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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.

 Jun 19, 2020
 #1
avatar+8304 
0

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).

 

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

.
 Jun 19, 2020
 #2
avatar+587 
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Thank youuu!!

AnimalMaster  Jun 19, 2020
 #3
avatar+8304 
0

Np ;)

MaxWong  Jun 19, 2020

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