Lesson 6-2 Exponential Functions

Objective: To identify and solve exponential functions

Exponential Functions

Exponential Functions have the form:

A constant raised to a variable power.

f(x) = bx where b >0 & b ≠ 1

Numb3rs – season 1 episode 8 identity crisis (~10 min)

Graphs of Exponential Functions

Enter into calculator and graph: f(x) = 2x

[2nd] [table] to see list of data points

Notice how quickly exponential functions grow.

Properties of Exponential Functions

If bu = bv then u = v When b > 0 and b≠ 1

The graph of f(x) = bx always passes through the points (0,1) and (1,b) The graph of f(x) = bx is the reflection about the y-axis of the graph of f(x)=The graph of f(x) = bx has the horizontal asymptote y = 0.

xx

bb

1

Properties of Exponential Functions

The domain of f(x) = bx is the set of real numbers: the range is the set of positive numbers.f(x) = bx is increasing if b > 1; f(x) = bx is decreasing if 0< b< 1.

f(x) = bx is a one-to-one function since it passes the horizontal line test.

Graph the exponential Function:

f(x) = 4x

x y-2 1/16

-1 ¼0 11 42 163 64

Solving Exponential Equations310 = 35x

10 = 5xx = 2

27 = (x-1)7 if x > 12 = x-13 = x

33x = 9x-1

33x = 32(x-1)

3x = 2x -2x = -2

Practice28 = 2x+1

42x+1 = 411

8x+1 = 2

The Number e

e is an irrational numberIt is , as m gets larger and larger.It is approximately 2.71828

f(x)= ex is a natural exponential functiongraph f(x)= ex on the calculator

m

m

11

Warm up

Solve for x:

1. 53 − 2x =5−x

2. 32a=3−a

3. 31 − 2x= 243

Applications of exponential functions

Compound InterestContinuous CompoundingExponential Growth or decay (bacteria/ radiation half life)

Compound interestCompound interest means the each payment is calculated by including the interest previously earned on the investment.

Investing at 10% interest Compounded Annually

Year Investment at Start Interest Investment at End

0 (Now) $1,000.00 ($1,000.00 × 10% = ) $100.00 $1,100.00

1 $1,100.00 ($1,100.00 × 10% = ) $110.00 $1,210.00

2 $1,210.00 ($1,210.00 × 10% = ) $121.00 $1,331.00

3 $1,331.00 ($1,331.00 × 10% = ) $133.10 $1,464.10

4 $1,464.10 ($1,464.10 × 10% = ) $146.41 $1,610.51

5 $1,610.51

formula

If you have a bank account whose principal = $1000, and your bank compounds the interest twice a year at an interest rate of 5%, how much money do you have in your account at the year's end?

Solution

Continous Compounding

When n gets very large it approaches becoming continuous compounding. The formula is:

P = principal amount (initial investment)r = annual interest rate (as a decimal)t = number of yearsA = amount after time t

rtPeA

Example

An amount of $2,340.00 is deposited in a bank paying an annual interest rate of 3.1%, compounded continuously. Find the balance after 3 years.Solution

A = 2340 e(.031)(3)

A = 2568.06

Exponential Growth

A = Pert ...or... A = Pekt ...or... Q =ekt ...or... Q = Q0ekt

k is the growth constant

Bacteria Growth

In t hours the number of bacteria in a culture will grow to be approximately Q = Q0e2t where Q0 is the original number of bacteria. At 1 PM the culture has 50 bacteria. How many bacteria does it have at 4 PM? at noon?

Q = 50e2(3) Q = 50e2(-1)

Q = 50e6 Q = 50e-2

Q = 20,248 Q = 7

Practice

1. If you start a bank account with $10,000 and your bank compounds the interest quarterly at an interest rate of 8%, how much money do you have at the year’s end ? (assume that you do not add or withdraw any money from the account)2. An amount of $1,240.00 is deposited in a bank paying an annual interest rate of 2.85 %, compounded continuously. Find the balance after 2½ years.

Solution1.

2. A = 1240e(.0285)(2.5)

= $1,331.57

Exponential Decay

An artifact originally had 12 grams of carbon-14 present. The decay model A = 12e-0.000121t

describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in this artifact after 10,000 years?

A = 12e-0.000121t

A = 12e-0.000121(10,000)

A = 12e-1.21

A = 3.58