ , respectively.
  Next, we need determine how to modify $f (x) = {(\frac{1}{3})}^{x}$ to obtain $F (x) = 2 {(\frac{1}{3})}^{x - 1}$ . The key is to recognize the argument, or `inside' of the function is the exponent and the `outside' is anything outside the base of $\frac{1}{3}$ . Using these principles as a guide, we find $F (x) = 2 f (x - 1)$ .
  Per Theorem 5.11, we first add $1$ to the $x$ -coordinates of the points on the graph of $y = f (x)$ , shifting the graph to the right $1$ unit. Next, multiply the $y$ -coordinates of each point on this new graph by $2$ , vertically stretching the graph by a factor of $2$ .
  Looking point by point, we have $open parenthesis minus 1, 3 close parenthesis → open parenthesis 0, 3 close parenthesis → open parenthesis 0, 6 close parenthesis$ , $open parenthesis 0, 1 close parenthesis → open parenthesis 1, 1 close parenthesis → open parenthesis 1, 2 close parenthesis$ , and $(1, \frac{1}{3}) \to (2, \frac{1}{3}) \to (2, \frac{2}{3})$ . The horizontal asymptote, $y equals 0$ remains unchanged under the horizontal shift and the vertical stretch since $2 \cdot 0 = 0$ .
  Below we graph $y = f (x) = {(\frac{1}{3})}^{x}$ on the left $y = F (x) = 2 {(\frac{1}{3})}^{x - 1}$ on the right.
  Figure 7.7 $y = f (x) = {(\frac{1}{3})}^{x}$
  
  Figure 7.8 $y = F (x) = 2 {(\frac{1}{3})}^{x - 1}$
  As always we can check our answer by verifying each of the points $open parenthesis 0, 6 close parenthesis$ , $open parenthesis 1, 2 close parenthesis$ , $(2, \frac{2}{3})$ is on the graph of $F (x) = 2 {(\frac{1}{3})}^{x - 1}$ by checking $F ⁢ open parenthesis 0 close parenthesis equals 6$ , $F ⁢ open parenthesis 1 close parenthesis equals 2$ , and $F (2) = \frac{2}{3}$ .
  We can check the end behavior as well, that is, $lim_{x \to - \infty} F (x) = \infty$ and $lim x → ∞ F ⁢ open parenthesis x close parenthesis equals 0$ . We leave these calculations to the reader.
2. Since the base of the exponential in $G (t) = 2 - e^{- t}$ is $e$ , we start with the graph of $g (t) = e^{t}$ .
  Note that since $e$ is an irrational number, we will use the approximation $e approximately equals 2.718$ when plotting points. However, when it comes to tracking and labeling said points, we do so with exact coordinates, that is, in terms of $e$ .
  We choose points corresponding to $t = - 1$ , $t equals 0$ , and $t equals 1$ : $(- 1, e^{- 1}) \approx (- 1, 0.368)$ , $open parenthesis 0, 1 close parenthesis$ , and $open parenthesis 1, e close parenthesis approximately equals open parenthesis 1, 2.718 close parenthesis$ , respectively.
  Next, we need to determine how the formula for $G (t) = 2 - e^{- t}$ can be obtained from the formula $g (t) = e^{t}$ . Rewriting $G (t) = - e^{- t} + 2$ , we find $G (t) = - g (- t) + 2$ .
  Following Theorem 5.11, we first multiply the $t$ -coordinates of the graph of $y = g (t)$ by $minus 1$ , effecting a reflection across the $y$ -axis. Next, we multiply each of the $y$ -coordinates by $minus 1$ which reflects the graph about the $t$ -axis. Finally, we add $2$ to each of the $y$ -coordinates of the graph from the second step which shifts the graph up $2$ units.
  Tracking points, we have $(- 1, e^{- 1}) \to (1, e^{- 1}) \to (1, - e^{- 1}) \to (1, - e^{- 1} + 2) \approx (1, 1.632)$ , $open parenthesis 0, 1 close parenthesis → open parenthesis 0, 1 close parenthesis → open parenthesis 0, minus 1 close parenthesis → open parenthesis 0, 1 close parenthesis$ , and $(1, e) \to (- 1, e) \to (- 1, - e) \to (- 1, - e + 2) \approx (- 1, - 0.718)$ . The horizontal asymptote is unchanged by the reflections, but is shifted up $2$ units $y = 0 \to y = 2$ .
  We graph $g (t) = e^{t}$ below on the left and the transformed function $G (t) = - e^{- t} + 2$ below on the left. As usual, we can check our answer by verifying the indicated points do, in fact, lie on the graph of $y = G (t)$ along with checking end behavior. We leave these details to the reader.
  Figure 7.9 $y = g (t) = e^{t}$
  
