InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Let `f: [-1 ,1] to Y : f(x) =.(x)/((x+2)), x ne -2 " and " Y=` range (`f`). Show that `f` is invertible and find `f^(-1)` |
|
Answer» we have `f(x_(1)) =f(x_(2)) rArr .(x_(1))/(x_(1)=2) =(x_(2))/(x_(2)+2)` `rArr x_(1)x_(2) +2x_(1) =x_(1)x_(2) +2x_(2)` `rArr 2(x_(1)-x_(2)) =0` `rArr x_(1) - x_(2)=0` `rArr x_(1) =x_(2)` `:.` f is one-one Since range `( f) =Y ` so f is onto Thus f is one-one onto and therefore invertible . Let `y in Y.` Then there exists `x in [ -1,1]` such that f(x) =y. Now ` y = f(x) rArr y= (x)/((x+2)) ` ` rArr x= (2y)/((1-y))` `rArr f^(-1) (y) = (2y)/((1-y))` Thus we define : `f^(-1) : [ -1,1] to Y : f^(-1) (y) = (2y)/((1-y)) , y ne 1` |
|