00:01
Thinking about this problem is just starting out with g of x is equal to the integral from 0 to x of f of t dt.
00:08
And from there, i know that the derivative of g is from the fundamental theorem of calculus, just going to be f of x.
00:18
And i don't have to worry about the chain rule because the derivative of x is just one, which you don't have to write down.
00:23
And they also tell us in the directions that f has a derivative.
00:27
So therefore, its derivative would equal the second derivative of g.
00:33
So because this is true, i can conclude that both a is true, that g is differentiable.
00:40
Let me put a little therefore, since f prime is differentiable, i guess i should say it this way.
00:46
Since f is differential, then g is differential.
00:52
Now, because differentiability implies continuity, we can then, conclude that g is continuous because if it's already differentiable must be continuous.
01:06
Now we also have this other statement that f of 1 is equal to 0.
01:12
So let's think about what that means.
01:15
Now i guess what we also know is they say in the directions that f has a positive derivative...