00:01
Suppose that f and g are continuous on the closed interval from a to b and differential on a to b.
00:06
Suppose that f of a equals g of a and f prime of x is less than equal to g prime of x for a less than x less than b.
00:13
And we are being asked to prove this statement right here, f of b is less than g of b.
00:18
Okay, so the first thing we have to do in this problem is we're going to assume what the statement says, h equals fg.
00:29
So we're just going to rewrite that h of x equals f of x minus g of x.
00:41
So now since we are already told that these two functions are both continuous and differential, we can apply the mean value theorem right away because those are the conditions required for the mean value theorem.
00:58
So we're going to apply it to h of x.
01:01
We'll get h prime of c equals h of b minus h of a all over b minus a.
01:17
And then we can rewrite this because we have h of x.
01:21
Since we assume h of x is equal to f of x minus g of x, we can actually rewrite the left side and the right side.
01:27
We can write f prime of c minus g, primacy and since h of b we can also rewrite that as f of b minus minus g of b and this is one group and then minus and we can do the same thing with f of a minus g of a and just remember these are just substitution we are doing these are all coming from the initial assumption we made earlier.
02:11
Okay now that we have have this statement we are also told that f of a is equal to g of a so we can automatically um since f of one the same number minus the number is zero we can just we can just put a big zero right here because that is the same thing we're told that and then what i'm going to do here is i'm going to bring this b minus a on this side so i'm going to rewrite this we're going to bring this to another the page, we're going to have b minus a time f prime of c minus g prime of c equal f of b minus g of a...