  Figure 7.10 $y = G (t) = - e^{- t} + 2$
Since we are told to assume the base of the exponential function is $2$ , we assume the function $F (x)$ is the result of the transforming the graph of $f (x) = 2^{x}$ using Theorem 5.11. This means we are tasked with finding values for $a$ , $b$ , $h$ , and $k$ so that $F (x) = a f (b x - h) + k = a \cdot 2^{b x - h} + k$ .
Since the horizontal asymptote to the graph of $y = f (x) = 2^{x}$ is $y equals 0$ and the horizontal asymptote to the graph $y = F (x)$ is $y equals 4$ , we know the vertical shift is $4$ units up, so $k equals 4$ .
Next, looking at how the graph of $F$ approaches the vertical asymptote, it stands to reason the graph of $f (x) = 2^{x}$ undergoes a reflection across $x$ -axis, meaning $a less than 0$ . For simplicity, we assume $a = - 1$ and set see if we can find values for $b$ and $h$ that go along with this choice.
Since $open parenthesis minus 1, 0 close parenthesis$ and $open parenthesis 0, minus 4 close parenthesis$ on the graph of $F (x) = - 2^{b x - h} + 4$ , we know $F (- 1) = 0$ and $F (0) = - 4$ . From $F (- 1) = 0$ , we have $- 2^{- b - h} + 4 = 0$ or $2^{- b - h} = 4 = 2^{2}$ . Hence, $- b - h = 2$ is one solution.⁶
Next, using $F (0) = - 4$ , we get $- 2^{- h} + 4 = - 4$ or $2^{- h} = 8 = 2^{3}$ . From this, we have $- h = 3$ so $h = - 3$ . Putting this together with $- b - h = 2$ , we get $- b + 3 = 2$ so $b equals 1$ .
Hence, one solution to the problem is $F (x) = - 2^{x + 3} + 4$ . To check our answer, we leave it to the reader verify $F (- 1) = 0$ , $F (0) = - 4$ , $lim_{x \to - \infty} F (x) = 4$ , $lim_{x \to \infty} F (x) = - \infty$ .
Since we made a simplifying assumption ( $a = - 1$ ), we may well wonder if our solution is the only solution. Indeed, we started with what amounts to three pieces of information and set out to determine the value of four constants. We leave this for a thoughtful discussion in Exercise.

Our next example showcases an important application of exponential functions: economic depreciation.

Example 2

The value of a car can be modeled by $V (t) = 25 {(0.8)}^{t}$ , where $t greater than or equal to 0$ is number of years the car is owned and $V (t)$ is the value in thousands of dollars.

Find and interpret $V ⁢ open parenthesis 0 close parenthesis$ , $V ⁢ open parenthesis 1 close parenthesis$ , and $V ⁢ open parenthesis 2 close parenthesis$ .
Find and interpret the average rate of change of $V$ over the intervals $open bracket 0, 1 close bracket$ and $open bracket 0, 2 close bracket$ and $open bracket 1, 2 close bracket$ .
Find and interpret $\frac{V (1)}{V (0)}$ , $\frac{V (2)}{V (1)}$ and $\frac{V (2)}{V (0)}$ .
For $t greater than or equal to 0$ , find and interpret $\frac{V (t + 1)}{V (t)}$ and $\frac{V (t + k)}{V (t)}$ .
Find and interpret $\frac{V (1) - V (0)}{V (0)}$ , $\frac{V (2) - V (1)}{V (1)}$ , and $\frac{V (2) - V (0)}{V (0)}$ .
For $t greater than or equal to 0$ , find and interpret $\frac{V (t + 1) - V (t)}{V (t)}$ and $\frac{V (t + k) - V (t)}{V (t)}$ .
Graph $y = V (t)$ starting with the graph of $y = V (t)$ and using transformations.
Interpret the horizontal asymptote of the graph of $y = V (t)$ .
Using a graphing utility, determine how long it takes for the car to depreciate to (a) one half its original value and (b) one quarter of its original value. Round your answers to the nearest hundredth.

