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Exponential Functions and Properties

MATH162
Reference: Stewart 6.2  •  Chapter: 6  •  Section: 2

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Exponential Functions and Properties

Why Exponential Functions Matter

Population growth, radioactive decay, compound interest, viral spread—these phenomena share a common pattern: the rate of change is proportional to the current amount. The mathematical tool that captures this is the exponential function.

Unlike polynomial functions where $x$ is raised to a power (like $x^2$ or $x^3$), in an exponential function the variable is in the exponent: $f(x) = b^x$. This seemingly small change produces dramatically different behavior.

Consider: if you fold a piece of paper 50 times (doubling its thickness each time), the final thickness would be about 17 million miles—enough to reach the sun! That's exponential growth.

Prerequisite Map

Prerequisites
Function DefinitionOne-to-One FunctionsGraphing Basics
This skill
Exponential Functions

Quick Reference

Concept Formula/Property
Exponential Function $f(x) = b^x$ where $b > 0$, $b \neq 1$
Domain $\mathbb{R}$ (all real numbers)
Range $(0, \infty)$ (always positive)
Key Point Always passes through $(0, 1)$ since $b^0 = 1$
Increasing/Decreasing $b > 1$: increasing; $0 < b < 1$: decreasing

Key Concepts

Definition of Exponential Functions

$$\boxed{f(x) = b^x \text{ is an \textbf{exponential function} where } b > 0 \text{ and } b \neq 1}$$

The constant $b$ is called the base. The variable $x$ is the exponent.

Crucial distinction:

These behave very differently! Exponential functions eventually outgrow any polynomial.

What Does $b^x$ Mean for All Real $x$?

We build up the meaning step by step:

Type of Exponent Meaning Example
Positive integer $n$ $b^n = b \cdot b \cdot \ldots \cdot b$ ($n$ factors) $2^3 = 2 \cdot 2 \cdot 2 = 8$
Zero $b^0 = 1$ $5^0 = 1$
Negative integer $-n$ $b^{-n} = \frac{1}{b^n}$ $2^{-3} = \frac{1}{8}$
Rational $\frac{p}{q}$ $b^{p/q} = \sqrt[q]{b^p} = (\sqrt[q]{b})^p$ $8^{2/3} = (\sqrt[3]{8})^2 = 4$
Irrational Limit of rational approximations $2^{\sqrt{3}} \approx 3.322$

For irrational exponents like $2^{\sqrt{3}}$, we use the fact that $\sqrt{3} = 1.732050808...$:

$$2^{\sqrt{3}} = \lim_{r \to \sqrt{3}} 2^r \quad \text{where } r \text{ is rational}$$

The values $2^{1.7}, 2^{1.73}, 2^{1.732}, ...$ converge to a unique real number.

Laws of Exponents

$$\boxed{\begin{aligned} &1. \quad b^{x+y} = b^x \cdot b^y \\[0.3em] &2. \quad b^{x-y} = \frac{b^x}{b^y} \\[0.3em] &3. \quad (b^x)^y = b^{xy} \\[0.3em] &4. \quad (ab)^x = a^x \cdot b^x \end{aligned}}$$

These laws, familiar from algebra with rational exponents, extend to all real exponents.

Graph Behavior

     y                              y                           y
     |     /                        |                           |
     |    /                         |------ y = 1               | \
     |   /                          |                           |  \
     |  /                           |                           |   \
     |_/________________________    |_______________________    |____\________________
     |                         x    |                      x    |                     x
     (0,1)                          (0,1)                       (0,1)

     b > 1 (increasing)             b = 1 (constant)            0 < b < 1 (decreasing)

Key observations:

Limit Behavior

$$\boxed{\begin{aligned} &\text{If } b > 1: \quad \lim_{x \to \infty} b^x = \infty \quad \text{and} \quad \lim_{x \to -\infty} b^x = 0 \\[0.5em] &\text{If } 0 < b < 1: \quad \lim_{x \to \infty} b^x = 0 \quad \text{and} \quad \lim_{x \to -\infty} b^x = \infty \end{aligned}}$$

In either case, the $x$-axis is a horizontal asymptote.

Comparing Bases

Base Behavior Example Uses
$b > 1$ Exponential growth Population, compound interest
$b = 1$ Constant function $y = 1$ Not interesting!
$0 < b < 1$ Exponential decay Radioactive decay, cooling

Larger bases grow faster: $10^x$ grows faster than $2^x$ for $x > 0$.

