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If $f:R\rightarrow R$ is defined as $f(x)=x^2+1$ then $f^{-1}(x)$ = ?

$\begin{array}{1 1} \large\frac{1}{x^2+1} \\ \sqrt {x-1} \\ f^{-1}(x) \text{does not exist because f is not 1-1.} \\f^{-1}(x) \text{ does not exist because f is not onto.} \end{array}$

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1 Answer

  • For $f^{-1}$ to exist $f$ should be bijective function.( i.e., both 1-1 and onto.)
$\therefore f$ is not 1-1 function
Hence $f^{-1}(x)$ does not exist.
answered May 24, 2013 by rvidyagovindarajan_1

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