Question

A little help

Answer 1

what's your question?

Answer 2

The function f is defined as \[f=\]\[f={(x,y)|y=(2x+1)/(x-3)}\] where x=/= 3
Find the value of K so that the inverse of f will be \[f ^{-1}={(x,y)|y=(3x+1)/(x-K)}\] where x=/= K

Answer 3

x=/= 3 what does this notation mean?

Answer 4

"doesn't equal"

Answer 5

\[\neq\]

Answer 6

ok

Answer 7

but the inverse of a function should be in terms of the other variable. isn't it?

Answer 8

You can plug in a value for your function for instance. when x = 1 y will be equal to -2. So you have this point for your function (1,-2)

Answer 9

y= (2x+1)/(x-3)

Answer 10

y(x-3)=2x+1

Answer 11

Then you go to your inverse function and you switch the variables around so now your point on the inverse function will be (-2,1) So for the second function (inverse) plug in -2 for x and set the function equal to 1. Then you have an unknown K that you can solve for

Answer 12

So that you don't have to do much algebra manipulation

Answer 13

I'll try that

Answer 14

cool let me know how it goes

Answer 15

yx-2x = 3y+1

Answer 16

x(y-2) = 3y+1

Answer 17

x= (y-2)/(3y+1)

Answer 18

so isnt f^-1(x) given to you?

Answer 19

the question tells f(x) is given

Answer 20

yes but there is an unknown variable K that you have to solve for

Answer 21

exactly

Answer 22

so then what did you get for f^-1(x)?

Answer 23

the unknown in the f^-1

Answer 24

I'm supposed to manipulate it somehow and finish with y= (3x+1)/(x-2), K=2

Answer 25

x= (y-2)/(3y+1

Answer 26

that's the f^-1

Answer 27

for f^-1(x), i got y=(1+3x)/(x-2)

Answer 28

Yeah, that's the answer, I just don't get how to find it though

Answer 29

then you got the solution

Answer 30

would it help for me to show how i got it @ebnor?

Answer 31

I'd appreciate that

Answer 32

solve \[f(f^{-1}(x))=x\] for k

you will get 2

Answer 33

sorry bad drawing

Answer 34

You pick a random value for x. For example x = 2 and you plug it into the original function and you get: f(2) = [2(2) + 1]/2-3] = -5.
So you go to your inverse function that's already given and you plug in -5 for x. So you get:
f^-1(-5) = [3(-5) + 1]/[-5 - K] = 2
\[\ \frac{-14}{-5-K} \ = 2\]
\[-14 = -10 - 2K\]
\[-2K = -4\]
\[K = 2\]
The concept behind this solution is that for the inverse function you switch the domain and range of original function..is it clear now?

Answer 35

Ahh.. right! Thanks for explaining it to me, I get it now

Answer 36

Great! You're welcome!