1. Unit 5 Review
By: Hasitha and Sejal

2. Inverse Functions

3. What is an inverse function?
Reflection of original graph over y=x
The y and x coordinates of the original graph are switched
Sometimes you need to restrict the domain

4. How to Check if a Function is Inverse
Graphing
Composition
Algebraically

5. Graphing Graph the first function Graph the second function
Check if it has reflected over y=x

6. Composition Plug in g(x) into f(g(x)) and f(x) into g(f(x))
If they both equal x, then your function is an inverse of the original

7. Algebraically First, put x by itself Then, switch x and y
This gives you your new inverse function
To make sure that is the inverse, plug in a value (x=4) to the original equaiton and then plug in that answer (the y value) to the inverse. The answer to the inverse is the original x value.

8. Exponential Growth and Decay

9. What is Exponential Growth and Decay?
Exponential growth and decay is when the growth/decay rate of the value of a mathematical function is proportional to the function's current value
The exponential function is y=ab
The exponential growth function is y=a(1+r)
The exponential decay function is y=a(1-r) x x x
a: initial value
r: rate in decimal
x: time

10. Example of Exponential Growth and Decay
Find a bank account balance that starts with $500, has an annual rate of 6% and has been left in the account for 10 years.
First, find out whether the function is growth or decay.
This function is growth, so find the important parts of the function: y=a(1+r)
The initial value is $500
The rate in decimal is 0.06
The time is 10 years
Now plug it into the formula and solve x 10
y=500(1+0.06)
The answer is about $895.42

11. What is Half Life?
Amount of time it takes for a substance to diminish in half
Formula: y=a(½)
x/h
a: initial value
x: time
h: half life

12. Example of Half-Life y=12(½) The answer is about 5.51-mg of Hg-197
Hg-197 is used in kidney scans. It has a half-life of hours. Write the exponential decay function for a 12-mg sample. Find the amount remaining after 72 hours.
First, let’s identify the different parts of the formula: y=a(½ )
The initial value or a is the 12-mg sample
The time or x is 72 hours
The half-life or h is hours
Now that we have identified the important parts of the problem, we plug it into the formula and solve.
x/h
y=12(½)
72/
The answer is about 5.51-mg of Hg-197

13. What is Compound Interest?
Interest calculated on the initial value (principal) and also on the interest of previous deposits or loans
Formula: y= P(1+ r/n) nt
P: principal
r: rate in decimal
n: number of times compounded
t: time in years

14. Example of Compound Interest
Marty invested $5000 at an annual interest of 6.9%, compounded monthly. How much will he have in the account after 10 years?
Find the important parts of the formula y=P(1+r/n)
The principal is $5000
The rate in decimal is 0.069
The number of times compounded is 12
The time in years is 10 years
Now, plug it into the formula and solve nt
y=5000( / 12)
12(10)
The answer is about $

15. What is Continuously Compounded Interest?
Continuously compounded interest is when your principal is constantly earning interest
Formula: A=Pe rt
A: amount
P: principal
e: value on calculator (2nd ln)
r: rate in decimal
t: time

16. Example of Continuously Compounded Interest
If you invest $500,000 in an account playing 12% compounded continuously, how much will you have after 15 years?
Find the important parts of the formula: A=Pe
The principal is $500,000
The rate in decimal is 0.12
The time is 15 years
Now plug it into the formula and solve
The answer is about $3,024,823.73 rt
0.12(15)
A=500,000e

17. Logarithms

18. What are Logarithms?
Logarithms are the power to which a number must be raised in order to get some other number
Logarithmic and exponential functions are inverses of each other
A log is usually written in the form of y=log
b = x x b y

19. Converting Logarithms to Exponents
Change to exponent form
log6(216) = 3
63 = 216
log4(1024)=5
45 = 1024
Change to log form
5 = 25
2 = log 25 2 5

20. Natural and Common Log Common logs are logs with the base 10.
If a log does not have a specified base, then the base is most likely 10.
log10=1
log100=2
log1000=3
Natural logs are logs to the base e of a number
ln(x)
ln(e )=log (e )
ln(e )=y
e =e
ln(2)=
4.5
4.5 e
4.5 y
4.5
y=4.5
0.69

21. Properties of Logarithms
Product Property- multiplication turns to addition
logb(mn) = logb(m) + logb(n)
Quotient Property- division turns to subtraction
logb(m/n) = logb(m) – logb(n)
Power Property- exponents turn to multiplier (goes to front of function)
logb(mn) = n · logb(m)

22. Expanding using Properties of Logs
log (8x /5)
First, use the quotient property to expand 8x and 5
log2(8x4) – log2(5)
Then use the product property to expand 8x
log2(8) + log2(x4) – log2(5)
Lastly, use the power property to expand x
log2(8) + 4log2(x) – log2(5)
log2(8) is also equal to 3, so the final answer would be: 4 2 4 4 4
3+ 4log2(x) – log2(5)

23. Solving Logarithmic Equations
Solve:
log40-logx=log8
log40/x=log8
40/x=8
log3+logx=log15
log3x=log15
3x=15
2logx=log16
logx =log16
x =16
x=5
x=5 2 2
x=4

24. Arithmetic and Geometric Series

25. What are Sequences and Series?
A sequence is an ordered list of numbers
1,3,5,7,9
A series is the sum of the terms in the sequence

26. Arithmetic Series
Sum of Finite Arithmetic Series- is the sum of a limited number of terms
S =n/2(a +a )
Sum of Infinite Arithmetic Series- is the sum of an infinite number of terms 1 n

27. Example of Arithmetic Series=?
The first thing you need to do in these situations is find out the number of terms you have. In this example it is an obvious answer but in other problems you will have to use the explicit formula by plugging in the terms you know. So, the number of terms in this problem is 5.
Next, we note that that the first term is 2 and the last term is 8.
By plugging this in, we get:
S = 5/2 ( 2 + 8)
S = 25
So, the sum of this series is 25.

28. Geometric Series
Sum of Finite Geometric Series- sum of a finite number of terms
S =a (1-r )/1-r
Sum of Infinite Geometric Series when I r I is less than 1 - sum of an infinite number of terms
S = a /1-r
When I r I is less than 1 it converges
When I r I is greater than 1 it diverges n n 1 n 1

29. Example of Finite Geometric Series
/2 + ¾ = ?
The first thing we do in a geometric series is find the r or common ratio. In this problem, the answer would be ½ .
Next, we find the number of terms which is 4 in this case
Then, we plug it into the formula and solve.
6(1-.5^4 )/1-0.5
The answer is 11.25

30. Example of Infinite Geometric Series
/2 + ¾……
Since there is no end to this series, we need to figure out if it converges or diverges. The common ratio is ½, as we know from the previous example. Because the absolute vaue of the ratio is less than 1 we can say the series converges, allowing us to find the sum.
Now, we plug in the information into the formula and solve.
6/1-0.5
The answer is 12.

31. THANKS FOR WATCHING!!!!!!