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$f(x) =
\begin{cases}
k(x) &\text{if }x>3, \\
x^2-6x+12&\text{if }x\leq3.
\end{cases}
$

 

Find the function $k(x)$ such that $f$ is its own inverse.

 

---

 

I found inverse of $x^2 - 6x + 12$ as $3 \pm \sqrt{x - 3}$, but will that make the function inverse of itself? Hints appreciated.

 May 23, 2021
 #1
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0

Since you want f(x) to be its own inverse, k(x) = x^2 - 6x + 12.

 May 23, 2021
 #2
avatar+51 
+1

What!? No way... I dont think that is the right answer... The inverse of a function that is itself can't be itself! :P

Do you have a better idea? :)

 May 24, 2021
 #3
avatar+51 
0

AHHHHH


How do you find the inverse of its own function???

 May 26, 2021

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