I actually don't know how to get the inverse, mathematically......maybe some other people on the site know a trick or two and could show you....!!!
Strictly speaking, we would have to have a restricted domain on the original function since it isn't one-to-one.....
However, here's what WolframAlpha has .......(the red graph is the original and the blue one is the inverse)
We would want to write it as:
f-1(x) = (x+2)/2 ±(1/2)√(x)√(x+4)
I actually don't know how to get the inverse, mathematically......maybe some other people on the site know a trick or two and could show you....!!!
Strictly speaking, we would have to have a restricted domain on the original function since it isn't one-to-one.....
However, here's what WolframAlpha has .......(the red graph is the original and the blue one is the inverse)
We would want to write it as:
f-1(x) = (x+2)/2 ±(1/2)√(x)√(x+4)