Exponential Functions

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characteristics

The formula template for the basic exponential function looks like

\[ A \, R^x \, \text { with } \, A, R \in \mathbb {R} \, | \, R > 0, R \ne 1 \]

We can always view this as a composition and let the Chain Rule tells us the behavior.

We can also read the coefficients, where $A$ and $R$ control the behavior.

We have four combinations

$A>0$ and $R>1$
$A>0$ and $R<1$
$A<0$ and $R>1$
$A<0$ and $R<1$

For graphs of basic exponential functions, the horizontal axis is a horizontal asymptote in one direction or the other.

When $R > 1$ we have a growing function.

\[ \lim _{x \to -\infty } A \, R^x = 0 \, \text { and } \, \lim _{x \to \infty } A \, R^x = \pm \infty \]

The sign of $A$ dictates if the unbounded growth is positive or negative.

When $R<1$, the horizontal axis is still a horizontal asymptote, just in the other direction. The function now decays.

\[ \lim _{x \to -\infty } A \, R^x = \pm \infty \, \text { and } \, \lim _{x \to \infty } A \, R^x = 0 \]

The sign of $A$ dictates if the function decays through positive or negative values.

All four graphs share a common structure.

All have the horizontal axis as an asymptote.
In one direction, the graph approaches the asymptote. In the other direction, the graph moves away from the asymptote.

These are the important aspects or characteristics that we use when shifting and stretching graphs of exponential functions.

Exponential Behavior

Basic exponential functions, $a \cdot r^x$, are either increasing functions or decreasing functions.

$\blacktriangleright $ Base Greater than $1$: $r > 1$

greater positive exponents mean multiplying by the base more, which results in larger values.
greater negative exponents mean multiplying by the reciprocal of the base more, which results in smaller values.

The coefficient in front, $a$, tells us if this larger/smaller value is larger positively or negatively.

$a > 0$ and $r > 1$ : increasing function
$a < 0$ and $r > 1$ : decreasing function

$\blacktriangleright $ Base Less than $1$: $r < 1$

greater positive exponents mean multiplying by the base more, which results in smaller values.
greater negative exponents mean multiplying by the reciprocal of the base more, which results in larger values.

The coefficient in front, $a$, tells us if this larger/smaller value is larger positively or negatively.

$a > 0$ and $r < 1$ : decreasing function
$a < 0$ and $r < 1$ : increasing function

Exponential Function

Analyze $f(x) = \frac {1}{3} \, 2^{x+5}$

Categorizing: $f(x) = \frac {1}{3} \, 2^{x+5}$ is an exponential function since it matches our official template, $A r^{B \, x + C}$.

Domain

$f$ is an exponential function, which tells us that its domain is $(-\infty , \infty )$.

Zeros

$f$ is an exponential function, therefore it has no zeros.

Continuity

$f$ is an exponential function, therefore it is continuous.

Behavior (Increasing and Decreasing)

The base is $2$, which is greater than $1$.
The leading coeffcient is $\frac {1}{3} > 0$.
The leading coeffcient of the linear exponent is $1 > 0$.

That tells us that $f$ is increasing and positive.

Or, work with the Chain Rule.

\[ f(x) = \frac {1}{3} \, 2^{x+5} = (L_{out} \circ E \circ L_{in})(x) \]

where

$E(t) = 2^x$, an increasing core exponential function.
$L_{out}(u) = \frac {1}{3} \, u$, an increasing linear function.
$L_{in}(v) = v + 5$, an increasing linear function.

The Chain Rule gives us

\[ inc \circ inc \circ inc = inc \]

End-Behavior

$f$ is an exponential function, therefore the end-behavior of one side is $0$ and the other is unbounded. Since $f$ is positive and increasing, we have

\[ \lim \limits _{x \to -\infty } f(x) = 0 \]

\[ \lim \limits _{x \to \infty } f(x) = \infty \]

Local Maximum and Minimum

$f$ is an exponential function, therefore it has no local extrema.

Global Maximum and Minimum

$f$ is an exponential function, therefore it has no global extrema.

Range

$f$ is continuous
$f$ is increasing and positive
$\lim \limits _{x \to -\infty } f(x) = 0$
$\lim \limits _{x \to \infty } f(x) = \infty $

The range is $(0, \infty )$.

Graphing

For graphs of exponential functions, the horizontal axis is the horizontal asymptote.

The “inside”, representing the domain, is $x+5$. This equals $0$, when $x=-5$.

At $x=-5$, we have our one anchor point for the graph. The point is $\left (-5, \frac {1}{3} \right )$, which is $\frac {1}{3}$ above the horizontal asymptote, $y = 0$.

Graph of $y = f(x)$.

Our graph agrees with our analysis.

Exponential Function

Analyze $B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t}$

Categorizing: $B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t}$ is an exponential function since it matches our official template, $A r^{B \, x + C}$.

Idea

First, observe that the base $\frac {2}{3} < \answer {1}$.

Our base is less than $1$. Therefore, as its exponent gets large and positive, we multiply by more $\frac {2}{3}$’s and the overall values get smaller.

Except, the variable, $t$, in the exponent is multiplied by $-1$. Therefore, we need $t$ to get large and negative in order for the exponent to get large and positve.

