Question

g(t) = 1/t +1 - find g^-1(t)

Answer 1

i got y = t-1

Answer 2

but that cant be right

Answer 3

u should get 1/(t-1)

Answer 4

\[t=\frac{ 1 }{ y }+1\]

Answer 5

I don't think you want that "y" in there...

Answer 6

y=1/t+1
y-1 =1/t
t=1/(y-1)
now replace

Answer 7

but cant u just replace y for t at the beginning to do inverses? or is that harder?

Answer 8

same thing

Answer 9

yeah, you're right... as you wrote it 2 up, looks good to me

Answer 10

the final form shouldn't have the y, though, should it? Should just be function of t, right?

Answer 11

right.. the answer would be \[f ^{-1}(x) = \frac{ 1 }{ t-1 }\] i am having serious issues with this for whatever reason.

Answer 12

what issues ? didn't u get how it came ?

Answer 13

embarrasingly enough i cannot figure out the fraction.. lol

Answer 14

t=1/y+1
t-1 =1/y
y=1/(t-1)

Answer 15

y = 1/t +1 . so i subtract 1 ofrom each side.. y-1 = 1/t .. now? multiply y to each side? thats where i get confused.

Answer 16

i mean multiply t to each side to get rid of the fraction?

Answer 17

take reciprocal

Answer 18

if a = 1/b, then 1/a = b

Answer 19

did u skip a step between t-1 = 1/y and your answer?

Answer 20

just taking reciprocals.........

Answer 21

meaning.. from t-1 = 1/y... i should multiply both sides by the reciprocal. y/1?

Answer 22

if a = 1/b, then 1/a = b
so
if t-1 = 1/y
then
y=1/(t-1)

Answer 23

so u dont multiply both sides by the reciprocal?

Answer 24

u want to multiply or divide something ??
1. multiply both sides by y
2. then divide both sides by t-1

Answer 25

sorry i just dont have that immeidate association with math theorems

Answer 26

so a=1/b then 1/a = b makes sense.. but ...

Answer 27

y(t-1) = 1 then divide both sides by t-1. awesome thanks.

Answer 28

i m glad that u got it :)