We find $V (0) = 25 {(0.8)}^{0} = 25 \cdot 1 = 25$ , $V (1) = 25 {(0.8)}^{1} = 25 \cdot 0.8 = 20$ and $V (2) = 25 {(0.8)}^{2} = 25 \cdot 0.64 = 16$ . Since $t$ represents the number of years the car has been owned, $t equals 0$ corresponds to the purchase price of the car. Since $V (t)$ returns the value of the car in thousands of dollars, $V ⁢ open parenthesis 0 close parenthesis equals 25$ means the car is worth $$ 25, 000$ when first purchased. Likewise, $V ⁢ open parenthesis 1 close parenthesis equals 20$ and $V ⁢ open parenthesis 2 close parenthesis equals 16$ means the car is worth $$ 20, 000$ after one year of ownership and $$ 16, 000$ after two years, respectively.
Recall to find the average rate of change of $V$ over an interval $[a, b]$ , we compute: $\frac{V (b) - V (a)}{b - a}$ . For the interval $open bracket 0, 1 close bracket$ , we find $\frac{V (1) - V (0)}{1 - 0} = \frac{20 - 25}{1} = - 5$ , which means over the course of the first year of ownership, the value of the car depreciated, on average, at a rate of $$ 5000$ per year.
For the interval $open bracket 0, 1 close bracket$ , we compute $\frac{V (2) - V (0)}{2 - 0} = \frac{16 - 25}{2} = - 4.5$ , which means over the course of the first two years of ownership, the car lost, on average, $$ 4500$ per year in value.
Finally, we find for the interval $open bracket 1, 2 close bracket$ , $\frac{V (2) - V (1)}{2 - 1} = \frac{16 - 20}{1} = - 4$ , meaning the car lost, on average, $$ 4000$ in value per year between the first and second years.
Notice that the car lost more value over the first year ( $$ 5000$ ) than it did the second year ( $$ 4000$ ), and these losses average out to the average yearly loss over the first two years ( $$ 4500$ per year.)⁷
We compute: $\frac{V (1)}{V (0)} = \frac{20}{25} = 0.8$ , $\frac{V (2)}{V (1)} = \frac{16}{20} = 0.8$ , and $\frac{V (2)}{V (0)} = \frac{16}{25} = 0.64$ .
The ratio $\frac{V (1)}{V (0)} = 0.8$ can be rewritten as $V ⁢ open parenthesis 1 close parenthesis equals 0.8 ⁢ V ⁢ open parenthesis 0 close parenthesis$ which means that the value of the car after $1$ year, $V ⁢ open parenthesis 1 close parenthesis$ is $0.8$ times, or $80 %$ the initial value of the car, $V ⁢ open parenthesis 0 close parenthesis$ .
Similarly, the ratio $\frac{V (2)}{V (1)} = 0.8$ rewritten as $V ⁢ open parenthesis 2 close parenthesis equals 0.8 ⁢ V ⁢ open parenthesis 1 close parenthesis$ means the value of the car after $2$ years, $V ⁢ open parenthesis 2 close parenthesis$ is $0.8$ times, or $80 %$ the value of the car after one year, $V ⁢ open parenthesis 1 close parenthesis$ .
Finally, the ratio $\frac{V (2)}{V (0)} = 0.64$ , or $V ⁢ open parenthesis 2 close parenthesis equals 0.64 ⁢ V ⁢ open parenthesis 0 close parenthesis$ means the value of the car after $2$ years, $V ⁢ open parenthesis 2 close parenthesis$ is $0.64$ times, or $64 %$ of the initial value of the car, $V ⁢ open parenthesis 0 close parenthesis$ .
Note that this last result tracks with the previous answers. Since $V ⁢ open parenthesis 1 close parenthesis equals 0.8 ⁢ V ⁢ open parenthesis 0 close parenthesis$ and $V ⁢ open parenthesis 2 close parenthesis equals 0.8 ⁢ V ⁢ open parenthesis 1 close parenthesis$ , we get $V (2) = 0.8 V (1) = 0.8 (0.8 V (0)) = 0.64 V (0)$ . Also note it is no coincidence that the base of the exponential, $0.8$ has shown up in these calculations, as we'll see in the next problem.
Using properties of exponents, we find
$\frac{V (t + 1)}{V (t)} = \frac{25 {(0.8)}^{t + 1}}{25 {(0.8)}^{t}} = {(0.8)}^{t + 1 - t} = 0.8$
Rewriting, we have $V (t + 1) = 0.8 V (t)$ . This means after one year, the value of the car $V ⁢ open parenthesis t plus 1 close parenthesis$ is only $80 %$ of the value it was a year ago, $V (t)$ .
Similarly, we find
$\frac{V (t + k)}{V (t)} = \frac{25 {(0.8)}^{t + k}}{25 {(0.8)}^{t}} = {(0.8)}^{t + k - t} = {(0.8)}^{k}$
which, rewritten, says $V (t + k) = V (t) {(0.8)}^{k}$ . This means in $k$ years' time, the value of the car $V (t + k)$ is only $open parenthesis 0.8 close parenthesis to the power k$ times what it was worth $k$ years ago, $V (t)$ .
These results shouldn't be too surprising. Verbally, the function $V (t) = 25 {(0.8)}^{t}$ says to multiply $25$ by $0.8$ multiplied by itself $t$ times. Therefore, for each additional year, we are multiplying the value of the car by an additional factor of $0.8$ .
We compute $\frac{V (1) - V (0)}{V (0)} = \frac{20 - 25}{25} = - 0.2$ , $\frac{V (2) - V (1)}{V (1)} = \frac{16 - 20}{20} = - 0.2$ , and $\frac{V (2) - V (0)}{V (0)} = \frac{16 - 25}{25} = - 0.36$ .
The ratio $\frac{V (1) - V (0)}{V (0)}$ computes the ratio of difference in the value of the car after the first year of ownership, $V ⁢ open parenthesis 1 close parenthesis minus V ⁢ open parenthesis 0 close parenthesis$ , to the initial value, $V ⁢ open parenthesis 0 close parenthesis$ . We find this to be $minus 0.2$ or a $20 %$ decrease in value. This makes sense since we know from our answer to number, the value of the car after $1$ year, $V ⁢ open parenthesis 1 close parenthesis$ is $80 %$ of the initial value, $V ⁢ open parenthesis 0 close parenthesis$ . Indeed:
$\frac{V (1) - V (0)}{V (0)} = \frac{V (1)}{V (0)} - \frac{V (0)}{V (0)} = \frac{V (1)}{V (0)} - 1,$
and since $\frac{V (1)}{V (0)} = 0.8$ , we get $\frac{V (1) - V (0)}{V (0)} = 1 - 0.8 = - 0.2$ .
Likewise, the ratio $\frac{V (2) - V (1)}{V (1)} = - 0.2$ means the value of the car has lost $20 %$ of its value over the course of the second year of ownership.
Finally, the ratio $\frac{V (2) - V (0)}{V (0)} = - 0.36$ means that over the first two years of ownership, the car value has depreciated $36 %$ of its initial purchase price. Again, this tracks with the result of number which tells us that after two years, the car is only worth $64 %$ of its initial purchase price.
Using properties of fractions and exponents, we get:
$\frac{V (t + 1) - V (t)}{V (t)} = \frac{25 {(0.8)}^{t + 1} - 25 {(0.8)}^{t}}{25 {(0.8)}^{t}} = \frac{25 {(0.8)}^{t + 1}}{25 {(0.8)}^{t}} - \frac{25 {(0.8)}^{t}}{25 {(0.8)}^{t}} = 0.8 - 1 = - 0.2,$
so after one year, the value of the car $V ⁢ open parenthesis t plus 1 close parenthesis$ has lost $20 %$ of the value it was a year ago, $V (t)$ .
Similarly, we find:
$\frac{V (t + k) - V (t)}{V (t)} = \frac{25 {(0.8)}^{t + k} - 25 {(0.8)}^{t}}{25 {(0.8)}^{t}} = \frac{25 {(0.8)}^{t + 1}}{25 {(0.8)}^{t}} - \frac{25 {(0.8)}^{t}}{25 {(0.8)}^{t}} = {(0.8)}^{k} - 1,$
so after $k$ years' time, the value of the car $V (t)$ has decreased by $({(0.8)}^{k} - 1) \cdot 100 %$ of the value $k$ years ago, $V (t)$ .
To graph $y = 25 {(0.8)}^{t}$ , we start with the basic exponential function $f (t) = {(0.8)}^{t}$ . Since the base $b equals 0.8$ satisfies $0 less than b less than 1$ , the graph of $y = f (t)$ is decreasing. We plot the $y$ -intercept $open parenthesis 0, 1 close parenthesis$ and two other points, $open parenthesis minus 1, 1.25 close parenthesis$ and $open parenthesis 1, 0.8 close parenthesis$ , and label the horizontal asymptote $y equals 0$ .
To obtain the graph of $y = 25 {(0.8)}^{t} = 25 f (t)$ , we multiply all of the $y$ values in the graph by $25$ (including the $y$ value of the horizontal asymptote) in accordance with Theorem 5.9 to obtain the points $open parenthesis minus 1, 31.25 close parenthesis$ , $open parenthesis 0, 25 close parenthesis$ and $open parenthesis 1, 20 close parenthesis$ . The horizontal asymptote remains the same, since $25 \cdot 0 = 0$ . Finally, we restrict the domain to $open bracket 0, ∞ close parenthesis$ to fit with the applied domain given to us.
Figure 7.11 $y = f (t) = {(0.8)}^{t}$
$\overset{\overset{vertical scale by a factor of 25}{\overset{}{\to}}}{multiply each y -coordinate by 25}$
Figure 7.12 $y = V (t) = 25 f (t)$ , $t greater than or equal to 0$
We see from the graph of $V$ that its horizontal asymptote is $y equals 0$ . This means as the car gets older, its value diminishes to $0$ .
We know the value of the car, brand new, is $$ 25, 000$ , so when we are asked to find when the car depreciates to one half and one quarter of this value, we are trying to find when the value of the car dips to $$ 12, 500$ and $$ 6, 125$ , respectively. Since $V (t)$ is measured in thousands of dollars, we this translates to solving the equations $V ⁢ open parenthesis t close parenthesis equals 12.5$ and $V ⁢ open parenthesis t close parenthesis equals 6.125$ .
Since we have yet to develop any analytic means to solve equations like $25 {(0.8)}^{t} = 12.5$ (since $t$ is in the exponent here), we are forced to approximate solutions to this equation numerically⁸ or use a graphing utility. Choosing the latter, we graph $y = V (t)$ along with the lines $y equals 12.5$ and $y equals 6.125$ and look for intersection points.
We find $y = V (t)$ and $y equals 12.5$ intersect at (approximately) $open parenthesis 3.10612.5 close parenthesis$ which means the car depreciates to half its initial value in (approximately) $3.11$ years. Similarly, we find the car depreciates to one-quarter its initial value after (approximately) $6.21$ years.⁹
Figure 7.13