$\left ( \frac {2}{3} \right )^{3-t}$ decays when $t$ becomes more negative.
$\left ( \frac {2}{3} \right )^{3-t}$ grows when $t$ becomes more positive.

The exponential stem of $B(t)$ is $\left ( \frac {2}{3} \right )^{-t}$, which is a transformed version of the basic exponential function model $M(t) = \left ( \frac {2}{3} \right )^{t}$.

When $t < 0$, then $-t > 0$ and we get $\left ( \frac {2}{3} \right )^{-t} = \left ( \frac {2}{3} \right )^{positive}$ and the stem is becoming smaller, approaching $0$.

\[ \lim \limits _{t \to -\infty } \left ( \frac {2}{3} \right )^{-t} = 0 \]

When $t > 0$, then $-t < 0$ and we get $\left ( \frac {2}{3} \right )^{-t} = \left ( \frac {2}{3} \right )^{negative}$ and the stem is becoming larger.

\[ \lim \limits _{t \to \infty } \left ( \frac {2}{3} \right )^{-t} = \infty \]

Finally, all of that is multiplied by $-2$, which switches all of the behavior.

Graph of $y = B(t)$.

With this thinking, we can create an algebraic analysis.

explanation

Domain

$B$ is an exponential function, therefore its domain is $(-\infty , \infty )$.

Zeros

$B$ is an exponential function, therefore it has no zeros.

Continuity

$B$ is an exponential function, therefore it is continuous.

End-Behavior

$B$ is an exponential function, therefore on one side the end-behavior is $0$ and unbounded on the other side.

Since the leading coefficient is $-2 < 0$, we know $B$ will be unbounded negatively. We just need to figure out which side.

the base of $B$ is $\frac {2}{3} < 1$
the leading coefficient of $B$ is $-2$, which is negative.
the leading coefficient of the linear exponent is $-1$, which is negative.

This makes $B$ a decreasing function with neagtive values.

\[ \lim \limits _{t \to -\infty } B(t) = 0 \]

\[ \lim \limits _{t \to \infty } B(t) = -\infty \]

Behavior (Increasing and Decreasing)

$B(t) = -2 \, \left ( \frac {2}{3} \right )^{3-t}$, which can be viewed as a composition.

Let $E(x) = \left ( \frac {2}{3} \right )^x$, a decreasing core exponential function.
Let $L_{out}(u) = -2\,u$, a decreasing linear function.
Let $L_{in}(v) = 3-v,u$, a decreasing linear function.

The Chain Rule gives $dec \circ dec \circ dec = dec$.

This makes $B$ a decreasing function.

Local Maximum and Minimum

$B$ is an exponential function, therefore it has no local extrema.

Global Maximum and Minimum

$B$ is an exponential function, therefore it has no global extrema.

Range

$B$ is continuous
$B$ is increasing and positive.
$\lim \limits _{t \to -\infty } B(t) = 0$
$\lim \limits _{t \to \infty } B(t) = \infty $

The range is $(-\infty , 0)$.

Our analysis agrees with the graph.

Exponential Function

Analyze $K(f) = 3^{5-f}$

Idea

The formula for $K$ matches which template?

$A \, x + B$ $A \, x^2 + B \, x + C$ $A \, |B \, x + C | + D$ $A \, r^{B \, x + C}$ $A \, r^{B \, x + C} + D$ $A \, \ln (B \, x + C) + D$

The exponent gets big and positive when $f$ gets big and positivenegative.

The graph will grow to the rightleft.

The graph will approach the asymptote to the rightleft.

Our one anchor point moves to $\left (\answer {5}, \answer {1}\right )$.

The graph will become unbounded updown.

Graph of $y = K(f)$.

With this thinking, we can create an algebraic analysis.

explanation

Domain

The natural or implied domain of $K$ is $\mathbb {R}$, because $K$ is an exponential function.

Zeros

There are no zeros, because $K$ is an exponential function.

Continuity

$K$ is continuous, because $K$ is an exponential function.

Behavior (Increasing and Decreasing)

The base is greater than $1$ and the leading coeffcients are of opposite sign. That tells us that $K$ is decreasing.

Or, use the Chain Rule.

$K(f) = 3^{5-f}$ can be viewed as a composition.

Let $E(x) = 3^x$, an increasing core exponential function.
Let $L_{in}(v) = 5-v,u$, a decreasing linear function.

The Chain Rule gives $inc \circ dec = dec$.

This makes $B$ a decreasing function.

End-Behavior

The leading coefficient is $1$, which is positive, which tells us that $K$ is a positive function. $K$ only has positive values.

We just found out that $K$ is a decreasnig function.

The end-behavior of a positive decreasing exponential function is

\[ \lim \limits _{f \to -\infty } K(f) = \infty \]

\[ \lim \limits _{f \to \infty } K(f) = 0 \]

Global Maximum and Minimum

Exponential functions do not have global maximum or minimum values.

Local Maximum and Minimum

Exponential functions do not have local maximum or minimum values.

Range

$K$ is continuous
$K$ is decreasing and positive.
$\lim \limits _{f \to -\infty } K(f) = \infty $
$\lim \limits _{f \to \infty } K(f) = 0$

The range is $(0, \infty )$.

This all agrees with the graph.

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2026-05-30 21:59:48