00:01
Okay, so let's start by showing that f is one to one.
00:04
So assume there's two functions, f x1, which is equal to f x2.
00:12
So that gives us the square roots of x1 minus 2 is equal to the square roots of x2 minus 2.
00:20
Now let's take the square root of both sides, and we get x1 minus 2 is equal to x2 minus 2.
00:29
And now adding two to both sides, we see that x1 is equal to x2.
00:36
2.
00:37
So we see that when we do have a point that is the same, our function's output is the same as well.
00:43
So this means that our function is one to one.
00:46
And now we're asked to find f inverse prime of a, which is equal to 1 over f prime and f inverse as a.
00:57
So let's start by finding f inverse as a, and that's 2...