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Section 3.1 Intro to Exponential Functions

Subsection 3.1.1 Exponential Functions

Definition 3.1.1.

An exponential function with base \(b\) is the function given by,

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

where \(b > 0, b \neq 1\text{.}\)

This defines a function because, for all \(x \in \mathbb{R}\text{,}\) \(b^x\) is a (unique) real number.

Some exponential functions include,

\begin{equation*} f(x) = 2^x \qquad f(x) = 5^x \qquad f(x) = 1.12^x \qquad f(x) = \brac{\frac{1}{2}}^x \qquad f(x) = 0.83^x \end{equation*}

Some functions that are not exponential functions include,

\begin{align*} f(x) \amp = x^2 \amp\amp \text{polynomial function, $x$ is not an exponent}\\ f(x) \amp = 1^x \amp\amp \text{the base $b$ cannot be 1}\\ f(x) \amp = (-2)^x \amp\amp \text{the base $b$ must be positive}\\ f(x) \amp = x^x \amp\amp \text{the base is not a constant} \end{align*}

Recall that a polynomial function \(f(x) = x^n\) has the variable \(x\) as the base of a power, and the exponent is a constant. Exponential functions, on the other hand are functions of the form \(f(x) = b^x\text{,}\) where the variable \(x\) is an exponent, and the base is a constant.

Subsection 3.1.2 Graphing Basic Exponential Functions

Consider the graph of \(f(x) = 2^x\text{.}\) First, create a table of values,

\begin{equation*} \begin{array}{c|c} x \amp f(x) \\ \hline -2 \amp 2^{-2} = 1/4 \\ -1 \amp 2^{-1} = 1/2 \\ 0 \amp 2^0 = 1 \\ 1 \amp 2^1 = 2 \\ 2 \amp 2^2 = 4 \end{array} \end{equation*}

In summary,

  1. Create a table of values for \(x = -2, -1, 0, 1, 2\text{,}\) and plot the points.

  2. Connect the points with a smooth curve.

Graph \(f(x) = \brac{\frac{1}{2}}^x\text{.}\)

Subsection 3.1.3 Properties of Exponential Functions

Consider an exponential function \(f(x) = b^x\text{.}\) First, its domain is all real numbers, because for every number \(x\text{,}\) \(b^x\) is defined (there is no division by zero, or roots).

  • The domain of \(f\) is all real numbers, and \(f\) is continuous on \(\mathbb{R}\text{.}\)

  • The \(y\)-intercept is \((0,1)\text{,}\) because \(f(0) = b^0 = 1\) for any \(b > 0\text{.}\)

Subsection 3.1.4 Transforming Exponential Functions

Using the theory of transformations, the parent function \(f(x) = b^x\) can be transformed using stretches, reflections, and shifts. In particular, recall that,

\begin{equation*} y = f(x) \longrightarrow y = a f\brac{b(x - h)} + k \end{equation*}

Because the base of the exponential function already uses the letter \(b\text{,}\) we will use \(c\) to represent the horizontal stretch. Then, for \(f(x) = b^x\text{,}\) the function becomes,

\begin{equation*} f(x) = a \cdot b^{c(x - h)} + k \end{equation*}

However, this can be rewritten as,

\begin{equation*} f(x) = a \cdot (b^c)^{x - h} + k \end{equation*}

Then, the exponential function can simply be thought of as base \(b^c\) with no horizontal stretch.

Also, the horizontal shift doesn't matter as much, because,

\begin{align*} a \cdot b^{x - h} + k \amp = a \cdot b^x \cdot b^{-h} + k\\ \amp = \brac{\frac{a}{b^h}} \cdot b^x + k \end{align*}

The horizontal shift can be considered as a vertical stretch. Then, typically, we write a transformed exponential function in the form,

\begin{equation*} f(x) = a \cdot b^x + k \end{equation*}

Properties of the transformed function,

  • To determine the \(x\)-intercept, set \(f(x) = 0\) and solve for \(x\text{.}\) If there is no solution (because it requires taking the logarithm of a negative number), then there is no \(x\)-intercept.

