Key Property: The variable is in the exponent, not the base!

Basic Exponential Functions

The parent function is $f(x) = b^x$ where $b > 0$ and $b \neq 1$.

Two Cases:

1. Exponential Growth when $b > 1$

   • Example: $f(x) = 2^x$
   • The function increases as $x$ increases
   • Rises steeply to the right

2. Exponential Decay when $0 < b < 1$

   • Example: $f(x) = \left(\frac{1}{2}\right)^x$

Properties of Exponential Functions

For $f(x) = ab^x$ where $a > 0$ and $b > 0, b \neq 1$:

1. Domain: All real numbers $(-\infty, \infty)$
2. Range: $(0, \infty)$ if $a > 0$, or $(-\infty, 0)$ if $a < 0$
3. y-intercept: $(0, a)$ since $f(0) = ab^0 = a$
4. Horizontal Asymptote: $y = 0$ (the x-axis)
5. Always positive (if $a > 0$) - never crosses the x-axis

Exponential Growth and Decay Models

Growth Model

$A(t) = A_0(1 + r)^t$

where:

• $A(t)$ = amount after time $t$
• $A_0$ = initial amount
• $r$ = growth rate (as a decimal)
• $t$ = time

Decay Model

$A(t) = A_0(1 - r)^t$

where:

• $r$ = decay rate (as a decimal)

Continuous Compounding

$A(t) = A_0 e^{rt}$

where $e \approx 2.71828$ (Euler's number)

Key Patterns

• Doubling: If something doubles, multiply by 2
• Tripling: If something triples, multiply by 3
• Half-life: If something halves, multiply by $\frac{1}{2}$
• Growth by 5%: Multiply by $(1 + 0.05) = 1.05$
• Decay by 5%: Multiply by $(1 - 0.05) = 0.95$

Transformations

Just like other functions, exponential functions can be transformed:

• $f(x) = ab^x + k$: vertical shift by $k$
• $f(x) = ab^{x-h}$: horizontal shift by $h$
• $f(x) = -ab^x$: reflection over x-axis

Step 2: Determine the base for doubling every 3 hours.

Since it doubles every 3 hours, after 3 hours we have $2 \times 200 = 400$.

We need to find $b$ such that $200b^3 = 400$.

$b^3 = 2$ $b = 2^{1/3} = \sqrt[3]{2}$

Step 3: Write the function.

$A(t) = 200 \cdot 2^{t/3}$

OR equivalently: $A(t) = 200 \cdot (\sqrt[3]{2})^t$

Step 4: Verify.

• At $t = 0$: $A(0) = 200 \cdot 2^{0/3} = 200$ ✓
• At $t = 3$: $A(3) = 200 \cdot 2^{3/3} = 200 \cdot 2 = 400$ ✓
• At $t = 6$: $A(6) = 200 \cdot 2^{6/3} = 200 \cdot 4 = 800$ ✓

Answer: $A(t) = 200 \cdot 2^{t/3}$ bacteria

Here $k = -0.0315$, so the decay rate is $0.0315 = 3.15\%$ per year.

Part (b): Half-life is when $A(t) = \frac{A_0}{2}$:

$\frac{A_0}{2} = A_0 e^{-0.0315t}$

$\frac{1}{2} = e^{-0.0315t}$

$\ln(0.5) = -0.0315t$

$t = \frac{\ln(0.5)}{-0.0315} = \frac{-0.693}{-0.0315} \approx 22.0$ years

Part (c): With $A_0 = 200$ grams and $t = 50$ years:

$A(50) = 200 \cdot e^{-0.0315(50)}$

$A(50) = 200 \cdot e^{-1.575}$

$A(50) = 200 \cdot 0.2067$

$A(50) \approx 41.3$ grams

Since the base is greater than 1, this represents exponential growth.

Part b) $g(x) = 3(0.8)^x$

Base: $b = 0.8 < 1$ (but $0.8 > 0$)

Since the base is between 0 and 1, this represents exponential decay.

Part c) $h(x) = 2e^{-0.5x}$

Rewrite: $h(x) = 2(e^{-0.5})^x$

The base is $e^{-0.5} \approx 0.606 < 1$

Since the base is between 0 and 1, this represents exponential decay.

Alternative approach for part c): The negative exponent indicates decay.

Answers:

• a) Growth
• b) Decay
• c) Decay