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Let \(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\) Find the function \(k(x)\) such that \(f\) is its own inverse.

DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jul 3, 2018
 #1
avatar+117 
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I still need help on this problem.

DanielCai  Jul 14, 2018
 #2
avatar+92874 
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\(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\)

 

Mmm.

The first thing I notice is that f(2)=2  so this point will be its own inverse (it lies on the line y=x)

That helps becasue it gives me less things I have to think about.

 

\(\text{This will work if k(x) is the inverse of   } \\f(x)=2+(x-2)^2 \;\; where\;\ x\le 2\)

 

So I need to find the inverse:

 

let\(y=2+(x-2)^2    \qquad    x\le 2, \quad y\ge 2\\ \text{I need to make x the subject}\\ y-2=(x-2)^2\\ \pm\sqrt{y-2}=x-2\\ \pm\sqrt{y-2}+2=x\\ \text{But x has to be less then 2 so it must be the neg square root}\\ x=-\sqrt{y-2}+2 \qquad    x\le 2, \quad y\ge 2\\ \text{Now for the inverse swap the x and y values}\\ y=-\sqrt{x-2}+2 \qquad x\ge2, \quad y\le 2 \)

 

so

 

\(f(x) = \begin{cases} -\sqrt{x-2}+2 &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\\ \text{will be its own inverse}\)

 

so

 

\(g(x)=-\sqrt{x-2}+2 \qquad\text{where}\;\;x>2\)

 

Here is the graph:

https://www.desmos.com/calculator/0kxvg4l6ei

Melody  Jul 14, 2018

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