00:03
Okay, f of x is equal to x squared minus 4x.
00:07
We want to use the limit of the difference quotient to find f prime of 3.
00:14
So f prime of 3 is going to be the limit as h approaches 0 of f evaluated at 3 plus h minus f evaluated at 3, divided by h.
00:40
For this function f of x f prime of 3 will equal the limit of this expression as h approaches 0 so now we just have to go ahead and find out what is f of 3 plus h what is f of 3 substitute those in and then try to do a little bit of algebra and then eventually take the limit f of 3 plus h f of x is x squared minus 4 x so f of 3 plus h is going to be 3 plus h squared, substituting 3 plus h n for x.
01:25
So 3 plus h squared minus 4 times x.
01:32
We're plugging in 3 plus h for x, so minus 4 times 3 plus h.
01:42
And then we need to subtract f of 3.
01:46
So substituting 3 in for x, we're going to subtract, well, f of 3 is going to be 3 squared, minus 4 times straight.
02:03
And that all has to get put over each.
02:13
So f prime of 3 will be the limit of this expression as h approaches 0.
02:29
So expanding this out, 3 plus h squared, is 9 plus 6h plus 8 squared, then minus 4 times 3, minus 12, minus 4x h.
02:41
Minus 3 squared, that subtract 9, and then minus minus 12 is plus 12.
02:47
All over h.
02:49
So our derivative f prime of 3 is going equal to limit of this function as h approaches 0...