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If $f:R\rightarrow R$ is defined as $f(x)=3x-5$, then $f^{-1}(x)$ = ?

(A) $\large\frac{x+5}{3}$
(B) $\large\frac{1}{3x-5}$
(C) does not exist because $f$ is not one to one.
(D) does not exist because $f$ is not onto.
Can you answer this question?
 
 

1 Answer

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$y=f(x)=3x-5$
$\Rightarrow\:x=\large\frac{y+5}{3}$$=f^{-1}(y)$
$\therefore\:f^{-1}(x)=\large\frac{x+5}{3}$
answered May 24, 2013 by rvidyagovindarajan_1
 

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