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Lesson 1: Exponential Functions

Estimated time: 35-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Introduction: What Are Exponential Functions?

In previous modules, we studied polynomial functions where the variable appears in the base and exponents are constants. For example, f(x) = x² and g(x) = x³ are polynomial functions. Now we introduce a fundamentally different type of function: exponential functions.

Exponential Function

An exponential function is a function of the form:

f(x) = a · b^x

where:

  • a is a nonzero constant (the initial value or coefficient)
  • b is a positive constant called the base (b > 0, b ≠ 1)
  • x is the variable (appears in the exponent)

Key Distinction: Polynomial vs. Exponential

  • Polynomial: f(x) = x³ (variable is the base, exponent is constant)
  • Exponential: f(x) = 3^x (variable is the exponent, base is constant)

This seemingly small difference creates dramatically different behavior!

Why b ≠ 1?

If b = 1, then f(x) = a · 1^x = a for all x, which is just a constant function (horizontal line), not an exponential function.

Exponential functions model many real-world phenomena including:

Section 1: Exponential Growth (b > 1)

When the base b is greater than 1, we have exponential growth. The function values increase rapidly as x increases.

Exponential Growth

If f(x) = a · b^x where b > 1, the function exhibits exponential growth.

Characteristics:

  • The function is always increasing (rises from left to right)
  • Domain: all real numbers (-∞, ∞)
  • Range: (0, ∞) if a > 0; (-∞, 0) if a < 0
  • y-intercept: (0, a) since f(0) = a · b⁰ = a · 1 = a
  • Horizontal asymptote: y = 0 (the x-axis)
  • No x-intercepts (never crosses the x-axis)

Example 1: Evaluating an Exponential Function

Let f(x) = 2^x. Evaluate f at the following values: x = -2, -1, 0, 1, 2, 3.

Solution

We substitute each value into the function:

  • f(-2) = 2^(-2) = 1/2² = 1/4 = 0.25
  • f(-1) = 2^(-1) = 1/2¹ = 1/2 = 0.5
  • f(0) = 2⁰ = 1
  • f(1) = 2¹ = 2
  • f(2) = 2² = 4
  • f(3) = 2³ = 8

Observation: Notice that as x increases by 1, the function value doubles. This is characteristic of exponential growth with base 2.

x f(x) = 2^x
-2 0.25
-1 0.5
0 1
1 2
2 4
3 8

Example 2: Graphing an Exponential Growth Function

Graph the function f(x) = 3^x and identify its key features.

Solution

Step 1: Create a table of values.

x f(x) = 3^x
-2 3^(-2) = 1/9 ≈ 0.11
-1 3^(-1) = 1/3 ≈ 0.33
0 3⁰ = 1
1 3¹ = 3
2 3² = 9

Step 2: Plot the points and connect with a smooth curve.

Graph Description:

  • The curve passes through (0, 1)
  • As x increases, the curve rises steeply to the right
  • As x decreases (moving left), the curve approaches the x-axis but never touches it
  • The curve is always above the x-axis (all y-values are positive)

Key Features:

  • Domain: (-∞, ∞)
  • Range: (0, ∞)
  • y-intercept: (0, 1)
  • Horizontal asymptote: y = 0
  • Increasing throughout its domain
  • No x-intercepts

Example 3: Population Growth Model

A city's population is modeled by P(t) = 50,000 · (1.05)^t, where t is the number of years since 2020.

(a) What was the population in 2020?
(b) What will the population be in 2030?
(c) What does the base 1.05 represent?

Solution

(a) Population in 2020:

In 2020, t = 0 (years since 2020).

P(0) = 50,000 · (1.05)⁰ = 50,000 · 1 = 50,000

The population in 2020 was 50,000 people.

(b) Population in 2030:

In 2030, t = 10 (years since 2020).

P(10) = 50,000 · (1.05)^10

P(10) = 50,000 · 1.62889...

P(10) ≈ 81,445 people

(c) Meaning of the base 1.05:

The base 1.05 = 1 + 0.05 = 1 + 5%

This means the population increases by 5% each year. We can write 1.05 as "100% + 5% growth rate."

Example 4: Comparing Growth Rates

Compare the growth of f(x) = 2^x and g(x) = 3^x. Which function grows faster?

