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.
The equation y = x is its own inverse.
Thats not an answer
the inverse of
\(y=-\frac{1}{2x}\qquad where \qquad x<0\\is\\ x=-\frac{1}{2y} \qquad where \qquad x>0\\ rearrange\\ y=-\frac{1}{2x} \qquad where \qquad x>0\\ \)
so
\(k(x)=-\frac{1}{2x}\)
