00:01
Okay, so we let f of x here be equal to the square root of x, and therefore we have that f prime of x is going to be equal to 1 over 2 square root of x.
00:13
So for the square root of 101, by the mean value theorem on the interval from 100 to 101, we have the square root of 101 minus 10 is equal to f prime of c of 1 of 1.
00:30
And then for some c between 100 and 101, we have that since f prime is decreasing, we have that 1 over 2 square root of 101 is less than f prime of c, which is less than 1 over 20.
00:47
So therefore we get here, 0 .049 is less than square to 101, which is minus 10, which is less than 0 .05.
00:56
Therefore, we get that 10 .049 is less than the square root of 101, which is less than 10 .05.
01:07
And then for the square root of 105, by the mean value theorem, squared 1 .05 minus 10 equals f prime of c of 5.
01:17
And then we get here that that 5 over 2 squared 105 is greater than 0 .24...