## Section2.3Exponential Functions

An exponential function is a function of the form $f(x)=a^x\text{,}$ where $a$ is a constant. Examples are $2^x\text{,}$ $10^x$ and $(1/2)^x\text{.}$ To more formally define the exponential function we look at various kinds of input values.

It is obvious that $\ds a^5=a\cdot a\cdot a\cdot a\cdot a$ and $\ds a^3=a\cdot a\cdot a\text{,}$ but when we consider an exponential function $\ds a^x$ we can't be limited to substituting integers for $x\text{.}$ What does $\ds a^{2.5}$ or $a^{-1.3}$ or $\ds a^\pi$ mean? And is it really true that $a^{2.5}a^{-1.3}=a^{2.5-1.3}\text{?}$ The answer to the first question is actually quite difficult, so we will evade it; the answer to the second question is “yes.”

We'll evade the full answer to the hard question, but we have to know something about exponential functions. You need first to understand that since it's not “obvious” what $\ds 2^x$ should mean, we are really free to make it mean whatever we want, so long as we keep the behavior that is obvious, namely, when $x$ is a positive integer. What else do we want to be true about $\ds 2^x\text{?}$ We want the properties of the previous two paragraphs to be true for all exponents: $\ds 2^x2^y=2^{x+y}$ and $\ds (2^x)^y=2^{xy}\text{.}$

After the positive integers, the next easiest number to understand is 0: $\ds 2^0=1\text{.}$ You have presumably learned this fact in the past; why is it true? It is true precisely because we want $\ds 2^a2^b=2^{a+b}$ to be true about the function $\ds 2^x\text{.}$ We need it to be true that $\ds 2^02^x=2^{0+x}=2^x\text{,}$ and this only works if $\ds 2^0=1\text{.}$ The same argument implies that $\ds a^0=1$ for any $a\text{.}$

The next easiest set of numbers to understand is the negative integers: for example, $\ds 2^{-3}=1/2^3\text{.}$ We know that whatever $\ds 2^{-3}$ means it must be that $\ds 2^{-3}2^{3}=2^{-3+3}=2^0=1\text{,}$ which means that $\ds 2^{-3}$ must be $1/2^3\text{.}$ In fact, by the same argument, once we know what $\ds 2^x$ means for some value of $x\text{,}$ $\ds 2^{-x}$ must be $\ds 1/2^{x}$ and more generally $a^{-x}=1/a^x\text{.}$

Next, consider an exponent $1/q\text{,}$ where $q$ is a positive integer. We want it to be true that $\ds (2^x)^y=2^{xy}\text{,}$ so $\ds (2^{1/q})^q=2\text{.}$ This means that $\ds 2^{1/q}$ is a $q$-th root of 2, $\ds 2^{1/q}=\root q\of{2\ }\text{.}$ This is all we need to understand that $2^{p/q}=(2^{1/q})^p=(\root q\of{2\ })^p$ and $a^{p/q}=(a^{1/q})^p=(\root q\of{a\ })^p\text{.}$

What's left is the hard part: what does $\ds 2^x$ mean when $x$ cannot be written as a fraction, like $\ds x=\sqrt{2\ }$ or $\ds x=\pi\text{?}$ What we know so far is how to assign meaning to $\ds 2^x$ whenever $x=p/q\text{.}$ If we were to graph $a^x$ (for some $a>1$) at points $x=p/q$ then we'd see something like this:

This is a poor picture, but it illustrates a series of individual points above the rational numbers on the $x$-axis. There are really a lot of “holes” in the curve, above $x=\pi\text{,}$ for example. But (this is the hard part) it is possible to prove that the holes can be “filled in”, and that the resulting function, called $\ds a^x\text{,}$ really does have the properties we want, namely that $\ds a^xa^y=a^{x+y}$ and $\ds (a^x)^y=a^{xy}\text{.}$ Such a graph would then look like this:

###### Three Types of Exponential Functions.

There are three kinds of exponential functions $f(x)=a^x$ with $x$ real depending on whether $a>1\text{,}$ $a=1$ or $0\lt a\lt 1\text{:}$

### Subsection2.3.1Properties of Exponential Functions

The first thing to note is that if $a\lt 0$ then problems can occur. Observe that if $a=-1$ then $(-1)^x$ is not defined for every $x\text{.}$ For example, $x=1/2$ is a square root and gives $(-1)^{1/2}=\sqrt{-1}$ which is not a real number.