Some remarks about Example 7.1.2 are in order. First the function in the previous example is called a `decay curve'. Increasing exponential functions are used to model `growth curves' and we shall see several different examples of those in Section 7.6.

Second, as seen in numbers and, $V (t + 1) = 0.8 V (t)$ . That is to say, the function $V$ has a constant unit multiplier, in this case, $0.8$ because to obtain the function value $V ⁢ open parenthesis t plus 1 close parenthesis$ , we multiply the function value $V (t)$ by $b$ . It is not coincidence that the multiplier here is the base of the exponential, $0.8$ .

Indeed, exponential functions of the form $f (x) = a \cdot b^{x}$ have a constant unit multiplier, $b$ . To see this, note

$\frac{f (x + 1)}{f (x)} = \frac{a \cdot b^{x + 1}}{a \cdot b^{x}} = b^{1} = b .$

Hence $f (x + 1) = f (x) \cdot b$ . This will prove useful to us in Section 7.6 when making decisions about whether or not a data set represents exponential growth or decay.

More generally, one can show (see Exercise ) for any real number $x sub 0$ that $f (x_{0} + Δ x) = f (x_{0}) b^{Δ x}$ . That is, to obtain $f ⁢ open parenthesis x sub 0 plus Δ ⁢ x close parenthesis$ from $f ⁢ open parenthesis x sub 0 close parenthesis$ , we multiply by $Δ x$ factors of the constant unit multiplier, $b$ . This is at the heart of what it means to be an exponential function.

If this discussion seems familiar, it should. For linear functions, $f (x) = m x + b$ , we can obtain the slope $m$ by computing $f (x + 1) - f (x)$ . To see this, note $f (x + 1) - f (x) = (m (x + 1) + b) - (m x + b) = m$ so that $f (x + 1) = f (x) + m$ . In this way, we see that the slope $m$ is the constant unit addend in that in order to obtain $f ⁢ open parenthesis x plus 1 close parenthesis$ , we add $m$ to the function value $f (x)$ .

This notion is solidified in the point-slope form of a linear function, Equation 1.1. For for any real numbers $x$ and $x sub 0$ , we have $f (x) = f (x_{0}) + m (x - x_{0})$ . If we let $x = x_{0} + Δ x$ , we get $f (x_{0} + Δ x) = f (x_{0}) + m Δ x$ . In other words, to obtain $f ⁢ open parenthesis x sub 0 plus Δ ⁢ x close parenthesis$ from $f ⁢ open parenthesis x sub 0 close parenthesis$ , we add $m$ times $Δ x$ .

