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Mathematics 22 Online
OpenStudy (anonymous):

FAN AND MEDAL PLEASE HELP Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses. f(x) x+a/b g(x)=cx−d Part 2. Show your work to prove that the inverse of f(x) is g(x). @ganeshie8 @amistre64 @uri

OpenStudy (anonymous):

btw its g(x)=cx-d

OpenStudy (amistre64):

lets start with how do you define an inverse?

OpenStudy (anonymous):

I think it is an opposite

OpenStudy (amistre64):

an inverse undoes .... is a good way to look at it, but more mathically: if f and g are inverses then: f(g(x)) = x AND g(f(x)) = x for all x

OpenStudy (anonymous):

but it says to insert numbers for them and then explain how they are inverses i just dont understand this lesson at all.

OpenStudy (amistre64):

well, we need to determine what values will make them inverses; so lets start by making the compositions and then see where it leads us

OpenStudy (anonymous):

okay

OpenStudy (amistre64):

is it f = (x+a)/b, or f = x + (a/b)

OpenStudy (anonymous):

the first one

OpenStudy (amistre64):

then f(g) = (g+a)/b = (cx-d+a)/b is an inverse at best if f(g) = x and g(f) = cf-d = c(x+a)/b - d is an inverse at best if g(f) = x since x=x, when does: (cx-d+a)/b = c(x+a)/b - d

OpenStudy (amistre64):

or we might be able to work it simpler one by one c(x+a)/b - d = x and (cx-d+a)/b = x either way we should be able to get some options worked out for us

OpenStudy (anonymous):

okay so now we plug in numbers?

OpenStudy (amistre64):

we can start plugging numbers sure, like b=1 might be a good start

OpenStudy (anonymous):

so now its c(x+a)/1-d=x and (cx-d)/1=x

OpenStudy (amistre64):

or work it all out such that (cx-d+a)/b = c(x+a)/b - d cx-d+a = c(x+a) - bd cx-d+a = cx+ca - bd -d+a = ca - bd

OpenStudy (anonymous):

so nvm its -d+a=ca-1d?

OpenStudy (amistre64):

well, from this, if we let b=1 and c=1 a-d = 1a - 1d is true for any values of a and d

OpenStudy (anonymous):

oh okay so is that the answer for part 1 and 2

OpenStudy (amistre64):

well, part 1 says to find some abcd values that make f and g inverses. part 2 says to prove that those values work

OpenStudy (amistre64):

show that f(g) = x and g(f) = x for c=1, b=1, a=a, d=d

OpenStudy (anonymous):

okay

OpenStudy (amistre64):

of course pick your favorite a and d values since they seem to be unimportant overall

OpenStudy (anonymous):

6

OpenStudy (amistre64):

so a=6, and d=?

OpenStudy (anonymous):

4

OpenStudy (amistre64):

that should work fine :)

OpenStudy (amistre64):

so: f = (x+6)/1, g=1x-4 f(g) = (x-4)+6 doesnt equal x so we might have to try this again. might be a typo up there

OpenStudy (anonymous):

what do you mean?

OpenStudy (anonymous):

Dang it i have to go in like 30 mins

OpenStudy (amistre64):

f = (x+a)/b g = cx - d ---------------------- f(g) = (cx-d+a)/b g(f) = c(x+a)/b - d --------------------- inverses are defined as: f(g) = x g(f) = x (cx-d+a)/b = x c(x+a)/b - d = x cx +a -d = bx cx +ca - bd = bx let b=1 cx +a -d = x cx +ca - d = x let c=1 x +a - d = x x +a - d = x a-d = 0 if a=d is where we went wrong

OpenStudy (amistre64):

if a=6, then d=6

OpenStudy (amistre64):

the process of determining abcd is actually proves they are inverses, but they seem to want you to go about it in a backwards kind of way.

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