Let \[f:R\to R\] be given by \[f(x)={{(x+1)}^{2}}-1,x\ge -1.\] Then, \[{{f}^{-1}}(x),\] is

Does not exist because f is not one-one

Does not exist because f is not onto

Let \[f:R\to R\] be given by \[f(x)={{(x+1)}^{2}}-1,x\ge -1.\] Then, \

1.	Let \[f:R\to R\] be given by \[f(x)={{(x+1)}^{2}}-1,x\ge -1.\] Then, \[{{f}^{-1}}(x),\] is
A.	\[-1+\sqrt{x+1}\]
B.	\[-1-\sqrt{x+1}\]
C.	Does not exist because f is not one-one
D.	Does not exist because f is not onto
Answer» B. \[-1-\sqrt{x+1}\]

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Let \[f:R\to R\] be given by \[f(x)={{(x+1)}^{2}}-1,x\ge -1.\] Then, \[{{f}^{-1}}(x),\] is

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