Taking inspiration from linear functions, we define the `point-base' form of an exponential function below.

Definition

The point-base form of the exponential function $f (x) = b^{x}$ is

$f (x) = f (x_{0}) b^{x - x_{0}}$

Just as the point-slope form of a linear function is helpful in building linear models, the point-base form of an exponential function will prove useful in building exponential models.

Next, while we saw in Example 7.1.2 number, exponential functions, unlike linear functions, do not have a constant rate of change. However, in numbers and, we see that in some cases, they do have a constant relative rate of change. We define this notion below.

Definition

Let $f$ be a function defined on the interval $[a, b]$ where $f ⁢ open parenthesis a close parenthesis not equal to 0$ .

The relative rate of change of $f$ over $[a, b]$ is defined as:

$\frac{Δ [f (x)]}{f (a)} = \frac{f (b) - f (a)}{f (a)} .$

For exponential functions of the form $f (x) = a \cdot b^{x}$ , we compute the relative rate of change over the interval $open bracket x, x plus 1 close bracket$ and find it is constant:

$\frac{f (x + 1) - f (x)}{f (x)} = \frac{f (x + 1)}{f (x)} - \frac{f (x)}{f (x)} = b - 1,$

where we are using the fact that $\frac{f (x + 1)}{f (x)} = b$ .

One way to interpret this result is when comparing $f (x)$ to $f ⁢ open parenthesis x plus 1 close parenthesis$ , the exponential function grows (if $b greater than 1$ ) or decays (if $b less than 1$ ) by $(b - 1) \cdot 100 %$ . In our example, $V (t) = 25 {(0.8)}^{t}$ so $b equals 0.8$ and, as we saw, the relative rate of change from $V (t)$ to $V ⁢ open parenthesis t plus 1 close parenthesis$ was $0.8 - 1 = - 0.2$ , meaning the value of the car over the course of one year depreciates by $20 %$ .

We close this section with another important application of exponential functions, Newton's Law of Cooling.

Example 3

According to Newton's Law of Cooling ¹⁰ the temperature of coffee $T (t)$ (in degrees Fahrenheit) $t$ minutes after it is served can be modeled by $T (t) = 70 + 90 e^{- 0.1 t}$ .

Find and interpret $T ⁢ open parenthesis 0 close parenthesis$ .
Sketch the graph of $y = T (t)$ using transformations.
Find and interpret $lim t → ∞ T ⁢ open parenthesis t close parenthesis$ .

Since $T (0) = 70 + 90 e^{- 0.1 (0)} = 160$ , the temperature of the coffee when it is served is $160 degrees ⁢ F$ .
To graph $y = T (t)$ using transformations, we start with the basic function, $f (t) = e^{t}$ . As in Example 7.1.1, we track the points $(- 1, e^{- 1}) \approx (- 1, 0.368)$ , $open parenthesis 0, 1 close parenthesis$ , and $open parenthesis 1, e close parenthesis approximately equals open parenthesis 1, 2.718 close parenthesis$ , along wth the horizontal asymptote $y equals 0$ through each of transformations.
To use Theorem 5.11, we rewrite $T (t) = 70 + 90 e^{- 0.1 t} = 90 e^{- 0.1 t} + 70 = 90 f (- 0.1 t) + 70$ . Following Theorem 5.11, we first divide the $t$ -coordinates of each point on the graph of $y = f (t)$ by $minus 0.1$ which results in a horizontal expansion by a factor of $10$ as well as a reflection about the $y$ -axis.
Next, we multiply the $y$ -values of the points on this new graph by $90$ which effects a vertical stretch by a factor of $90$ . Last but not least, we add $70$ to all of the $y$ -coordinates of the points on this second graph, which shifts the graph upwards $70$ units.
Tracking points, we have $(- 1, e^{- 1}) \to (10, e^{- 1}) \to (10, 90 e^{- 1}) \to (10, 90 e^{- 1} + 70) \approx (10, 103.112)$ , $open parenthesis 0, 1 close parenthesis → open parenthesis 0, 1 close parenthesis → open parenthesis 0, 90 close parenthesis → open parenthesis 0, 160 close parenthesis$ , and $(1, e) \to (- 10, e) \to (- 10, 90 e) \to (- 10, 90 e + 70) \approx (- 10, 314.62)$ . The horizontal asymptote $y equals 0$ is unaffected by the horizontal expansion, reflection about the $y$ -axis, and the vertical stretch. The vertical shift moves the horizontal asymptote up $70$ units, $y = 0 \to y = 70$ . After restricting the domain to $t greater than or equal to 0$ , we get the graph below on the right.
Figure 7.14 $y = f (t) = e^{t}$

Figure 7.15 $y = T (t)$
We can determine $lim t → ∞ T ⁢ open parenthesis t close parenthesis$ two ways. First, we can employ the `number sense' developed in Chapter 3.
That is, as $t → ∞$ , We get $T (t) = 70 + 90 e^{- 0.1 t} \approx 70 + 90 e^{very big (-)}$ . Since $e greater than 1$ , $e^{very big (-)} \approx very small (+)$ The larger $t$ becomes, the smaller $e^{- 0.1 t}$ becomes, so the term $90 e^{- 0.1 t} \approx very small (+)$ . Hence, $T (t) = 70 + 90 e^{- 0.1 t} \approx 70 + very small (+) \approx 70$ .
Alternatively, we can look to the graph of $y = T (t)$ . We know the horizontal asymptote is $y equals 70$ which means as $t → ∞$ , $T ⁢ open parenthesis t close parenthesis approximately equals 70$ .
In either case, we find that as time goes by, the temperature of the coffee is cooling to $70 degrees$ Fahrenheit, ostensibly room temperature.

