Question

Will medal
Please help.
g(x) = (x)/x-5
(i) find g^-1(x)
(ii) find g(6) and g^-1(1)

Answer 1

@butterflydreamer

Answer 2

@sammixboo @ganeshie8

Answer 3

@fishejac000

Answer 4

yes

Answer 5

oh gosh xD I always get confused when i do inverse functions.
Firstly, let's start off with our function:
\[g(x) = \frac{ x }{ x - 5 }\]
Rename g(x) as "y"
\[y = \frac{ x }{ x - 5}\]
Multiply both sides by (x - 5)
\[y (x - 5) = x\]
Expand .
\[xy - 5y = x\]
move all the x's to one side.
\[xy - x = 5y\]
Factorise out the "x".
\[x ( y - 1) = 5y\]
Make "x" the subject.
\[x = \frac{ 5y }{ y - 1}\]
Now to find the inverse, simply swap the x's with y's and vice versa :)
Then rename the function as it's inverse \[g ^{-1} (x)\]

Answer 6

so i get \[\left(\begin{matrix}5x \\ x-1\end{matrix}\right)\]

Answer 7

er.. if that's a fraction then yes LOL.
\[g^{-1} (x) = \frac{ 5x }{ x-1 } \]

Answer 8

yes, what about the (ii) one

Answer 9

For part (II).
We know that \[g(x) = \frac{ x }{ x - 5 }\] so to find g(6) sub x = 6 and we get:
\[g(6) = \frac{ 6 }{ 6 - 5 } = ?\]
Do the same for g^-1(x) but sub x = 1
\[g^{-1} (x) = \frac{ 5x }{ x-1} \rightarrow g^{-1} (1) = \frac{ 5(1) }{ 1 - 1 } = ?\]

Answer 10

5

Answer 11

how about the g^-1(1)

Answer 12

5
give me my medal

Answer 13

?? take a look at what i typed in my prev. post :)
i showed you both g(6) and g^-1 (1)