Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (help_people):

Help Please!!! Begging you? I do not understand this and I really need help. Will fan and medal anybody who helps me (Algebra)

OpenStudy (help_people):

@TheSmartOne

OpenStudy (help_people):

Task 1 Part 1. 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 c x d Part 2. Show your work to prove that the inverse of f(x) is g(x). Part 3. Show your work to evaluate g(f(x)). Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include 5 values for each function. Graph the line y = x on the same graph.

OpenStudy (help_people):

Part 1\[f(x) = x+a/b\] (to clarify all those numbers) \[g(x)=cx-d\]

OpenStudy (help_people):

@Nnesha

OpenStudy (amistre64):

what is f(g)?

OpenStudy (amistre64):

f(x) = x + a/b f(g) is just f composed of g, instead of composed of x f(g) = g + a/b but we have a definition for g, use it f(g) = (cx-d) + a/b now the thing is, what defines an inverse?

OpenStudy (help_people):

@amistre64 may we go part by part please?

OpenStudy (amistre64):

i thought we did :) lets start with what defines an inverse .... what can you tell me?

OpenStudy (help_people):

like f(g(x))=g(f(x))?

OpenStudy (amistre64):

yeah, but somehtings missing f(g(x)) = g(f(x)) = x ^^

OpenStudy (amistre64):

so lets set it up what is f(g)? this is what i worked up to start with cn work up g(f)? or ask me question if you get lost

OpenStudy (help_people):

ok i understand that is the inverse

OpenStudy (help_people):

So for part 1 i can put in any numbers right?

OpenStudy (help_people):

@amistre64

OpenStudy (amistre64):

no, we are working it out to determine what numbers we can use. f(g) = cx-d + a/b what is g(f)?

OpenStudy (help_people):

g(f)=b/a+z-cx?

OpenStudy (amistre64):

g(f) = c(f) - d ; replace f with its definition g(f) = c(x + a/b) - d g(f) = cx + ca/b - d not sure where z is coming from

OpenStudy (help_people):

ok thank you srry did not mean to put z (meant to pout d)

OpenStudy (amistre64):

now since they have to equal x .... they have to equal each other at best f(g) = g(f) cx - d + a/b = cx + ca/b - d since both side have a -d, it cancels out and can be anything it wants to be; whats your favorite d value?

OpenStudy (help_people):

idk 2?

OpenStudy (amistre64):

2 is fine; let d=2 cx + a/b = cx + ca/b now can you tell me, what c value we can use to make this true? give me an educated guess.

OpenStudy (help_people):

4?

OpenStudy (amistre64):

4 might not be good notice that if c=1 1(x) + a/b = 1(x) + 1(a/b)

OpenStudy (help_people):

ok so d=4 c=1

OpenStudy (amistre64):

well, you said d=2 so lets not go chagning things ... d=4 is fine but it helps to stay consistent

OpenStudy (amistre64):

now a and b can be anything you like, what makes your favorite fraction?

OpenStudy (help_people):

so d=2 c =1 fine a = 3 b= 4

OpenStudy (amistre64):

that should work like a charm

OpenStudy (help_people):

ok so lets fit it in may you do it please my mother is calling me

OpenStudy (help_people):

f(x) =3(1)-5 g(x)=1 +5/3

OpenStudy (help_people):

@amistre64 so part 1 is done

OpenStudy (amistre64):

f(x) = x + a/b g(x) = cx - d fill in the values f(x) = x + (3)/(4) g(x) = (1)x - (2) now i have to see what we did wrong ..... since this didnt work out that well

OpenStudy (help_people):

ok thank you and we did nothing wrong :)

OpenStudy (amistre64):

thats it tho, we cant do part 2 if part 1 isnt valid. f(g) = cx - d + a/b g(f) = c(x+a/b) - d cx - d + a/b = c(x+a/b) - d when c=1 x - d + a/b = x+a/b - d this should be true for all abd values but: f(x) = x + 3/4 and g(x) = x - 2 are not inverses so we would need to reevaluate the issue let a/b = d ... or if b=1, then a=d

OpenStudy (amistre64):

let a = d = 2, ;et c = b = 1

OpenStudy (help_people):

wait can you just show me the new equation please?

OpenStudy (help_people):

for part 1, because you are typing bad and it does not make sense

OpenStudy (amistre64):

if working with me is incompatible, then you might what to find/wait for someone else. feel free to repost the question if you want so that others would be better attracted to it.

OpenStudy (help_people):

srry i did not mean to say in that kind of way i meant to say repost i also type bad i just kind of yeah.... @amistre64 please help me so let's con.?

OpenStudy (help_people):

@amistre64

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!