+118658 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}\\