Exercises

In Exercises -, sketch the graph of $g$ by starting with the graph of $f$ and using transformations. Track at least three points of your choice and the horizontal asymptote through the transformations. State the domain and range of $g$ .

$f (x) = 2^{x}$ , $g (x) = 2^{x} - 1$
$f (x) = {(\frac{1}{3})}^{x}$ , $g (x) = {(\frac{1}{3})}^{x - 1}$
$f (x) = 3^{x}$ , $g (x) = 3^{- x} + 2$
$f (x) = 10^{x}$ , $g (x) = 10^{\frac{x + 1}{2}} - 20$
$f (t) = {(0.5)}^{t}$ , $g (t) = 100 {(0.5)}^{0.1 t}$
$f (t) = {(1.25)}^{t}$ , $g (t) = 1 - {(1.25)}^{t - 2}$
$f (t) = e^{t}$ , $g (t) = 8 - e^{- t}$
$f (t) = e^{t}$ , $g (t) = 10 e^{- 0.1 t}$

In Exercises, -, the graph of an exponential function is given. Find a formula for the function in the form $F (x) = a \cdot 2^{b x - h} + k$ .

Points: $(- 2, - \frac{5}{2})$ , $(- 1, - 2)$ , $open parenthesis 0, minus 1 close parenthesis$ , Asymptote: $y = - 3$ .
Figure 7.16
Points: $open parenthesis minus 1, 1 close parenthesis$ , $open parenthesis 0, 2 close parenthesis$ , $(1, \frac{5}{2})$ , Asymptote: $y equals 3$ .
Figure 7.17
Points: $(\frac{5}{2}, \frac{1}{2})$ , $open parenthesis 3, 1 close parenthesis$ , $(\frac{7}{2}, 2)$ , Asymptote: $y equals 0$ .
Figure 7.18
Points: $(- \frac{1}{2}, 6)$ , $open parenthesis 0, 3 close parenthesis$ , $(\frac{1}{2}, \frac{3}{2})$ , Asymptote: $y equals 0$ .
Figure 7.19
Find a formula for each graph in Exercises - of the form $G (x) = a \cdot 4^{b x - h} + k$ . Did you change your solution methodology? What is the relationship between your answers for $F (x)$ and $G (x)$ for each graph?
In Example 7.1.1 number, we obtained the solution $F (x) = - 2^{x + 3} + 4$ as one formula for the given graph by making a simplifying assumption that $a = - 1$ . This exercises explores if there are any other solutions for different choices of $a$ .
1. Show $G (x) = - 4 \cdot 2^{x + 1} + 4$ also fits the data for the given graph, and use properties of exponents to show $G (x) = F (x)$ . (Use the fact that $4 equals 2 squared$ …)
2. With help from your classmates, find solutions to Example 7.1.1 number using $a = - 8$ , $a = - 16$ and $a = - \frac{1}{2}$ . Show all your solutions can be rewritten as: $F (x) = - 2^{x + 3} + 4$ .
3. Using properties of exponents and the fact that the range of $2 to the power x$ is $open parenthesis 0, ∞ close parenthesis$ , show that any function of the form $f (x) = - a \cdot 2^{b x - h} + k$ for $a greater than 0$ can be rewritten as $f (x) = - 2^{c} 2^{b x - h} + k = - 2^{b x - h + c} + k$ . Relabeling, this means every function of the form $f (x) = - a \cdot 2^{b x - h} + k$ with four parameters ( $a$ , $b$ , $h$ , and $k$ ) can be rewritten as $f (x) = - 2^{b x - H} + k$ , a formula with just three parameters: $b$ , $H$ , and $k$ . Conclude that every solution to Example 7.1.1 number reduces to $F (x) = - 2^{x + 3} + 4$ .

In Exercises -, write the given function as a nontrivial decomposition of functions as directed.

