Let
\(f(x) = \begin{cases} k(x) &\text{if }x>0, \\ -\frac1{2x}&\text{if }x< 0\\ 0&\text{if }x=0. \end{cases} \)
Find the function $k(x)$ such that $f(x)$ is its own inverse function.
For all x < 0,
\(f(f(x)) = x\text{ (given in the problem)}\\ f\left(-\dfrac1{2x}\right) = x\)
But -1/(2x) > 0, so \(k\left(-\dfrac1{2x}\right) = x\)
Now, let t = -1/(2x). x = -1/(2t).
\(k(t) = -\dfrac1{2t}\\ \boxed{k(x) = -\dfrac1{2x}}\)
Thank youuu!!
Np ;)
+865 
