Question

Will medal
please help
find the inverse of
g(x) = (1)/x-1 + 3

Answer 1

\[\left(\begin{matrix}\\ \frac{ 1}{ x-1 }\end{matrix}\right) + 3\]

Answer 2

\[\frac{ 1 }{ x-1} + 3\]

Answer 3

if this more clear

Answer 4

g(x) is same as y
so you can replace g(x) with y \[\huge\rm \color{ReD}{y}=\frac{ 1 }{ \color{blue}{x}-1 } +3\]
to find inverse `switch x and y`
\[\huge\rm \color{blue}{x}=\frac{ 1 }{ \color{red}{y}-1 } +3\]
now solve for y

Answer 5

ok, x - 3 = \[\frac{ 1 }{ y-1 }\]

Answer 6

is that right

Answer 7

@IrishBoy123

Answer 8

@HELPMEPLZ!!

Answer 9

still on inverse, this is too much for me

Answer 10

sorry i was afk
yes that's right \[\huge\rm x-3=\frac{ 1 }{ y-1}\]
now how would you cancel out y-1 from right side ?

Answer 11

divde it?

Answer 12

write y in terms of x and then jst switch y and x :)
function which u get is inverse

Answer 13

its already divided by 1 (1/y-1 )to cancel it out you should do opposite of division

Answer 14

multiply

Answer 15

yes right multiply both sides by y-1

Answer 16

so 1x-3

Answer 17

multiply `both sides` by y-1 \[\huge\rm (y-1) (x-3)=\frac{ 1 }{\cancel{y-1} } \times \cancel{(y-1)}\]
we can cancel out y-1
\[\large\rm (y-1)(x-3)=1\]
now again to solve for y move `x-3` to the right side

Answer 18

ohh, then how to get y by itself?

Answer 19

how would you cancel x-3 from left side ?

Answer 20

divide

Answer 21

right so divide both sides by x-3 let me know what you get

Answer 22

it's multiplication( y-1 ) times (x-3) so yes you should divide

Answer 23

y-1 = 1/x-3

Answer 24

right whats nexxt ?

Answer 25

y = 1/x-3 + 1

Answer 26

right now we should get the common denominator