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.

qwertyzz Apr 16, 2020

#1

#4**+1 **

When x=2

f(x)= 2 +0 = 2

(2,2) is on the line y=x.

The inverse of a function is its reflection over the line y=x

you can find it by swapping the x with the y (although occasionally there can be a problem.)

so

if a function is

y=2+(x-2)^2

then the inverse will be

x=2+(y-2)^2

which simplifies as follows

\(x=2+(y-2)^2\\ x=2+(y^2-4y+4)\\ x=y^2-4y+6\\ x-6=y^2-4y\\ x-6+4=y^2-4y+4\\ x-2=(y-2)^2\\ \pm\sqrt{x-2}=y-2\\ y=2\pm\sqrt{x-2}\\\)

Ok, I have hit a slight problem. Which one will I choose? It is not going to be both.

x>2 and x-2 must be bigger than 0 so that is always true, so that didn't help.

Lets consider the original function some more

\(f(x) =(x-2)^2 +2 \qquad where \qquad x\leq2\)

This is the left side of a concave up parabola with the vertex at (2,2)

So it is always going to be above the line y=x.

So the inverse will always be below the line y=x

This will only be true of the negative case.

This can also be seen from the graph x=(y-2)^2+2 (part of the working above)

which is a sideways parabola and you want the bottom half so you want the neg square root.

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

check:

Here is the graph: https://www.desmos.com/calculator/9ntcun0vs6

Given that there were complications, someone may be able to show you a more straight forward method. Not sure about that.

__LaTex__:

x=2+(y-2)^2\\

x=2+(y^2-4y+4)\\

x=y^2-4y+6\\

x-6=y^2-4y\\

x-6+4=y^2-4y+4\\

x-2=(y-2)^2\\

\pm\sqrt{x-2}=y-2\\

y=2\pm\sqrt{x-2}\\

Melody Apr 16, 2020