Question

The functions

Answer 1

f and g are defined for \(x\in\mathbb{R}\) (excluding -1,0 and 1) by

Answer 2

\(f:x\rightarrow\frac{1+x}{1-x}\)

Answer 3

and \(g:x\rightarrow\frac{1}{x}\)

Answer 4

Show that \((f~.~g)^{-1}=f~.~g\)

Answer 5

Show that \((f~.~g)^{-1}=f~.~g\)

Answer 6

This is my working...but i will shorten it...
\(f^{-1}(x)=\frac{x-1}{x+1}\)

Answer 7

\(g^{-1}(x)=\frac{1}{x}\)

Answer 8

Not sure what to do next...
Show that \((f~.~g)^{-1}=f~.~g\)

Answer 9

Do u hv any ideas? @Zarkon

Answer 10

is your dot multiplication?

Answer 11

the dot is suppose to be o

Answer 12

\[f\circ g\]

f\circ g

Answer 13

yep

Answer 14

it is either multiplication or function composition.

Answer 15

it is function composition...
sorry...

Answer 16

\[(f\circ g)(x)=f(g(x))\]

Answer 17

we just need to choose either LHS or RHS
for proving?

Answer 18

work out both sides and see if they are equal

Answer 19

of the original

Answer 20

\[(f\circ g)(x)=f(g(x))=f\left(\frac{1}{x}\right)=\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\]

Answer 21

then simplify and find its inverse

Answer 22

i got...
\(=\frac{x+1}{x-1}\)

Answer 23

yes

Answer 24

for the inverse,it would be...

Answer 25

\((f\circ g)(x)^{-1}=\frac{1}{x-2}\)

Answer 26

is it correct?

Answer 27

jas

Answer 28

i just need to show the working on both sides just like that,right?

Answer 29

\(\frac{1}{x-2}=\frac{x+1}{x-1}\)

Answer 30

Was talking to a colleague ...here we go

\[y=\frac{x+1}{x-1}\]

\[x=\frac{y+1}{y-1}\]

\[(y-1)x=y+1\]

\[yx-x=y+1\]

\[yx-y=x+1\]

\[y(x-1)=x+1\]

\[y=\frac{x+1}{x-1}\]

Answer 31

so the inverse is \[\frac{x+1}{x-1}\]

which is what we wanted

Answer 32

Thank you! @Zarkon ^_^

Answer 33

Thank you! @Zarkon ^_^