###### Exponential Function Properties.
• Only defined for positive $a\text{:}$ $a^x$ is only defined for all real $x$ if $a>0$

• Always positive: $a^x>0\text{,}$ for all real $x$

• Exponent rules: If $a,b > 0$ and $x,y$ real numbers, then

1. $\ds{a^xa^y=a^{x+y}}$

2. $\ds{\frac{a^x}{a^y}=a^{x-y}}$

3. $\ds{\left(a^x\right)^y=a^{xy}=a^{yx}=\left(a^y\right)^x}$

4. $\ds{a^xb^x=(ab)^x}$

• Long-term behaviour: If $a>1\text{,}$ then $a^x\to\infty$ as $x\to\infty$ and $a^x\to 0$ as $x\to-\infty\text{.}$

The last property can be observed from the graph. If $a>1\text{,}$ then as $x$ gets larger and larger, so does $a^x\text{.}$ On the other hand, as $x$ gets large and negative, the function approaches the $x$-axis, that is, $a^x$ approaches $0\text{.}$

###### Example2.13. Reflection of Exponential.

Determine an equation of the function after reflecting $y=2^x$ about the line $x=-2\text{.}$

Solution

First reflect about the $y$-axis to get $y=2^{-x}\text{.}$ Now shift by $2\times 2=4$ units to the left to get $y=2^{-(x+4)}\text{.}$ Side note: Can you see why this sequence of transformations is the same as reflection in the line $x=-2\text{?}$ Can you come up with a general rule for these types of reflections?

###### Example2.14. Determine the Exponential Function.

Determine the exponential function $f(x)=ka^x$ that passes through the points $(1,6)$ and $(2,18)\text{.}$

Solution

We substitute our two points into the equation to get:

\begin{equation*} x=1,y=6\to6=ka^1 \end{equation*}
\begin{equation*} x=2,y=18\to18=ka^2 \end{equation*}

This gives us $6=ka$ and $18=ka^2\text{.}$ The first equation is $k=6/a$ and subbing this into the second gives: $18=(6/a)a^2\text{.}$ Thus, $18=6a$ and $a=3\text{.}$ Now we can see from $6=ka$ that $k=2\text{.}$ Therefore, the exponential function is

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

There is one base that is so important and convenient that we give it a special symbol. This number is denoted by $e=2.71828\ldots$ (and is an irrational number). Its importance stems from the fact that it simplifies many formulas of Calculus and also shows up in other fields of mathematics.

###### Example2.15. Domain of Function with Exponential.

Find the domain of $\ds f(x)=\frac{1}{\sqrt{e^x+1}}\text{.}$

Solution

For domain, we cannot divide by zero or take the square root of negative numbers. Note that one of the properties of exponentials is that they are always positive! Thus, $e^x+1>0$ (in fact, as $e^x>0$ we actually have that $e^x+1$ is at least one). Therefore, $e^x+1$ is never zero nor negative, and gives no restrictions on $x\text{.}$ Thus, the domain is $\R\text{.}$

### Subsection2.3.2Special Base $e$

A question of interest in calculus is the following: What base of an exponential function has the property that at the point $(0,1)$ the slope of the tangent line is one? We will answer this informally here. Consider the function $f(x)=a^x$ and a tangent line at the point $(0,1)\text{.}$ If $a=2\text{,}$ the slope of the tangent line is approximately $0.7\text{,}$ see the left graph in Figure 2.4. If $a=3$ the slope of the tangent line is approximately $1.1\text{,}$ see the right graph in Figure 2.4. It turns out that when the base is

\begin{equation*} 2.71828182845904523536028747135266249775724709369995..\text{.} \end{equation*}

the slope of the tangent line is exactly equal to one! This number is denoted by e=2.71828... and is an irrational number. It is sometimes called Euler's constant named after the mathematician Leonhard Euler. Its importance stems from the fact that it simplifies many formulas of Calculus and also shows up in other fields of mathematics.

##### Exercises for Section 2.3.

Determine an equation of the function $y=a^x$ passing through the point $(3,8)\text{.}$

$y=2^x$
Solution

We wish to find an equation of the form $y = a^{x}$ which passes through the point $\left(3, 8\right)\text{.}$ Making the appropriate subsitution, we find

\begin{equation*} \begin{split} 8 \amp = a^{3} \\ 8^{\frac{1}{3}} \amp = a \\ 2 \amp = a \end{split} \end{equation*}

Which leads to the desired equation, $y = 2^{x}\text{.}$

Find the $y$-intercept of $f(x)=4^x+6\text{.}$

$y=7$

Find the $y$-intercept of $f(x)=2\left(\frac{1}{2}\right)^x\text{.}$

$y=2$
Find the domain of $\ds{y=e^{-x}+e^{\frac{1}{x}}}\text{.}$
$x\neq 0$
The domain of $e^{-x}$ is $(-\infty,\infty)\text{.}$ The domain of $e^{1/x}$ is $(-\infty,0)\cup(0,\infty)\text{.}$ Therefore, the domain of $y = e^{-x}+e^{1/x}$ is $(-\infty,0)\cup(0,\infty)$ — that is, all real numbers $x$ such that $x \neq 0\text{.}$