Common Mistakes

Mistake Correct Understanding
Thinking $b^0 = 0$ $b^0 = 1$ for any $b > 0$
Confusing $b^{x+y}$ with $b^x + b^y$ $b^{x+y} = b^x \cdot b^y$ (multiply, don't add)
Writing $(ab)^x = a^x b$ $(ab)^x = a^x \cdot b^x$ (both get exponent)
Thinking $b^x$ can be negative $b^x > 0$ always (with $b > 0$)
Confusing $2^x$ and $x^2$ Exponential vs. power function—very different!

Practice Problems

Level 1 Evaluating Exponential Expressions

Evaluate without a calculator: (a) $4^{3/2}$ (b) $27^{-2/3}$ (c) $9^0 + 9^1$

Thought Process

For rational exponents, use $b^{p/q} = (\sqrt[q]{b})^p$. For negative exponents, take the reciprocal. For $b^0$, the answer is always 1.

Show Answer

(a) $4^{3/2} = (\sqrt{4})^3 = 2^3 = \boxed{8}$

(b) $27^{-2/3} = \frac{1}{27^{2/3}} = \frac{1}{(\sqrt[3]{27})^2} = \frac{1}{3^2} = \boxed{\frac{1}{9}}$

(c) $9^0 + 9^1 = 1 + 9 = \boxed{10}$

Level 2 Simplifying with Laws of Exponents

Simplify using the laws of exponents: $$\frac{3^{x+2} \cdot 3^{2x}}{3^{x-1}}$$

Thought Process

Use $b^m \cdot b^n = b^{m+n}$ for the numerator, then $\frac{b^m}{b^n} = b^{m-n}$ for the fraction.

Show Answer

First, combine the numerator: $$3^{x+2} \cdot 3^{2x} = 3^{(x+2) + 2x} = 3^{3x+2}$$

Then divide: $$\frac{3^{3x+2}}{3^{x-1}} = 3^{(3x+2) - (x-1)} = 3^{3x+2-x+1} = \boxed{3^{2x+3}}$$

Level 2 Graph Transformations

Starting with the graph of $y = 2^x$, describe the transformations needed to obtain the graph of $y = 2^{-x} - 1$. Then identify the horizontal asymptote.

Thought Process

Break down the transformation: replacing $x$ with $-x$ reflects about the $y$-axis. Subtracting 1 shifts down. The asymptote shifts with vertical translations.

Show Answer

Step 1: $y = 2^x \to y = 2^{-x}$ This reflects the graph about the $y$-axis.

Step 2: $y = 2^{-x} \to y = 2^{-x} - 1$ This shifts the graph down by 1 unit.

Horizontal asymptote: The original asymptote $y = 0$ shifts down to $\boxed{y = -1}$.

Level 3 Finding an Exponential Function from Data

A scientist measures a quantity at two times and records:

  • At $t = 2$: the value is 12
  • At $t = 5$: the value is 96

Find constants $A$ and $b$ so that $Q(t) = A \cdot b^t$ models this data.

Thought Process

Goal: Find the exponential function through two given points.

Strategy: Set up two equations from the two data points. The key insight is that dividing one equation by the other eliminates $A$, letting us solve for $b$ first.

Step-by-step:

  1. Write equations: $Q(2) = Ab^2 = 12$ and $Q(5) = Ab^5 = 96$
  2. Divide to eliminate $A$: $\frac{Ab^5}{Ab^2} = \frac{96}{12} = 8$
  3. Simplify: $b^3 = 8$, so $b = 2$
  4. Substitute back to find $A$
Show Answer

From the two data points: $$Q(2) = Ab^2 = 12$$ $$Q(5) = Ab^5 = 96$$

Divide the second equation by the first: $$\frac{Ab^5}{Ab^2} = \frac{96}{12}$$ $$b^3 = 8$$ $$b = 2 \quad \text{(taking positive root since } b > 0\text{)}$$

Substitute back: $A \cdot 2^2 = 12 \implies A \cdot 4 = 12 \implies A = 3$

Answer: $\boxed{Q(t) = 3 \cdot 2^t}$

Verification: $Q(2) = 3 \cdot 4 = 12$ ✓ and $Q(5) = 3 \cdot 32 = 96$ ✓

Physical interpretation: This represents exponential growth with a doubling factor—the quantity doubles each time $t$ increases by 1.

Level 4 Evaluating Limits

Evaluate: $$\lim_{x \to \infty} \frac{5^x - 3^x}{5^x + 3^x}$$

Thought Process

When comparing exponentials with different bases as $x \to \infty$, the larger base dominates. Factor out the dominant term ($5^x$) from both numerator and denominator.

