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Let `f: R ->R`be defined as `f(x) = 10 x + 7`. Find the function `g: R ->R`such that`gof=fog=1_R` |
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Answer» In `F: R rarr R`, f(x) = 10 x +7 Let x,y `in`R and f(x)=f(y) `rArr 10 x +7 =10 y +7` `rArr 10x = 10 y ` `rArr x=y` `therefore` f is one - one Again let f(x) =y where `y in R` `rArr 10x +7 =y rArr x(y-7)/(10)i R` `therefore` f is onto Now `(fog )(x) = f((y-7)/(7))` `=10 ((y-7)/(10))+7=y` `therefore go f =I_R "and" fog =I_R` `therefore g: R rarr R ` is defined as g(y) `=(y-7)/(10)` |
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