If f: N rightarrow N and f(x) = x + 3,then f^ | vedclass

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(B) For a function $f: N \rightarrow N$ to have an inverse $f^{-1}$,the function must be a bijection (both one-one and onto). Given $f(x) = x + 3$,whe

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does not exist

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(B) For a function $f: N ightarrow N$ to have an inverse $f^{-1}$,the function must be a bijection (both one-one and onto). Given $f(x) = x + 3$,where $N$ is the set of natural numbers. For the function to be onto,the range must be equal to the codomain. The range of $f(x) = x + 3$ for $x \in N$ is $\{4, 5, 6, \dots \}$. The codomain is $N = \{1, 2, 3, 4, 5, 6, \dots \}$. Since the range $\{4, 5, 6, \dots \} 
eq N$,the function is not onto. Therefore,$f^{-1}(x)$ does not exist.

If $f: N \rightarrow N$ and $f(x) = x + 3$,then $f^{-1}(x) =$ . . . . . . .

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