objective: to identify and solve exponential functions
TRANSCRIPT
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Lesson 6-2 Exponential Functions
Objective: To identify and solve exponential functions
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Exponential Functions
Exponential Functions have the form:
A constant raised to a variable power.
f(x) = bx where b >0 & b ≠ 1
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Numb3rs – season 1 episode 8 identity crisis (~10 min)
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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.
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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
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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.
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Graph the exponential Function:
f(x) = 4x
x y-2 1/16
-1 ¼0 11 42 163 64
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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
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Practice28 = 2x+1
42x+1 = 411
8x+1 = 2
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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
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Warm up
Solve for x:
1. 53 − 2x =5−x
2. 32a=3−a
3. 31 − 2x= 243
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Applications of exponential functions
Compound InterestContinuous CompoundingExponential Growth or decay (bacteria/ radiation half life)
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Compound interestCompound interest means the each payment is calculated by including the interest previously earned on the investment.
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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
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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?
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Solution
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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
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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
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Exponential Growth
A = Pert ...or... A = Pekt ...or... Q =ekt ...or... Q = Q0ekt
k is the growth constant
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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
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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.
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Solution1.
2. A = 1240e(.0285)(2.5)
= $1,331.57
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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