00:01
Okay, so we are told to let f of x equals 1 over x and let g of x be a piecewise defined function if fx greater than 0 is 1 over x.
00:09
If it's less than 0, it's 1 plus 1 of x.
00:12
And then we are being asked to show that if f prime of x, show that f prime of x equals g prime, we're all x in the domain.
00:19
And then the question asks us, can we conclude from corollary 7 that f minus g is constant? so we're going to go ahead and first prove the statement, f prime of x equals g prime of x.
00:31
So since x, when x is greater than 0, we know that f of x equal g of x because when g of x is greater than 0 is 1 over x and f of x is 1 over x.
00:42
So we know that f of x is equal to g of x.
00:48
And then we can also conclude that f prime of x is also equal to g prime of x.
00:57
For x less than 0, we know that f of x is 1 over x and its derivative.
01:11
So d over d x is 1 over x, i'm sorry, negative 1 over x squared.
01:24
And then for g of x, g prime of x is just the derivative of 1 plus 1 over x...