00:01
We wish to find the best possible bounds for the function f of x is equal to 1 of x squared.
00:11
And this is for x in between negative 1 and positive 2.
00:19
So to find the best possible bounds, we're looking for the highest value that this function can take and the lowest value that it can take within our domain.
00:31
So in other words, we're looking for the global maximum and the global minimum.
00:37
So to do that, we need to first find the derivative of that.
00:42
So f prime of x is equal to 1 over 1 plus x squared and then chain rule we need to multiply by 2x.
00:53
So this is equal to 2x over 1 plus x squared.
01:00
So now critical value, we need to multiply by 2x.
01:01
So now critical value, we need to 2x by 2x.
01:01
So this is equal to 2x over happen either where f prime of x doesn't exist or f prime of x is equal to zero.
01:08
So since our denominator is always going to be positive, it can never equal to zero, which means f prime of x exists everywhere.
01:19
So which means our only critical points could be found when we set f prime of x equal to zero...