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Let $$f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}$$ Find the function $k(x)$ such that $f$ is its own inverse.

I have no idea where to start. Any help would be appreciated.

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    $\begingroup$ Start by plotting the known part out first, then make a mirror image of that about $y=x$. That should fill the missing portion. $\endgroup$ Apr 26, 2018 at 18:13

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This is the given part of your function. Reflect that about $y=x$ will give you the inverse of that part, as well as fill in the missing part, such that $f(x)=f^{-1}(x)$. That is, the rule of $k(x)$ is the rule of the inverse of $2+(x-2)^2$.

enter image description here

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  • $\begingroup$ Thank you. You are right. $\endgroup$
    – TheLeogend
    Apr 26, 2018 at 18:27

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