Solution

Let's create a table comparing values:

x f(x) = 2^x g(x) = 3^x
0 1 1
1 2 3
2 4 9
3 8 27
4 16 81
5 32 243

Analysis:

  • Both functions pass through (0, 1)
  • For x > 0, g(x) > f(x) and the difference increases dramatically
  • g(x) = 3^x grows faster because it has a larger base
  • At x = 5, g(x) is more than 7 times larger than f(x)!

General Rule: For exponential growth functions with the same coefficient a, the function with the larger base grows faster.

Section 2: Exponential Decay (0 < b < 1)

When the base b is between 0 and 1, we have exponential decay. The function values decrease rapidly as x increases.

Exponential Decay

If f(x) = a · b^x where 0 < b < 1, the function exhibits exponential decay.

Characteristics:

  • The function is always decreasing (falls from left to right)
  • Domain: all real numbers (-∞, ∞)
  • Range: (0, ∞) if a > 0; (-∞, 0) if a < 0
  • y-intercept: (0, a)
  • Horizontal asymptote: y = 0 (the x-axis)
  • No x-intercepts

Alternative Form for Decay

Exponential decay can also be written as f(x) = a · (1/b)^x where b > 1, or equivalently as f(x) = a · b^(-x).

For example: f(x) = (1/2)^x = 2^(-x)

Example 5: Evaluating an Exponential Decay Function

Let f(x) = (1/2)^x. Evaluate f at x = -2, -1, 0, 1, 2, 3.

Solution

We substitute each value:

  • f(-2) = (1/2)^(-2) = 2² = 4
  • f(-1) = (1/2)^(-1) = 2¹ = 2
  • f(0) = (1/2)⁰ = 1
  • f(1) = (1/2)¹ = 1/2 = 0.5
  • f(2) = (1/2)² = 1/4 = 0.25
  • f(3) = (1/2)³ = 1/8 = 0.125

Observation: As x increases by 1, the function value is cut in half. This is exponential decay with base 1/2.

x f(x) = (1/2)^x
-2 4
-1 2
0 1
1 0.5
2 0.25
3 0.125

Example 6: Graphing Exponential Decay

Graph f(x) = (0.5)^x and identify its key features.

Solution

Step 1: Create a table of values (same as Example 5).

Step 2: Plot and connect the points.

Graph Description:

  • The curve passes through (0, 1)
  • As x increases (moving right), the curve approaches the x-axis but never touches it
  • As x decreases (moving left), the curve rises steeply upward
  • The curve is always above the x-axis

Key Features:

  • Domain: (-∞, ∞)
  • Range: (0, ∞)
  • y-intercept: (0, 1)
  • Horizontal asymptote: y = 0
  • Decreasing throughout its domain
  • No x-intercepts

Note: This graph is a reflection of y = 2^x across the y-axis, since (1/2)^x = 2^(-x).

Example 7: Radioactive Decay

A radioactive substance has a mass of 100 grams. Its mass decreases according to the formula A(t) = 100 · (0.95)^t, where t is measured in years.

(a) What is the initial mass?
(b) What will the mass be after 10 years?
(c) What percentage of the substance decays each year?

Solution

(a) Initial mass:

At t = 0:

A(0) = 100 · (0.95)⁰ = 100 · 1 = 100 grams

(b) Mass after 10 years:

A(10) = 100 · (0.95)^10

A(10) = 100 · 0.59874...

A(10) ≈ 59.87 grams

(c) Decay percentage:

The base is 0.95 = 1 - 0.05 = 100% - 5%

This means 5% of the substance decays each year (the mass retains 95% of its previous value each year).

Example 8: Half-Life Problem

The half-life of a radioactive element is the time it takes for half of the substance to decay. If a substance has a half-life of 5 years, write an exponential function to model the amount remaining, starting with an initial amount of A₀.

Solution

Understanding half-life:

After 5 years (one half-life), the amount remaining is A₀/2.

After 10 years (two half-lives), the amount remaining is A₀/4.

After 15 years (three half-lives), the amount remaining is A₀/8.

General formula:

The number of half-lives is t/5 (where t is in years).

After each half-life, we multiply by 1/2.

A(t) = A₀ · (1/2)^(t/5)

Or equivalently: A(t) = A₀ · (0.5)^(t/5)

Verification:

  • At t = 0: A(0) = A₀ · (1/2)⁰ = A₀
  • At t = 5: A(5) = A₀ · (1/2)¹ = A₀/2
  • At t = 10: A(10) = A₀ · (1/2)² = A₀/4

Section 3: Transformations of Exponential Functions

Just like with other functions, we can apply transformations to exponential functions. The parent functions y = b^x can be shifted, reflected, and stretched.

