InterviewSolution
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InterviewSolution
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Let `f : R to R : f (x) =10x +7.` Find the function `g : R to R : g o f = f o g =I_(g)` |
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Answer» Clearly `g = f^(-1)` Now `f(x_(1)) =f(x_(2)) rArr 10x_(1) +7 =10x_(2) +7` `rArr x_(1) =x_(2)` `:. ` f is one-one Now `y =f(x) rArr y = 10x +7` `rArr x= ((y-7))/(10)` Clearly for each `y in R` (codomain of f) there exists `x in R` such that `f(x) =f ((y-7)/(10)) ={ 10.((y-7)/(10))+7 } =y` ` :.` f is onto. Thus f is one-one onto and therefore `f^(-1)` exists we define `f^(-1) : R to R : f^(-1) (y) = (y-7)/(10)` Hence `g : R to R : g (y) = (y-7)/(10) ` [using (i)] |
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