1. Objective: To identify and solve exponential functions

2. Exponential Functions have the form: A constant raised to a variable power. f(x) = b x where b >0 & b ≠ 1

3. Numb3rs – season 1 episode 8 identity crisis (~10 minNumb3rs – season 1 episode 8 identity crisis (~10 min)

4. Enter into calculator and graph: f(x) = 2 x [2nd] [table] to see list of data points Notice how quickly exponential functions grow.

5. If b u = b v then u = v When b > 0 and b≠ 1 The graph of f(x) = b x always passes through the points (0,1) and (1,b) The graph of f(x) = b x is the reflection about the y-axis of the graph of f(x)= The graph of f(x) = b x has the horizontal asymptote y = 0.

6. The domain of f(x) = b x is the set of real numbers: the range is the set of positive numbers. f(x) = b x is increasing if b > 1 ; f(x) = b x is decreasing if 0< b< 1. f(x) = b x is a one-to-one function since it passes the horizontal line test.

7. f(x) = 4 x xy -2 1/16 ¼ 01 14 216 364

8. 3 10 = 3 5x 10 = 5x x = 2 2 7 = (x-1) 7 if x > 1 2 = x-1 3 = x 3 3x = 9 x-1 3 3x = 3 2(x-1) 3x = 2x -2 x = -2

9. 2 8 = 2 x+1 4 2x+1 = 4 11 8 x+1 = 2

10. e is an irrational number It is, as m gets larger and larger. It is approximately 2.71828 f(x)= e x is a natural exponential function graph f(x)= e x on the calculator

11. Solve for x: 1. 5 3 − 2x =5 −x 2. 3 2a =3 −a 3. 3 1 − 2x = 243

12. Compound Interest Continuous Compounding Exponential Growth or decay (bacteria/ radiation half life)

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

14. Year Investment at StartInterestInvestment at End 0 (Now)$1,000.00 ($1,000.00 × 10% = ) $100.00 $1,100.00 1 ($1,100.00 × 10% = ) $110.00 $1,210.00 2 ($1,210.00 × 10% = ) $121.00 $1,331.00 3 ($1,331.00 × 10% = ) $133.10 $1,464.10 4 ($1,464.10 × 10% = ) $146.41 $1,610.51 5

15. 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?principal interestan interest rate

17. 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 years A = amount after time t

18. 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

19. A = Pe rt...or... A = Pe kt...or... Q =e kt...or... Q = Q 0 e kt k is the growth constant

20. In t hours the number of bacteria in a culture will grow to be approximately Q = Q 0 e 2t where Q 0 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 = 50e 2(3) Q = 50e 2(-1) Q = 50e 6 Q = 50e -2 Q = 20,248 Q = 7

21. 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)interest rate 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.

22. 1. 2. A = 1240e (.0285)(2.5) = $1,331.57

23. 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