Question

Helene paid back $100 in Month 1 of her loan. In each month after that, Helene paid back $50. Write an explicit formula and a recursive formula that shows f(n), the total amount Helene had paid back by Month n.

Answer 1

what is your concept of the forms for these equations?

Answer 2

what do you mean?
recursive and explicit?

Answer 3

yes

Answer 4

recursive is f(x)= (n-1) something like that ?

Answer 5

recurse defines the current amount by adding to the last months amount;and it has to define a starting month value ...

(starting month) = K
(this month) = (last month) + (some payment)
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M_{1} = 100
M_{n} = M{n-1}+ 50

Answer 6

f(n) and f(n-1) i spose are the proper notation for this question

Answer 7

can you fix the notation?

Answer 8

so f(n)= f(n-1)+50?

Answer 9

and the starting value?

Answer 10

f(1)=100?

Answer 11

good, then thats our setup defined recursivly

f(n) = f(n-1) + 50, f(1) = 100

now we could easily start at 100, and add 50 a month for however many months ... right?

Answer 12

yes

Answer 13

out explicit formula simply does that, it starts at 100, and we add on 50 for however many months we pay ...

f(n) = 100 + 50(number of months)

now the number of months here has a little oddity to it, simply a correction factor.

f(1) = 100, only of 50*(number of months) is 0

if n represents the number of months, then how do we make that go to zero when n=1?

Answer 14

im confused, do we add 100 & 50 to get f(1)'s answer?

Answer 15

hmm, im not sure how your material wants you to define the explicit equation, but the 2 methods i have in my head are mathically equivalent.

we are adding up a bunch of 50s, a 50 for each month

50n is therefore the amount we have paid over n months, does this make sense?

in 3 months we have paid out at least 50(3)=150 dollars

Answer 16

yes that makes sense

Answer 17

then f(n) = 50n is almost right for us ... except for 1 case.

f(1) = 50(1) is not equal to 100. how much are we off by? how do we correct this?

Answer 18

50?

Answer 19

yeah, so lets add 50

f(n)= 50 + 50n would satisfy the results.

another form to write it in is:

f(n)= 100 + 50(n-1)

or

f(n)=50(n+1)

all of these are just algebraic manipulations of the same equation, they are all equivalent

Answer 20

can you help me with some more?

Answer 21

maybe

Answer 22

so it basically says
let f(x)=2x-1
g(x)=3x
&h(x)=x^2+1

Answer 23

and the first question says f(g(-3))

Answer 24

how would you determine g(-3) ?

Answer 25

g(x)=3x?

Answer 26

g(x)=3x

g(-3)....notice that x is replaced by -3 as the input value; so what is the value of g, when x=-3?

Answer 27

uhh 3x?

Answer 28

3x is not replacing x by -3 .... you have to know how to work a function.

Answer 29

when x=-3, what value do we get for the function of g?
\[\large g(\underbrace{\color{red}{x}}_{-3} ) = 3\underbrace{\color{red}{x}}_{-3}\]

Answer 30

No clue... -3?

Answer 31

then how do we read 3x? what does 3x mean?

Answer 32

g(-3)*3x ?

Answer 33

I don't know I'm sorry...

Answer 34

g is the name of a function, its value depends on the variable x:

g(x) reads:g of x

3x defines how g and x are related.

3x reads: 3, times x: x+x+x

if x=-3, then 3(-3) = -3-3-3, what is 3, times -3?

Answer 35

-9.

Answer 36

then g(-3)= -9

now the bummer of this is that we have to use the same logic to define f(-9)

Answer 37

the function f,defined by x: f(x) is related by the expression: 2x-1

if x is equal to -9, then what is the value of f(-9) ?

Answer 38

\[f(\color{red}{g(-3)})\implies f(\color{red}{-9})\implies 2(\color{red}{-9})-1\]

Answer 39

you mean 2*-9-1.. -19?

Answer 40

yep

Answer 41

so it's -19?

Answer 42

that was what we determined it to be, yes