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I am attempting the following question:

The following function is bijective:

$k: $ {$0, 1, 2, 3$} $\to$ {$3, 5, 7, 9$}

$k(x) = 2x + 3$

Find its invesrse $k^{-1}$, including stating the domain and codomain of $k^{-1}$

Now I am confident that $k^{-1}(x) = \frac {x-3} 2$

I am assuming the domain$ = $ {$3, 5, 7, 9$} and codomain$ = ${$0, 1, 2, 3$}, but that is just a guess based on the information provided in the question.

So is my inverse formula, domain and codomain correct? If so, how do you go about finding the domain and codomain of an inverse function?

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  • $\begingroup$ Correct........ $\endgroup$ – Mauro ALLEGRANZA Jan 6 '18 at 14:38
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Since your function is bijective the domain of the inverse function is the codomain of the function and the codomain of inverse function is the domain of the function. Your formula for the inverse function is correct. $$ y=2x+3$$ implies $$x=(y-3)/2$$ Thus the inverse function is $$f^{-1} (x)=(x-3)/2.$$

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For a function $k$ to be invertive it must be both injective and surjective, i.e, k must be bijective. Here $k:\{0,1,2,3\}\to\{3,5,7,9\}$ given by $k(x)=2x+3$, let $y=f(x),$ such that $y\in\{3,5,7,9\}$ $$ y=2x+3\implies x=\frac{y-3}{2} $$ Inverse of $k$=$k^{-1}=\frac{y-3}{2}$ $$ k(0)=3, k(1)=5, k(2)=7, k(3)=9 $$ for the inverse

$$ k^{-1}(3)=0, k^{-1}(5)=1, k^{-1}(7)=2, k^{-1}(9)=3 $$ Thus, $$ k=2x+3=\{(0,3),(1,5),(2,7),(3,9)\}\\k^{-1}=\frac{y-3}{2}=\{(3,0),(5,1),(7,2),(9,3)\} $$

Since $k$ in invertible, thus bijective, So $$ \text{Domain}(k)=\text{Codomain}(k^{-1}) \text{ and} \\\text{Domain}(k^{-1})=\text{Codomain}(k) $$ it is correct.

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A function $f:X\to Y$ is invertible iff a function $g$ exists such that: $$g\circ f=\mathsf{id}_X\text{ and }f\circ g=\mathsf{id}_Y$$

Note that this can only be true if $g$ has same as the codomain as $\mathsf{id}_X$ which is $X$.

Also note that this can only be true if $g$ has same as the domain as $\mathsf{id}_Y$ which is $Y$.

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