Transformation Summary

Given the parent function f(x) = b^x:

  • Vertical shift: f(x) = b^x + k (shifts up k units if k > 0, down if k < 0)
  • Horizontal shift: f(x) = b^(x-h) (shifts right h units if h > 0, left if h < 0)
  • Vertical stretch/compression: f(x) = a · b^x (stretches if |a| > 1, compresses if 0 < |a| < 1)
  • Reflection over x-axis: f(x) = -b^x
  • Reflection over y-axis: f(x) = b^(-x)

Example 9: Vertical Shift

Graph f(x) = 2^x + 3 and compare it to the parent function g(x) = 2^x.

Solution

Analysis:

The "+3" outside the exponential shifts the entire graph up 3 units.

x g(x) = 2^x f(x) = 2^x + 3
-2 0.25 3.25
-1 0.5 3.5
0 1 4
1 2 5
2 4 7

Key Changes:

  • Domain: (-∞, ∞) - unchanged
  • Range: (3, ∞) - changed from (0, ∞)
  • Horizontal asymptote: y = 3 - changed from y = 0
  • y-intercept: (0, 4) - changed from (0, 1)
  • The graph has the same shape but shifted up 3 units

Example 10: Horizontal Shift

Graph f(x) = 2^(x-1) and identify the transformation from g(x) = 2^x.

Solution

Analysis:

The "-1" in the exponent shifts the graph 1 unit to the right.

x g(x) = 2^x f(x) = 2^(x-1)
-1 0.5 0.25
0 1 0.5
1 2 1
2 4 2
3 8 4

Key Changes:

  • Domain: (-∞, ∞) - unchanged
  • Range: (0, ∞) - unchanged
  • Horizontal asymptote: y = 0 - unchanged
  • y-intercept: (0, 0.5) - changed from (0, 1)
  • The point (0, 1) on g(x) has moved to (1, 1) on f(x)

Note: We can verify: f(x) = 2^(x-1) = 2^x · 2^(-1) = (1/2) · 2^x, which is a vertical compression by factor 1/2.

Example 11: Reflection Over the x-axis

Graph f(x) = -2^x. How does this compare to g(x) = 2^x?

Solution

Analysis:

The negative sign in front reflects the graph over the x-axis.

x g(x) = 2^x f(x) = -2^x
-2 0.25 -0.25
-1 0.5 -0.5
0 1 -1
1 2 -2
2 4 -4

Key Changes:

  • Domain: (-∞, ∞) - unchanged
  • Range: (-∞, 0) - changed from (0, ∞)
  • Horizontal asymptote: y = 0 - unchanged
  • y-intercept: (0, -1) - changed from (0, 1)
  • The function is now decreasing instead of increasing
  • All y-values are negative instead of positive

Example 12: Combined Transformations

Describe the transformations applied to f(x) = 2^x to obtain g(x) = -3 · 2^(x+1) + 2.

Solution

Let's break down g(x) = -3 · 2^(x+1) + 2 step by step:

Starting with f(x) = 2^x, the transformations are:

  1. Horizontal shift left 1 unit: 2^(x+1)
  2. Vertical stretch by factor 3: 3 · 2^(x+1)
  3. Reflection over x-axis: -3 · 2^(x+1)
  4. Vertical shift up 2 units: -3 · 2^(x+1) + 2

Key features of g(x):

  • Domain: (-∞, ∞)
  • Range: (-∞, 2)
  • Horizontal asymptote: y = 2
  • y-intercept: g(0) = -3 · 2¹ + 2 = -6 + 2 = -4, so (0, -4)
  • The function is decreasing (due to the reflection)

Example 13: Writing Equations from Graphs

An exponential function has a horizontal asymptote at y = -1, passes through the point (0, 3), and is increasing. Write a possible equation for this function.

Solution

Step 1: Identify the vertical shift.

The horizontal asymptote is at y = -1, so we have a vertical shift of k = -1.

Form: f(x) = a · b^x - 1

Step 2: Use the point (0, 3) to find a.

f(0) = 3

a · b⁰ - 1 = 3

a · 1 - 1 = 3

a = 4

Step 3: Choose appropriate base b.