For $f (x) = e^{- x} + 1$ , find functions $g$ and $h$ so that $f = g + h$ .
For $f (x) = e^{2 x} - x$ , find functions $g$ and $h$ so that $f = g - h$ .
For $f (t) = t^{2} e^{- t}$ , find functions $g$ and $h$ so that $f = g h$ .
For $r (x) = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$ , find functions $f$ and $g$ so $r = \frac{f}{g}$ .
For $k (x) = e^{- x^{2}}$ , find functions $f$ and $g$ so that $k equals g ∘ f$ .
For $s (x) = \sqrt{e^{2 x} - 1}$ , find functions $f$ and $g$ so $s equals g ∘ f$ .
The amount of money in a savings account, $A (t)$ , in dollars, $t$ years after an initial investment is made is given by: $A (t) = 500 {(1.05)}^{t}$ , for $t greater than or equal to 0$ .
1. Find and interpret $A ⁢ open parenthesis 0 close parenthesis$ , $A ⁢ open parenthesis 1 close parenthesis$ , and $A ⁢ open parenthesis 2 close parenthesis$ .
2. Find and interpret the relative rate of change of $A$ over the intervals $open bracket 0, 1 close bracket$ , $open bracket 1, 2 close bracket$ , $open bracket 0, 2 close bracket$ .
3. Find, simplify, and interpret the relative rate of change of $A$ over the $open bracket t, t plus 1 close bracket$ . Assume $t greater than or equal to 0$ .
4. Use a graphing utility to estimate how long until the savings account is worth $$ 1500$ . Round your answer to the nearest year.
Based on census data,¹¹ the population of Lake County, Ohio, in 2010 was 230,041 and in 2015, the population was 229,437.
1. Show the percentage change in the population from 2010 to 2015 is approximately $minus 0.263 %$ .
2. If this percentage change remains constant, predict the population of Lake County in 2020.
3. Assuming this percentage change per five years remains constant, find an expression for the population $P (t)$ of Lake County where $t$ is the number of five year intervals after 2010. (So $t equals 0$ corresponds to 2010, $t equals 1$ corresponds to $2015$ , $t equals 2$ corresponds to $2020$ , etc.)
  HINT: Definitions 7.2 and 7.3 and ensuing discussion on that page is useful here.
4. Use your answer to to predict the population of Lake County in the year 2017.
5. Let $A (t)$ represent the population of Lake County $t$ years after 2010 where the we approximate the percentage change in population per year as $- \frac{0.263 %}{5} = - 0.0526 %$ . Find a formula for $A (t)$ and compare your predictions with $A (t)$ to those given by $P (t)$ . In particular, what population does each model give for the year 2050? Discuss any discrepancies with your classmates.
Show that the average rate of change of a function over the interval $open bracket x, x plus 2 close bracket$ is average of the average rates of change of the function over the intervals $open bracket x, x plus 1 close bracket$ and $[x + 1, x + 2]$ . Can the same be said for the average rate of change of the function over $open bracket x, x plus 3 close bracket$ and the average of the average rates of change over $open bracket x, x plus 1 close bracket$ , $[x + 1, x + 2]$ , and $[x + 2, x + 3]$ ? Generalize.
If $f (x) = b^{x}$ where $b greater than 0$ , $b not equal to 1$ , show $f (x_{0} + Δ x) = f (x_{0}) b^{Δ x}$ .
Which is larger: $e^{π}$ or $π^{e}$ ? How do you know? Can you find a proof that doesn't use technology?
1. Use properties of exponential functions to show that if $f (x) = e^{x}$ , then
  $f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = lim_{h \to 0} e^{x} (\frac{e^{h} - 1}{h})$
2. Numerically and graphically investigate the limit: $lim h → 0$ $\frac{e^{h} - 1}{h}$ .
3. Use parts and to show to your surprise and delight that the derivative of $e^{x}$ is … $e^{x}$ .
4. Write the equations of the tangent lines to $y = e^{x}$ at the following points:
  1. $open parenthesis 0, 1 close parenthesis$
  2. $open parenthesis 1, e close parenthesis$
  3. $(- 1, e^{- 1})$

