Question

Please help.

Answer 1

Show that a function in the form \[x \rightarrow \frac{ x+a }{ x-1} \] , x is all real numbers, , is self- inverse for all values of the constant a

Answer 2

is that true?

Answer 3

we can try it and see

Answer 4

\[f(x)=\frac{x+2}{x-1}\\
f\circ f(x)=f(\frac{x+2}{x-1})=\frac{\frac{x+2}{x-1}+2}{\frac{x+2}{x-1}-1}\] lets see if we get \(x\) back

Answer 5

get rid of annoying compound fraction by multiplying top and bottom by \(x-1\) and get
\[\frac{x+2+2(x-1)}{x+2-(x-1)}\]

Answer 6

multiply out, get
\[\frac{x+2+2x-2}{x+2-x+1}=\frac{3x}{3}=x\]
I'LL BE DAMNED learn something new every day!!!

Answer 7

:) @satellite73 , brilliant. But just a question so when it a function its self inverse would be the same then it should be x at the end or what ?

Answer 8

"self inverse" i guess means
\(f^{-1}(x)=f(x)\)
or equivalently
\[f^{-1}(f(x))=f(f^{-1}(x))=x\] or
\[f^{-1}\circ f(x)=f\circ f^{-1}(x)=x\]

Answer 9

i am not exactly sure what your question is asking though

Answer 10

in any case i gave myself an easier problem, i used \(2\) instead of \(c\) but if you mimic what i did and replace \(2\) by \(c\) you will arrive at the same conclusion

Answer 11

yeah yeah , what you have answered , has satisfied my doubt :)
very thankful