Since the function is increasing, we need b > 1. Any base greater than 1 works, but let's use b = 2.

Answer: f(x) = 4 · 2^x - 1

Verification:

  • Horizontal asymptote: y = -1
  • f(0) = 4 · 2⁰ - 1 = 4 - 1 = 3
  • Function is increasing since b = 2 > 1

Note: Other valid answers include f(x) = 4 · 3^x - 1, f(x) = 4 · 10^x - 1, etc.

Section 4: Compound Interest and the Number e

One of the most important applications of exponential functions is calculating compound interest on investments and loans.

Compound Interest Formula

A = P(1 + r/n)^(nt)

where:

  • A = final amount (principal + interest)
  • P = principal (initial amount invested or borrowed)
  • r = annual interest rate (as a decimal)
  • n = number of times interest is compounded per year
  • t = time in years

Common Compounding Periods

  • Annually: n = 1
  • Semiannually: n = 2
  • Quarterly: n = 4
  • Monthly: n = 12
  • Daily: n = 365

Example 14: Quarterly Compounding

You invest $5,000 in an account that pays 6% annual interest, compounded quarterly. How much will you have after 5 years?

Solution

Given information:

  • P = $5,000
  • r = 6% = 0.06
  • n = 4 (quarterly)
  • t = 5 years

Calculation:

A = P(1 + r/n)^(nt)

A = 5000(1 + 0.06/4)^(4·5)

A = 5000(1 + 0.015)^20

A = 5000(1.015)^20

A = 5000(1.34685...)

A ≈ $6,734.28

Answer: After 5 years, you will have approximately $6,734.28.

Interest earned: $6,734.28 - $5,000 = $1,734.28

Example 15: Comparing Compounding Frequencies

Compare the future value of $10,000 invested at 8% annual interest for 10 years with:

(a) Annual compounding
(b) Monthly compounding
(c) Daily compounding

Solution

Given: P = $10,000, r = 0.08, t = 10

(a) Annual compounding (n = 1):

A = 10,000(1 + 0.08/1)^(1·10)

A = 10,000(1.08)^10

A = 10,000(2.15892...)

A ≈ $21,589.25

(b) Monthly compounding (n = 12):

A = 10,000(1 + 0.08/12)^(12·10)

A = 10,000(1.00667)^120

A = 10,000(2.21964...)

A ≈ $22,196.40

(c) Daily compounding (n = 365):

A = 10,000(1 + 0.08/365)^(365·10)

A = 10,000(1.000219)^3650

A = 10,000(2.22544...)

A ≈ $22,254.39

Comparison:

Compounding Frequency Final Amount Interest Earned
Annual $21,589.25 $11,589.25
Monthly $22,196.40 $12,196.40
Daily $22,254.39 $12,254.39

Observation: More frequent compounding results in more interest earned, but the difference between monthly and daily is relatively small ($58).

Continuous Compounding and the Number e

What happens if we compound interest continuously (infinitely often)? This leads us to one of the most important constants in mathematics: e.

The number e ≈ 2.71828... is called Euler's number.

Continuous Compounding Formula:

A = Pe^(rt)

where e is Euler's number and the other variables are as before.

Example 16: Continuous Compounding

Using the same values from Example 15, calculate the future value with continuous compounding: P = $10,000, r = 8%, t = 10 years.

Solution

Using the continuous compounding formula:

A = Pe^(rt)

A = 10,000 · e^(0.08 · 10)

A = 10,000 · e^0.8

A = 10,000 · 2.22554...

A ≈ $22,255.41

Comparison with daily compounding:

  • Daily compounding: $22,254.39
  • Continuous compounding: $22,255.41
  • Difference: $1.02

Conclusion: Continuous compounding gives only slightly more than daily compounding, showing that there's a practical limit to how much extra interest can be earned by increasing the compounding frequency.

The Natural Exponential Function

The function f(x) = e^x is called the natural exponential function.

Properties:

  • Domain: (-∞, ∞)
  • Range: (0, ∞)
  • y-intercept: (0, 1) since e⁰ = 1
  • Horizontal asymptote: y = 0
  • Always increasing
  • Special property: The rate of growth equals the function value itself

Example 17: Time to Double an Investment

How long will it take for an investment to double if it earns 6% annual interest compounded continuously?

Solution

Set up:

We want A = 2P (double the principal).