  • The horizontal asymptote is \(y = k\text{.}\)

Subsection 3.1.5 Graphing a Transformed Exponential Function Using the Parent Function

Sketch a graph of \(f(x) = 3^{x+2} - 3\text{.}\)

Subsection 3.1.6 Restriction on the Base of an Exponential Function

Exponential functions \(f(x) = b^x\) require the base \(b\) to be positive and \(b \neq 1\text{.}\) This is for a few reasons.

  • First, if \(b = 1\text{,}\) then we would have \(f(x) = 1^x = 1\) for all \(x\text{.}\) In other words, we would get the constant function \(f(x) = 1\) which doesn't fit the characteristics of an exponential function.

  • If \(b = 0\text{,}\) similarly we would have,

    \begin{equation*} f(x) = 0^x = \begin{cases} 0 \amp x > 0 \\ \text{undefined} \amp x \lt 0 \end{cases} \end{equation*}
    This is because for \(x \lt 0\text{,}\) e.g. \(x = -2\text{,}\) we would have,
    \begin{equation*} f(-2) = 0^{-2} = \frac{1}{0^2} = \frac{1}{0} \end{equation*}
    which is undefined. Either way, this is not an exponential function.

  • Finally, if \(b \lt 0\text{,}\) then this would involve taking the power of a negative number. This creates some strange behaviour. Consider \(f(x) = (-2)^x\) for example. For positive integers \(x\text{,}\) \(f\) is defined. For example,

    \begin{equation*} (-2)^1 = -2 \qquad (-2)^2 = 4 \qquad (-2)^3 = 8 \end{equation*}

    and so on. For negative integers \(x\text{,}\) \(f\) is also defined,

    \begin{equation*} (-2)^{-1} = \frac{1}{-2} = -\frac{1}{2} \quad (-2)^{-2} = \frac{1}{(-2)^2} = \frac{1}{4} \quad (-2)^{-3} = \frac{1}{(-2)^3} = -\frac{1}{8} \end{equation*}

    However, for rational exponents, \(f\) can be undefined. Recall that for a rational exponent \(\frac{a}{b}\text{,}\) where \(a, b \in \mathbb{Z}\) are integers and \(b \neq 0\text{,}\) the expression \((-2)^{a/b}\) is defined as,

    \begin{equation*} (-2)^{a/b} = \sqrt[b]{(-2)^a} \end{equation*}

    which has some restrictions, as the even root of a negative number is undefined. First, we can assume that \(b > 0\text{,}\) so that the sign of \(\frac{a}{b}\) is determined by \(a\) (i.e. writing \(-2/3\) instead of \(2/-3\)).

    If \(b\) is odd, \((-2)^{a/b}\) is defined for all integers \(a\text{.}\) If \(b\) is even, then \(a\) must also be even for \((-2)^a\) to be positive and \(\sqrt[b]{(-2)^a}\) is defined. For example, for \(x = 1/2\) (\(a = 1\) odd and \(b = 2\) even), we have,

    \begin{equation*} f(1/2) = (-2)^{1/2} = \sqrt{-2} \end{equation*}

    which is undefined as a real number. Finally, for all irrational numbers \(x\text{,}\) \((-2)^x\) is undefined. Explaining this is a little too advanced, but it can be verified by trying \((-2)^x\) on your calculator for different irrational numbers, like \(\sqrt{2}, \sqrt{3}, \sqrt{6}\text{,}\) etc. These expressions will result in complex numbers.

    Putting it all together, \(f(x) = (-2)^x\) is defined for all \(x\) except if \(x\) is irrational (which is a lot of values), and if \(x = \frac{a}{b}\) is rational where \(a\) is odd and \(b\) is even. In other words, \(f\) is not continuous (it has infinitely many “holes”), so we don't consider it to be an exponential function.