Show Answer

Factor out $5^x$ from numerator and denominator: $$\frac{5^x - 3^x}{5^x + 3^x} = \frac{5^x(1 - (3/5)^x)}{5^x(1 + (3/5)^x)} = \frac{1 - (3/5)^x}{1 + (3/5)^x}$$

Since $0 < \frac{3}{5} < 1$, we have $(3/5)^x \to 0$ as $x \to \infty$.

Therefore: $$\lim_{x \to \infty} \frac{1 - (3/5)^x}{1 + (3/5)^x} = \frac{1 - 0}{1 + 0} = \boxed{1}$$

Insight: The $5^x$ term dominates both numerator and denominator, so the ratio approaches 1.

Level 5 Proving Exponential Dominance

Show that $2^x$ eventually exceeds $x^{10}$. That is, prove there exists some $N$ such that $2^x > x^{10}$ for all $x > N$.

Hint: Consider the ratio $\frac{2^x}{x^{10}}$ and what happens as $x \to \infty$.

Thought Process

This is a deep result: exponential growth beats polynomial growth, no matter how large the polynomial's degree. Consider the ratio and show it goes to infinity. You can use the fact that $\lim_{x \to \infty} \frac{2^x}{x^n} = \infty$ for any fixed $n$.

Show Answer

Consider the ratio $\frac{2^x}{x^{10}}$.

Take logarithms: $\ln\left(\frac{2^x}{x^{10}}\right) = x \ln 2 - 10 \ln x$

As $x \to \infty$:

  • $x \ln 2$ grows linearly (at rate $\ln 2 \approx 0.693$)
  • $10 \ln x$ grows logarithmically (much slower)

Since linear growth dominates logarithmic growth: $$\lim_{x \to \infty} (x \ln 2 - 10 \ln x) = \infty$$

Therefore $\ln\left(\frac{2^x}{x^{10}}\right) \to \infty$, which means $\frac{2^x}{x^{10}} \to \infty$.

This proves $2^x > x^{10}$ for all sufficiently large $x$.

Numerical check: At $x = 60$: $2^{60} \approx 1.15 \times 10^{18}$ while $60^{10} \approx 6.05 \times 10^{17}$. Indeed $2^{60} > 60^{10}$.

Big picture: Exponential functions eventually dominate any polynomial, no matter how high the degree. This is why exponential growth is so powerful (and dangerous in contexts like epidemics).

CCI-Style Conceptual Questions

Question 1: The function $f(x) = 3^x$ passes through which of the following points?

(A) $(0, 0)$ (B) $(0, 1)$ (C) $(1, 1)$ (D) $(3, 1)$

Answer

(B) For any exponential function $b^x$, we have $b^0 = 1$, so the graph passes through $(0, 1)$. It does NOT pass through the origin.

Question 2: Which grows faster as $x \to \infty$: $f(x) = 1000x^{100}$ or $g(x) = 1.001^x$?

(A) $f(x) = 1000x^{100}$ grows faster (B) $g(x) = 1.001^x$ grows faster (C) They grow at the same rate (D) It depends on the value of $x$

Answer

(B) Even though $1.001$ is barely larger than 1, the exponential $1.001^x$ eventually dominates any polynomial. Exponential growth always beats polynomial growth for large $x$, regardless of coefficients or degrees.

Question 3: If $f(x) = b^x$ where $0 < b < 1$, which statement is TRUE?

(A) $f$ is increasing and $f(x) > 0$ for all $x$ (B) $f$ is decreasing and $f(x) > 0$ for all $x$ (C) $f$ is decreasing and $f(x) < 0$ for some $x$ (D) $f$ has a horizontal asymptote at $y = 1$

Answer

(B) When $0 < b < 1$, the function is decreasing (each multiplication by a fraction less than 1 makes the value smaller). However, $b^x > 0$ always—exponential functions never produce negative values.

Mastery Checklist

Mental Model

The Multiplier Effect:

Think of $b^x$ as starting with 1 and multiplying by $b$ a total of $x$ times.

This is why $b^x > 0$ always: you start positive and multiply by positive numbers.

Real-world analogy: Compound interest at rate $r$ multiplies your money by $(1 + r)$ each year. After $t$ years: $A = A_0(1+r)^t$. The "multiplier effect" makes even small rates produce dramatic long-term growth.

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Last updated: 2026-01-23