kinda confused right now...

Question

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587

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+865

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.

AnimalMaster Jun 19, 2020

MaxWong · Answer 1 · 2020-06-19T06:59:33

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).

$k(t) = -\dfrac1{2t}\\ \boxed{k(x) = -\dfrac1{2x}}$