Answers

Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus 1, ∞ close parenthesis$ Points: $(- 1, - \frac{1}{2})$ , $open parenthesis 0, 0 close parenthesis$ , $open parenthesis 1, 1 close parenthesis$ Asymptote: $y = - 1$
Figure 7.20 $y = g (x) = 2^{x} - 1$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 0, ∞ close parenthesis$ Points: $open parenthesis 0, 3 close parenthesis$ , $open parenthesis 1, 1 close parenthesis$ , $(2, \frac{1}{3})$ Asymptote: $y equals 0$
Figure 7.21 $y = g (x) = {(\frac{1}{3})}^{x - 1}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 2, ∞ close parenthesis$ Points: $(1, \frac{7}{3})$ , $open parenthesis 0, 3 close parenthesis$ , $open parenthesis minus 1, 5 close parenthesis$ Asymptote: $y equals 2$
Figure 7.22 $y = g (x) = 3^{- x} + 2$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus 20, ∞ close parenthesis$ Points: $(- 1, - 19)$ , $open parenthesis 1, minus 10 close parenthesis$ , $open parenthesis 3, 80 close parenthesis$ Asymptote: $y = - 20$
Figure 7.23 $y = g (x) = 10^{\frac{x + 1}{2}} - 20$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 0, ∞ close parenthesis$ Points: $open parenthesis minus 10, 200 close parenthesis$ , $open parenthesis 0, 100 close parenthesis$ , $open parenthesis 10, 50 close parenthesis$ Asymptote: $y equals 0$
Figure 7.24 $y = g (t) = 100 {(0.5)}^{0.1 t}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, 1 close parenthesis$ Points: $open parenthesis 1, 0.2 close parenthesis$ , $open parenthesis 2, 0 close parenthesis$ , $open parenthesis 3, minus 0.25 close parenthesis$ Asymptote: $y equals 1$
Figure 7.25 $y = g (t) = 1 - {(1.25)}^{t - 2}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis minus ∞, 8 close parenthesis$ Points: $(1, 8 - e^{- 1}) \approx (1, 7.63)$ , $open parenthesis 0, 7 close parenthesis$ , $(- 1, 8 - e) \approx (1, 5.28)$ Asymptote: $y equals 8$
Figure 7.26 $y = g (t) = 8 - e^{- t}$
Domain of $g$ : $open parenthesis minus ∞, ∞ close parenthesis$ Range of $g$ : $open parenthesis 0, ∞ close parenthesis$ Points: $(10, 10 e^{- 1}) \approx (10, 3.68)$ $open parenthesis 0, 10 close parenthesis$ , $(- 10, 10 e) \approx (- 10, 27.18)$ Asymptote: $y equals 0$
Figure 7.27 $y = g (t) = 10 e^{- 0.1 t}$
$F (x) = 2^{x + 1} - 3$
$F (x) = - 2^{- x} + 3$
$F (x) = 2^{2 x - 6}$
$F (x) = 3 \cdot 2^{- 2 x}$
Since $2 = 4^{\frac{1}{2}}$ , one way to obtain the formulas for $G (x)$ is to use properties of exponents. For example, $F (x) = 2^{x + 1} - 3 = {(4^{\frac{1}{2}})}^{x + 1} - 3 = 4^{\frac{1}{2} (x + 1)} - 3 = 4^{\frac{1}{2} x + \frac{1}{2}} - 3$ . In order, the formulas for $G (x)$ are:
- $G (x) = 4^{\frac{1}{2} x + \frac{1}{2}} - 3$
- $G (x) = - 4^{- \frac{1}{2} x} + 3$
- $G (x) = 4^{x - 3}$
- $G (x) = 3 \cdot 4^{- x}$
¹² $g (x) = e^{- x}$ and $h ⁢ open parenthesis x close parenthesis equals 1$ .
$g (x) = e^{2 x}$ and $h (x) = x$ .
$g (t) = t^{2}$ and $h (t) = e^{- t}$ .
$f (x) = e^{x} - e^{- x}$ and $g (x) = e^{x} + e^{- x}$ .
$f (x) = - x^{2}$ and $g (x) = e^{x}$ .
$f (x) = e^{2 x} - 1$ and $g (x) = \sqrt{x}$ .
1. $A ⁢ open parenthesis 0 close parenthesis equals 500$ , so the principal is $$ 500$ . $A ⁢ open parenthesis 1 close parenthesis equals 525$ , so after $1$ year, there is $$ 525$ in the savings account. $A ⁢ open parenthesis 2 close parenthesis equals 551.25$ , so after $2$ years, there is $$ 551.25$ in the savings account.
2. The relative rate of change of $A$ over the intervals $open bracket 0, 1 close bracket$ and $open bracket 1, 2 close bracket$ is $0.05$ which means the savings account is growing by $5 %$ each year for those two years. Over the interval $open bracket 0, 2 close bracket$ , the relative rate of change is $0.1025$ meaning the account has grown by $10.25 %$ over the course of the first two years. Note this is greater than the sum of the two rates $5 % + 5 % = 10 %$ . This is due to the `compounding effect' and will be discussed in greater detail in Section 7.6.
3. The relative rate of change of $A$ over the $open bracket t, t plus 1 close bracket$ is $0.05$ . This means over the course of one year, the savings account grows by $5 %$ .
4. Graphing $y = A (t)$ and $y equals 1500$ , we find they intersect when $t approximately equals 22.5$ so it takes approximately $22 minus 23$ years for the savings account to grow to $$ 1500$ in value.
1. $\frac{229437 - 230041}{230041} \approx 0.263 %$ .
2. Since 2020 is five years after 2015, we expect the population to decrease by $0.263 %$ of 229437, or approximately 603 people. Hence, we approximate the population in 2020 as 228834.
3. $P (t) = 230041 {(1 - 0.00263)}^{t} = 230041 {(0.99737)}^{t}$ , $t greater than or equal to 0$ .
4. Since 2017 is 7 years after 2010, we set $t = \frac{7}{5} = 1.4$ and find $P ⁢ open parenthesis 1.4 close parenthesis approximately equals 229194$ . So the population is approximately 229, 194 in 2017.
5. $A (t) = 230041 {(1 - 0.0005626)}^{t} = 230041 {(0.999474)}^{t}$ , $t greater than or equal to 0$ . Since 2050 is 40 years after 2010, using the model $P (t)$ , we divide $\frac{40}{5} = 8$ and find $P ⁢ open parenthesis 8 close parenthesis approximately equals 225, 245$ . On the other hand, $A ⁢ open parenthesis 40 close parenthesis approximately equals 225, 250$ . This is more than roundoff error. There is a compounding effect which makes the functions $A (t)$ and $P (t)$ different. ¹³
1. 1. $y = x + 1$
  2. $y = e x$
  3. $y = e^{- 1} x + 2 e^{- 1}$

See the discussion of real number exponents in Section 4.2. ↩
or, as we defined real number exponents in Section 4.2, if $x$ is an irrational number … ↩
Meaning, graph some more examples on your own. ↩
Recall that this means the graph of $f$ has no sharp turns or corners. ↩
The digital world is comprised of bytes which take on one of two values: $0$ or `off' and $1$ or `on.' ↩
This is the only solution. Since $f (x) = 2^{x}$ , the equation $2^{- b - h} = 2^{2}$ is equivalent to the functional equation $f (- b - h) = f (2)$ . Since $f$ is one-to-one, we know this is true only when $- b - h = 2$ . ↩
It turns out for any function $f$ , the average rate of change over the interval $open bracket x, x plus 2 close bracket$ is the average of the average rates of change of $f$ over $open bracket x, x plus 1 close bracket$ and $[x + 1, x + 2]$ . See Exercise . ↩
Since exponential functions are continuous we could use the Bisection Method to solve $f (t) = 25 {(0.8)}^{t} - 12.5 = 0$ . See the discussion on page in Section 2.3 for more details. ↩
It turns out that it takes exactly twice as long for the car to depreciate to one-quarter of its initial value as it takes to depreciate to half its initial value. Can you see why? ↩
We will discuss this in greater detail in Section 7.6. ↩
See here . ↩
Answers for Exercises - vary. We list one solution for each problem. ↩
See number above or, for more, see Section 7.6. ↩

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.