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

#2**+1 **

\(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\)

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:

Melody
Jul 14, 2018