Using A = Pe^(rt) with r = 0.06:

2P = P · e^(0.06t)

Simplify:

Divide both sides by P:

2 = e^(0.06t)

Solve for t:

We need to solve this equation. For now, we can use trial and error or a calculator:

  • Try t = 10: e^(0.6) ≈ 1.822 (too small)
  • Try t = 12: e^(0.72) ≈ 2.054 (too large)
  • Try t = 11.5: e^(0.69) ≈ 1.994 (close)
  • More precisely: t ≈ 11.55 years

Answer: It will take approximately 11.55 years for the investment to double.

Note: In Lesson 2 on logarithms, we'll learn how to solve this algebraically: t = ln(2)/0.06 ≈ 11.55.

Example 18: Effective Annual Rate

A bank advertises an account with 5% annual interest compounded monthly. What is the effective annual rate (the actual percentage increase over one year)?

Solution

Method: Calculate how much $1 grows to in one year.

Using P = 1, r = 0.05, n = 12, t = 1:

A = 1(1 + 0.05/12)^(12·1)

A = (1 + 0.004167)^12

A = (1.004167)^12

A ≈ 1.05116

Interpretation:

$1 grows to $1.05116, which is an increase of $0.05116 or 5.116%.

Answer: The effective annual rate is approximately 5.116%.

Comparison:

  • Nominal (stated) rate: 5%
  • Effective annual rate: 5.116%
  • The difference (0.116%) comes from the monthly compounding

Check Your Understanding

Question 1: Which of the following is an exponential function?

(A) f(x) = x³
(B) f(x) = 3x
(C) f(x) = 3^x
(D) f(x) = x + 3

Answer: (C) f(x) = 3^x

Explanation: In an exponential function, the variable is in the exponent and the base is a constant. (A) is a polynomial, (B) is linear, and (D) is also linear.

Question 2: Evaluate f(x) = 5^x when x = -2.

Answer: f(-2) = 1/25 or 0.04

Solution: f(-2) = 5^(-2) = 1/5² = 1/25 = 0.04

Question 3: Does f(x) = (0.7)^x represent exponential growth or decay?

Answer: Exponential decay

Explanation: Since the base 0.7 is between 0 and 1, this is exponential decay. The function values decrease as x increases.

Question 4: What is the horizontal asymptote of f(x) = 2^x + 5?

Answer: y = 5

Explanation: The "+5" shifts the graph up 5 units, so the horizontal asymptote moves from y = 0 to y = 5.

Question 5: What transformation is applied to f(x) = 2^x to get g(x) = 2^(x-3)?

Answer: Horizontal shift right 3 units

Explanation: The "-3" in the exponent shifts the graph to the right by 3 units.

Question 6: A population grows from 500 to 800 in 5 years. If the growth is exponential with model P(t) = 500 · b^t, which statement is true?

(A) b < 1
(B) b = 1
(C) b > 1
(D) Cannot be determined

Answer: (C) b > 1

Explanation: The population is growing (increasing from 500 to 800), so we have exponential growth, which requires b > 1.

Question 7: You invest $2,000 at 4% annual interest compounded quarterly for 3 years. What values should you use in the formula A = P(1 + r/n)^(nt)?

Answer: P = 2000, r = 0.04, n = 4, t = 3

Explanation:

  • P = $2,000 (principal)
  • r = 4% = 0.04 (as a decimal)
  • n = 4 (quarterly means 4 times per year)
  • t = 3 years

The future value would be: A = 2000(1 + 0.04/4)^(4·3) = 2000(1.01)^12 ≈ $2,253.65

Question 8: Which base results in faster growth: f(x) = 2^x or g(x) = 5^x?

Answer: g(x) = 5^x grows faster

Explanation: For exponential growth functions with the same coefficient, the function with the larger base grows faster. Since 5 > 2, we have 5^x > 2^x for all x > 0.

Question 9: What is the y-intercept of f(x) = 3 · 2^x - 5?

Answer: (0, -2)

Solution: f(0) = 3 · 2⁰ - 5 = 3 · 1 - 5 = 3 - 5 = -2

The y-intercept is the point (0, -2).

Question 10: An account earns 6% annual interest compounded continuously. Using A = Pe^(rt), what is the value of r?

Answer: r = 0.06

Explanation: The interest rate must be expressed as a decimal, so 6% = 0.06. In the continuous compounding formula, we use r as a decimal, not a percentage.

Key Takeaways

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