@butterflydreamer

@sammixboo @ganeshie8

@fishejac000

4. anonymous

yes

5. butterflydreamer

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 $g ^{-1} (x)$

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

7. butterflydreamer

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

yes, what about the (ii) one

9. butterflydreamer

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 $g^{-1} (x) = \frac{ 5x }{ x-1} \rightarrow g^{-1} (1) = \frac{ 5(1) }{ 1 - 1 } = ?$

5

12. anonymous

5 give me my medal

13. butterflydreamer

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