00:01
Okay, so we're asked to show that our function f is one to one.
00:06
So let's say that f evaluated at x1 is equal to f evaluated at x2.
00:14
So we have 9 minus x squared is equal to 9 or x1, and this is 9 minus x2 squared.
00:24
And now let's solve for x.
00:26
So we can subtract 9 on both sides and then multiply by negative 1 on both sides.
00:32
We get x1 squared is equal to x2 squared.
00:38
And taking the square root to both sides, we get that plus minus x1 is equal to plus minus x2.
00:44
But note that x1 and x2, they're between 0 and 3.
00:48
So we're taking the positive value.
00:50
That is, we get x1 is equal to x2.
00:52
So this shows that our function is 1 to 1.
00:56
And now for parts b, let's find f inverse of, and then the derivative of that applied at a where a is equal to two.
01:11
So that gives us 1 over f prime and then f inverse evaluated at 2.
01:20
Okay.
01:23
So we're asked to find f inverse at 2.
01:25
Well, that means that our function, f of x, is equal to 2, right? since the domain is equal to our range.
01:35
Okay, so let's solve for x...